8/30/2020 Linear Algebra Linear Algebra ut io n Fourth Edition N ot fo r D is tri b Kunquan Lan Ryerson University https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/1 8/30/2020 Contents Contents ut io n Preface v 1 Euclidean spaces 1 1.1 Euclidean spaces 1 1.3 Projection and areas of parallelograms 23 1.4 Linear combinations and spanning spaces 31 2 Matrices 38 2.1 Matrices 38 fo r 2.2 Product of two matrices 45 D is tri b 1.2 Normand angle 12 2.3 Square matrices 56 2.4 Row echelon and reduced row echelon matrices 61 2.6 Elementary matrices 78 N ot 2.5 Ranks, nullities, and pivot columns 74 2.7 The inverse of a square matrix 85 3 Determinants 95 3.1 Determinants of 3 × 3 matrices 95 3.2 Determinants of n × n matrices 101 3.3 Evaluating determinants by row operations 105 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/4 8/30/2020 Contents 3.4 Properties of determinants 119 4 Systems of linear equations 125 4.1 Systems of linear equations in n variables 125 4.2 Gaussian Elimination 132 4.4 Inverse matrix method and Cramer’s rule 150 4.5 Consistency of systems of linear equations 156 5 Linear transformations 172 5.1 Linear transformations 172 5.2 Onto, one to one, and inverse transformations 178 5.3 Linear operators in R2 and R3 184 D is tri b 4.6 Linear combination, spanning space and consistency 166 ut io n 4.3 Gauss-Jordan Elimination 141 fo r 6 Lines and planes in R3 195 6.1 Cross product and volume of a parallelepiped 195 6.3 Equations of planes in R3 203 7 Bases and dimensions 216 7.1 Linear independence 216 N ot 6.2 Equations of lines in R3 200 7.2 Bases of spanning and solution spaces 221 7.3 Methods of finding bases of spanning spaces 230 7.4 Coordinates and orthonormal bases 237 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/4 8/30/2020 Contents 8 Eigenvalues and Diagonalizability 246 8.1 Eigenvalues and eigenvectors 246 8.2 Diagonalizable matrices 255 ut io n 9 Vector spaces 273 9.1 Subspaces of Rn 273 9.2 Vector Spaces 279 10.2 Polar and exponential forms 293 10.3 Products in polar and exponential forms 298 10.4 Roots of complex numbers 302 fo r A Solutions to the Problems 307 A.1 Euclidean spaces 307 D is tri b 10 Complex numbers 287 10.1 Complex numbers 287 A.2 Matrices 328 A.3 Determinants 358 A.5 Linear transformations 409 A.6 Planes and lines in R3 420 N ot A.4 Systems of linear equations 376 A.7 Bases and dimensions 428 A.8 Eigenvalues and Diagonalizability 441 A.9 Vector spaces 462 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/4 8/30/2020 Contents A.10 Complex numbers 468 N ot fo r D is tri b ut io n Index 479 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/4 8/30/2020 Preface Preface ut io n This textbook is intended as an aid for students who are attending a course on Linear Algebra in universities and colleges. It is suitable for students who are tackling Linear Algebra for the first time and even those who have little prior knowledge in mathematics. D is tri b This is a self-contained textbook that contains full solutions to the questions in the exercises at the end of the book. The aim is to present the core of Linear Algebra in a clear way and provide many new and simple proofs for some fundamental results. For example, we give a very simple proof for the well-known CauchySchwarz inequality. The goal is to enhance students’ abilities of analyzing, solving, and generalizing problems. N ot fo r The material covered in this textbook is well organized and can be easily understood by students. The main topics include vectors, matrices, determinants, linear systems, linear transformations in the Euclidean spaces, bases and dimensions of spanning spaces in the Euclidean spaces, eigenvalues and diagonalizability, subspaces of the Euclidean spaces, and general vector spaces. When we discuss linear transformations, bases, and dimensions, we restrict our attention to the Euclidean spaces, but the ideas and methods involved can be extrapolated to general vector spaces. This textbook is based on the third edition of Linear Algebra. Some material has been rewritten for greater clarity and some sections and chapters have been combined for better understanding. New examples and exercises have been added. One new chapter introducing vector spaces has also been added. In particular, subspaces of the Euclidean spaces are discussed in more detail. I would like to thank Tim Robinson, Melody Vincent, Jared Steuernol, and Corina Wilshire from Pearson for their support during production of this textbook. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/2 8/30/2020 Preface Kunquan Lan N ot fo r D is tri b ut io n February 20, 2020 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/2 8/30/2020 Chapter 1 Euclidean spaces ut io n Chapter 1 Euclidean spaces 1.1 Euclidean spaces an element in ℝ, that is, 2 is a real number. Let n ∈ ℕ and I n = Definition 1.1.1. D is tri b We denote by ℝ the set of real numbers and by ℕ the set of positive integers, that is, ℕ = {1, 2, ⋯ }. If x is an element in ℝ, we write x ∈ ℝ where the symbol ∈ means “belongs to.” For example, 2 ∈ ℝ means that 2 is {1, 2, ⋯ , n }. {(x , x , ⋯ , x ) : x 1 2 n i N ot ℝn = fo r The set of all elements with n-ordered numbers is called ℝ n-space (ℝ n for short) or an n-dimensional Euclidean space and denoted by ∈ ℝ } and i ∈ I n . In geometry, ℝ 1 represents the x-axis and we write ℝ 1 = ℝ; ℝ 2 represents the xy-plane and ℝ 3 is the xyzspace. There is no geometry to show ℝ n for n ≥ 4, but these spaces exist. For example, we can denote a particle moving in xyz-space by a 4-dimensional space {(x, y, x, t) : x, y, t ∈ ℝ}, where (x, y, z, t) denotes the position of the particle at time t. The element (x 1, x 2, ⋯, x n) is called a point in ℝ n and x 1, x 2, ⋯, x nare called the coordinates of the point in ℝ n. We use symbols like P(x 1, x 2, ⋯, x n) or P to denote the point (x 1, x 2, ⋯, x n) (see Figure 1.1 (I)). In https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/21 8/30/2020 Chapter 1 Euclidean spaces particular, we use the symbol O(0, 0, ⋯ , 0) or O to denote the origin (0, 0, ⋯, 0) of ℝ n. D is tri b ut io n Figure 1.1: (I), (II), (III), (IV) Definition 1.1.2. fo r Let P(x 1, x 2, ⋯, x n) be point in ℝ n. The directed line segment starting at the origin O and ending at the → point P is called a vector or n-vector, and is denoted by OP (see Figure 1.1 (II)). The distance → between the points O and P is said to be the length of the vector. If P is the origin, then OP is called the → zero vector and if P is not the origin, then OP is called a nonzero vector. ( ) of the vector for each i ∈ I n. N ot We denote a vector by → a, → u, and so on. If → a = x 1, x 2, ⋯, x n , then xi is called the ith component (ith entry) A vector → a in ℝ n can be written into either a column form, called a column vector (or n-column vector), or a row form, called a column vector (or n-column vector), that is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/21 8/30/2020 Chapter 1 Euclidean spaces () x1 → a= x2 or → a = ⋮ (x1, x2, ⋯, xn ). ut io n xn () D is tri b For example, (0, 0, 0) T is a column vector (or 3-vector), → a = (5) is a row (or column) vector (or 1-vector); → a = (1, 2) is a row vector (or 2-vector). We can write 1 2 3 = (1, 2, 3, 4). 4 fo r Note that two vectors with the same components written in different orders may not be the same. For example, the column vectors (1, 2, 3) T and (3, 2, 1) T are different vectors. A vector whose components are all zero is a zero vector while a nonzero vector has at least one nonzero component. For example, (0, 0, 0) is Let (1.1.1) N ot a zero vector in ℝ 3 while (1, 0, 0) and (1, 2, 3) are nonzero vectors in ℝ 3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/21 8/30/2020 Chapter 1 Euclidean spaces () () () 1 0 ⋮ → , e2 = 0 1 ⋮ 0 → , ⋯, e n = 0 0 ⋮ 1 ut io n → e1 = 0 → → → be vectors in ℝ . Then each vector has n components. These vectors e 1 , e 2 , ⋯, e n are called the standard n () () → and e 3 = () D is tri b → → → 1 0 vectors in ℝ n. In particular, e 1 = and e 2 = are the standard vectors in ℝ 2 and e 1 = 0 1 0 0 are the standard vectors in ℝ 3. 1 () () 1 → 0 , e = 2 0 0 1 0 N ot fo r The difference between a point P(x 1, x 2, ⋯, x n) and a vector → a = (x 1, x 2, ⋯, x n) T is that a point P has neither a direction nor a length while a vector → a has both a direction, which starts at the origin and ends at the point P, and a length, which is the distance between O and P. In Definition 1.1.2 , a vector is defined as a directed line segment starting at the origin. The following definition introduces the notion of a vector whose initial point is not necessarily at the origin. Definition 1.1.3. Let P(x 1, x 2, ⋯, x n) and Q(y 1, y 2, ⋯, y n) be points in ℝ n. Let R denote the point (y 1 − x 1, y 2 − x 2, ⋯, y n − x n). The directed line segment starting at P and ending at Q is called a vector and is defined by https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/21 8/30/2020 Chapter 1 Euclidean spaces (1.1.2) → → PQ = OR = (y 1 − x 1, y 2 − x 2, ⋯, y n − x n), see Figure 1.1 (III). ut io n → By Definition 1.1.3 , we see that the vectorPQ has different initial and terminating points than the vector → → OR, but its mathematical expression is the same as OR; both are (y 1 − x 1, y 2 − x 2, ⋯, y n − x n). Mathematically, Definition 1.1.3 D is tri b Hence, if you move a nonzero vector → a anywhere in the xy-plane without changing its direction, the resulting → vector is equal to a (see Figure 1.1 (IV)). The same vector may have different initial and terminal points. does not introduce any new vectors. This establishes a one-to-one relation between points and vectors in ℝ n that is, for each point P(x 1, x 2, ⋯, x n) in ℝ n, there is a unique → vector OP corresponding to the point P(x 1, x 2, ⋯, x n). Conversely, for each vector → a = (x 1, x 2, ⋯, x n), there Example 1.1.1. can N ot fo r is a unique point (x 1, x 2, ⋯, x n) corresponding to the vector → a. Hence, the space ℝ n defined in (1.1.1) be treated as the set of all vectors in ℝ n. → → 1. Let P(1, 2), Q(3, 4), P 1(0, − 2, 3), and Q 1(1, 0, 2). Find PQ and P 1Q 1. Solution By (1.1.2) → → , PQ = (3 − 1, 4 − 2) = (2, 2) and P 1Q 1 = (1 − 0, 0 − ( − 2), 2 − 3) = (1, 2, 1). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/21 8/30/2020 Chapter 1 Euclidean spaces Operations on vectors For real numbers, there are four common operations: addition, subtraction, multiplication, and division. In this section, we generalize the first two operations: addition and subtraction to ℝ n in a natural way, and introduce ut io n scalar multiplication and the dot product of two vectors in ℝ n. The dot product is a real number–not a vector. The multiplication and division of two vectors introduced in a natural way are not useful, so we do not introduce such multiplication and division operations for two vectors. However, in Section 6.1 , we introduce a cross product of two vectors in ℝ 3, which is a vector in ℝ 3. We start with equal vectors in ℝ n. D is tri b Definition 1.1.4. Two vectors are said to be equal if the following two conditions are satisfied: i. The number of components of the two vectors are the same. ii. The corresponding components of the two vectors are equal. → → → fo r If → a, b are equal, we write → a = b. Otherwise, we write → a ≠ b. → → In Definition 1.1.3 , we see that the two vectors PQ and OR in (1.1.2) different initial and terminating points. N ot Let m, n ∈ ℕ and let → a = (a 1, a 2, ⋯, a m) According to Definition 1.1.4 are equal although they have → → and b = (b 1, b 2, ⋯, b n). → ,→ a = b if and only if m = n, that is, → a and b must be in the same space, and → a i = b i for each i ∈ I n. Moreover, → a ≠ b if and only if either m ≠ n or there exists some i ∈ I n such that a i ≠ b i https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/21 8/30/2020 Chapter 1 Euclidean spaces Example 1.1.2. ( ) () x−y and b = z 2. Let → a= () () 1 x2 → and b = 1 3. → a= 2 → and b = 3 () 1 2 () 1 4 → 3 . Find all x, y, z ∈ ℝ such that → a = b. 1 → . Find all x ∈ ℝ such that → a ≠ b. → . Identify whether → a and b are equal. D is tri b 1. Let → a= 1 → ut io n x+y → → 4. Let P(1, 1), Q(3, 4), P 1( − 2, 3), and Q 1(0, 6) be points in ℝ . Show that PQ = P 1Q 1. 2 Solution → 1. Because the number of components of → a and b are equal, by Definition 1.1.4 → , in order to make → fo r a = b, we need the corresponding components of → a and b to be equal, that is, x, y, z satisfy the following system of equations: → N ot { x+y=1 x−y=3 z = 1. Solving the above system, we get x = 2, y = − 1 and z = 1. Hence, when x = 2, y = − 1, and → z = 1, → a = b. → 2. Because the number of components of → a and b are equal, by Definition 1.1.4 → , if x 2 ≠ 4, than → a ≠ b. Solving x 2 ≠ 4, we get x ≠ 2 and x ≠ − 2. Hence, when x ≠ 2 and x ≠ − 2, → a ≠ b. → https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/21 8/30/2020 Chapter 1 Euclidean spaces → 3. Because the number of components of → a and b are 3 and 2, respectively, it follows from Definition → that → a ≠ b. 4. By (1.1.2) → → PQ = P 1Q 1. → → , we have PQ = (3 − 1, 4 − 2) = (2, 3) and P 1Q 1 = (0 − ( − 2), 6 − 3) = (2, 3). Hence, ut io n 1.1.4 Definition 1.1.5. → Let → a = (a 1, a 2, ⋯, a n) and b = (b 1, b 2, ⋯, b n) We define the following operations in a natural way. → D is tri b Addition: → a + b = (a 1 + b 1, a 2 + b 3, ⋯, a n + b n). → Subtraction: → a − b = (a 1 − b 1, a 2 − b 3, ⋯, a n − b n). Scalar multiplication: k→ a = (kx 1, kx 2, ⋯, kx n) fo r Remark 1.1.1. The addition and subtraction of vectors are introduced in the same space ℝ n. If two vectors are in N ot different spaces, say, one is in ℝ m and another in ℝ n, where m ≠ n, then the addition and subtraction of the two vectors are not defined. For example, () 1 2 + () 3 3 is not defined. 5 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/21 8/30/2020 Chapter 1 Euclidean spaces When n = 2, the addition of two vectors is illustrated in Figure 1.2 Figure 1.2 → . If → a = (x 1, y 1) and b = (x 2, y 2) then (I) shows that → a + b = (x 1 + x 2 , y 1 + y 2). → Figure 1.2 → (II) shows that if we place the initial point of the vector b on the terminating point of the vector → ut io n a, then the vector → a + b is the vector starting at the initial point of the vector → a and ending at the terminating → → → → point of vector b. In other words, the vector OA plus the vector AB is equal to the vector OB. → → → → OA + AB = OB. (II) N ot Figure 1.2: (I), (II), (III) (III), which can be derived from Figure 1.2 fo r The subtraction of two vectors is illustrated in Figure 1.2 by the relation (1.1.3) . D is tri b (1.1.3) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/21 8/30/2020 Chapter 1 Euclidean spaces By Definition 1.1.5 , the following results on the vector operations can be easily proved, so we leave the proofs to the reader. Theorem 1.1.1. → Let → a, b, → c ∈ ℝ n and let α, β ∈ ℝ. Then → ut io n 1. → a+0=→ a. → 2. 0→ a = 0. → → 3. → a+b = b+→ a. → 4. ( a + b) + c = a + (b + → a). → → → → D is tri b → → 5. α(→ a + b) = α→ a + α b. → → 6. (α + β) a = α a + β→ a. → → 7. (αβ) a = α(β a). () () 1 Let → a= 2 3 → and b = N ot 0 6 . Compute 5 fo r Example 1.1.3. → i. → a + b; → ii. → a − b; iii. 5→ a; iv. − → a; → v. 2→ a − 4 b. Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/21 8/30/2020 Chapter 1 Euclidean spaces , we obtain () () ( ) () () () ( ) ( ) () ( ) ( ) () ( ) ( ) () () () ( ) ( ) → 0 5 0+5 8 . 5 1 3 1−3 −2 + 2 ii. → a−b= − 0 6 = = 5 1 iii. 5→ a=5 2 0 6 2+6 2−6 ( − 1)(1) = −4 . −5 10 . 0 = (5)(0) iv. − → a = ( − 1) 2 0 = 5 (5)(2) = = 0−5 (5)(1) 1 −1 ( − 1)(2) −2 . 0 = ( − 1)(0) 1 3 → v. 2→ a − 4b = 2 2 0 −4 6 5 2 4 = 0 Example 1.1.4. 12 − 24 − 10 = − 20 . − 20 20 () () () 1 Let → a= 4 ut io n 2 i. → a+b= 1+3 N ot → 3 fo r 1 D is tri b By Definition 1.1.5 3 → 2 , b= 0 6 5 4 and → c = → 6 . Compute → a − b + 2→ c. 7 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/21 8/30/2020 Chapter 1 Euclidean spaces Solution By Theorem 1.1.1 , we have [( ) ( )] ( ) ( ) 1 → → a − b + 2→ c = (→ a − b) + 2→ c = → 2 − 4 6 +2 6 7 5 6 8 . 9 = ut io n 0 3 Find → a if 1 5 [ ( )] ( ) 4→ a− 9 3 = 1 5 − 2→ a. Solution From the given equation, we see that → a ∈ ℝ 2 and ( ) [( ) ] ( ) 9 3 1 =5 5 4→ a + 10→ a= ( ) () ( ) 5 25 + 9 3 = 14 28 N ot Hence, − 2→ a = and 14→ a= ( ) 14 28 5 25 − 10→ a. fo r 4→ a− D is tri b Example 1.1.5. . Hence, → a= 1 14 ( ) () 14 28 = 1 2 . Parallel vectors Definition 1.1.6. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/21 8/30/2020 Chapter 1 Euclidean spaces → Two nonzero vectors → a and b in ℝ n are said to be parallel if there exists a number k ∈ ℝ such that → → a = k b. When k > 0, → a and b have the same direction and when k < 0, their directions are opposite. → ut io n For example, because (2, 4, 6) = 2(1, 2, 3), the two vectors (2, 4, 6) and (1, 2, 3) are parallel and have the same direction. Similarly, because ( − 2, − 4, − 6) = − 2(1, 2, 3), the two vectors ( − 2, − 4, − 6) and (1, 2, 3) are parallel and have the opposite direction. → If → a and b in ℝ n are parallel, we write → → aǁ b. D is tri b Example 1.1.6. → → Let P(1, − 2, − 3), Q(2, 0, − 1), P 1( − 2, − 7, − 4), and Q 1(1, − 1, 2). Show that PQǁQP. Solution → , PQ = (2 − 1, 0 − ( − 2), − 1 − ( − 3)) = (1, 2, 2) and fo r By (1.1.2) N ot → P 1Q 1 = (1 − ( − 2), − 1 − ( − 7), 2 − ( − 4)) = (3, 6, 6). → → It follows that P 1Q 1 = (3, 6, 6) = 3(1, 2, 2) = 3PQ. By Definition 1.1.6 → → , PQǁQP. Dot product The scalar multiplication, addition, and subtraction of two vectors in ℝ n result in vectors in ℝ n. We don’t → define the multiplication and division of → a and b in the natural way as (a 1b 1, a 2b 2, ⋯, a nb n) and https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/21 8/30/2020 Chapter 1 Euclidean spaces ( a1 b1 a2 an , b , ⋯, b 2 n ) because they are not useful. But we introduce the notion of the product of two vectors, called the dot product, which can be seen to be useful later. The dot product of two vectors is a real number– not a vector in ℝ n. In Section 6.1 , we shall introduce another product of two vectors in ℝ 3 called the cross Definition 1.1.7. → → ut io n product of two vectors, which is a vector in ℝ 3. → D is tri b Let → a = (a 1 , a 2, ⋯, a n) and b = (b 1 , b 2, ⋯, b n). The dot product of → a and b denoted by → a ⋅ b, is defined by (1.1.4) → a ⋅ b = a 1b 1 + a 2b 2 + ⋯ + a nb n. → → It is convenient to write the dot product of → a and b in the following row times column form: fo r (1.1.5) N ot () b1 → → a ⋅ b = (a 1, a 2, ⋯, a n) b2 ⋮ = a 1b 1 + a 2b 2 + ⋯ + a nb n. bn → → Hence, → a ⋅ b is the sum of the products of the corresponding components of → a and b. Sometimes, we → use sigma notation to write → a ⋅ b, that is, https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/21 8/30/2020 Chapter 1 Euclidean spaces (1.1.6) n → a⋅b= → ∑ a ib i. i=1 → → Let → a = ( − 1, 3, 5, − 7) and b = (5, − 4, 7, 0). Find → a ⋅ b. Solution → The following result follows directly from Definition 1.1.7 . Theorem 1.1.2. Let → a, b, → c ∈ ℝ n and let α ∈ ℝ. Then fo r → → 1. → a ⋅ 0 = 0. → D is tri b a ⋅ b = ( − 1)(5) + (3)( − 4) + (5)(7) + ( − 7)(0) = 18. → ut io n Example 1.1.7. → 2. → a⋅b = b⋅→ a. (Commutative law for dot product) → → → → → 4. (α→ a) ⋅ b = → a ⋅ (α b) = α(→ a ⋅ b). Example 1.1.8. N ot 3. → a ⋅ (b + → c) = → a⋅b+→ a⋅→ c. (Distributive law for dot product) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/21 8/30/2020 Chapter 1 Euclidean spaces () () () Let → a= 0 4 2 → , b= 2 5 0 1 , and → c = 1 2 0 → . Calculate → a ⋅ (b + → c). 1 Solution () () 2 → → 5 1 + (1, 0, 4, 2) 2 D is tri b a ⋅ (b + → c) = → a⋅b+→ a⋅→ c = (1, 0, 4, 2) → ut io n 1 0 1 = 4 + 3 = 7. 1 fo r Exercises 0 N ot 1. Write a 3-column zero vector and a 3-column nonzero vector, and a 5-column zero vector and a 5column nonzero vector. 2. Suppose that the buyer for a manufacturing plant must order different quantities of oil, paper, steel, and plastics. He will order 40 units of oil, 50 units of paper, 80 units of steel, and 20 units of plastics. Write the quantities in a single vector. 3. Suppose that a student’s course marks for quiz 1, quiz 2, test 1, test 2, and the final exam are 70, 85, 80, 75, and 90, respectively. Write his marks as a column vector. ( ) () x − 2y 4. Let → a= 2x − y 2z 2 → and b = → − 2 . Find all x, y, z ∈ ℝ that → a = b. 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/21 8/30/2020 Chapter 1 Euclidean spaces 5. → a= () 6. → a= ( ) |x| y 2 → and b = x−y 4 → () ( ) and b = 1 4 → . Find all x, y ∈ ℝ such that → a = b. 2 x+y → . Find all x, y ∈ ℝ such that → a − b is a nonzero vector. → → 7. Let P(1, x ), Q(3, 4), P 1(4, 5), and Q 1(6, 1). Find all possible values of x ∈ ℝ such that PQ = P 1Q 1. 8. A company with 553 employees lists each employee’s salary as a component of a vector → a in ℝ 553. If a 6% salary increase has been approved, find the vector involving → a that gives all the new salaries. () ut io n 2 9. Let → a= 88 D is tri b 110 denote the current prices of three items at a store. Suppose that the store announces a 40 () () () 10. Let → a= 2 → 2 , b= 3 3 −2 , → c = 0 0 −1 → i. 2→ a − b + 5→ c; → → ii. 4 a + αb − 2→ c ( ) ( ) () 9 11. Find x, y,and z such that , α ∈ ℝ. Compute N ot −1 fo r sale so that the price of each item is reduced by 20%. a. Find a 3-vector that gives the price changes for the three items. b. Find a 3-vector that gives the new prices of the three items. 4y 2z 3x + 8 −6 0 = 0 . 0 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/21 8/30/2020 Chapter 1 Euclidean spaces → 12. Let → u = (1, − 1, 0, 2), → v = ( − 2, − 1, 1, − 4), and w = (3, 2, − 1, 0). → a. Find a vector → a ∈ ℝ 4 such that 2→ u − 3→ v−→ a = w. b. Find a vector → a 2 → [2→ u − 3→ v+→ a] = 2w +→ u − 2→ a. ut io n 1 → 13. Determine whether → a ǁ b, → 1. → a = (1, 2, 3) and b = ( − 2, − 4, − 6). → 2. → a = ( − 1, 0, 1, 2) and b = ( − 3, 0, 3, 6). → D is tri b 3. → a = (1, − 1, 1, 2) and b = ( − 2, 2, 2, 3). () () () 16. Let → a= 2 2 → , b= 3 Compute → a. → a ⋅ b; → → b. a ⋅ c; → −2 , → c = 0 3 0 −1 , and α ∈ ℝ. N ot −1 fo r 14. Let x, y, a, b ∈ ℝ with x ≠ y. Let P(x, 2x), Q(y, 2y), P 1(a, 2a), and Q 1(b, 2b) be points in ℝ 2. Show → → that PQ ǁ P 1Q 1. → → 15. Let P(x, 0), Q(3, x 2), P 1(2x, 1), and Q 1(6, x 2). Find all possible values of x ∈ ℝ such that PQ ǁ P 1Q 1. c. b ⋅ → c; → d. → a ⋅ (b + → c). → 17. Let → u = (1, − 1, 0, 2), → v = ( − 2, − 1, 1, − 4), and w = (3, 2, − 1, 0). → → → → → → Compute u ⋅ v, u ⋅ w, and w ⋅ v. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/21 8/30/2020 Chapter 1 Euclidean spaces ut io n 18. Assume that the percentages for homework, test 1, test 2, and the final exam for a course are 10%, 25%, 25%, and 40%, respectively. The total marks for homework, test 1, test 2 and the final exam are 10, 50, 50, and 90, respectively. A student’s corresponding marks are 8, 46, 48, and 81, respectively. What is the student’s final mark out of 100? 19. Assume that the percentages for homework, test 1, test 2, and the final exam for a course are 14%, 20%, 20%, and 46%, respectively. The total marks for homework, test 1, test 2, and the final exam are 14, 60, 60, and 80, respectively. A student’s corresponding marks are 12, 45, 51, and 60, respectively. What is the student’s final mark out of 100? 20. A manufacturer produces three different types of products. The demand for the products is denoted by the vector → a = (10, 20, 30). The price per unit for the products is given by the vector → D is tri b b = ($200, $150, $100). If the demand is met, how much money will the manufacturer receive? 21. A company pays four groups of employees a salary. The numbers of the employees for the four groups are expressed by a vector → a = (5, 20, 40). The payments for the groups are expressed by a → N ot fo r vector b = ($100, 000, $80, 000, $60, 000). Use the dot product to calculate the total amount of money the company paid its employees. Figure 1.3: https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/21 Chapter 1 Euclidean spaces fo r D is tri b ut io n 8/30/2020 N ot 22. There are three students who may buy a Calculus or Algebra book. Use the dot product to find the total number of students who buy both Calculus and Algebra books. 23. There are n students who may buy a Calculus or Algebra book (n ≥ 2). Use the dot product to find the total number of students who buy both Calculus and Algebra books. 24. Assume that a person A has contracted a contagious disease and has direct contacts with four people: P1, P2, P3, and P4. We denote the contacts by a vector → a : = (a 11, a 12, a 13, a 14), where if the person A has made contact with the person Pj, then a 1j = 1, and if the person A has made no contact with the person Pj, then a 1j = 0. Now we suppose that the four people then have had a variety → of direct contacts with another individual B, which we denote by a vector b : = (b 11, b 21, b 31, b 41), https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/21 8/30/2020 Chapter 1 Euclidean spaces where if the person Pj has made contact with the person B, then b j1 = 1, and if the person Pj has made no contact with the person B, then b j1 = 0. N ot fo r D is tri b ut io n i. Find the total number of indirect contacts between A and B. ii. If → a = (1, 0, 1, 1) and → a = (1, 1, 1, 1), find the total number of indirect contacts between A and B. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 21/21 8/30/2020 1.2 Norm and angle 1.2 Norm and angle ut io n In this section, we shall give definitions of the lengths of vectors and derive formulas used to compute the distance of two points. Some important equalities and inequalities are studied. In particular, the CauchySchwarz Inequality is obtained and applied to derive the angle between two vectors. by (1.2.1) √ 2 → ), which is the length of the vector OP. Now, we generalize the formula (1.2.1) 2 fo r (see Figure 1.2 2 x 1 + x 2, D is tri b In ℝ 1, it is known that the distance between a real number x and zero 0 is |x|, the absolute value of x. In ℝ 2 by the Pythagorean theorem, the distance formula between a point P(x 1, x 2) and the origin O(0, 0) is given n of the Definition 1.2.1. N ot length of a vector from ℝ to ℝ , where the length of a vector is called a norm of the vector. Let → a = (x 1, x 2, ⋯, x n) ∈ ℝ n. The norm of the vector → a denoted by ǁ→ aǁ, is defined by (1.2.2) ǁ→ aǁ = Processing math: 100% √x 2 1 2 2 + x 2 + ⋯ + x n. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/21 8/30/2020 1.2 Norm and angle Geometrically, the norm ǁ→ aǁ represents the length of the vector → a. It is obvious that → a is a zero vector if and only if ǁ→ aǁ = 0. Example 1.2.1. Let → a = (1, 3, − 2). Find ǁ→ aǁ. ut io n Solution √12 + 32 + ( − 2) = √14. ǁ→ aǁ = Theorem 1.2.1. → Let → a, b ∈ ℝ n. Then i. ǁ→ aǁ 2 = → a⋅→ a. ii. ǁk→ aǁ = | kǁ | aǁ for k ∈ ℝ, → → → → → → → and ǁk→ aǁ = kǁ→ aǁ for k ≥ 0. fo r iii. ǁ→ a + bǁ 2 = ǁ→ aǁ 2 + 2(→ a ⋅ b) + ǁ bǁ 2. iv. ǁ→ a − bǁ 2 = ǁ→ aǁ 2 − 2(→ a ⋅ b) + ǁ bǁ 2. → → → N ot v. (→ a − b) ⋅ (→ a + b) = ǁ→ aǁ 2 − ǁ bǁ 2. Proof We only prove (iii). By the result (i) and Theorem 1.1.2 → D is tri b The following result gives the basic properties of norms of vectors. → → , we obtain → → → → ǁ→ a + bǁ 2 = (→ a + b) ⋅ (→ a + b) = → a⋅→ a+→ a⋅b+b⋅→ a+b⋅b → → aǁ 2 + 2(→ a ⋅ b) + ǁ bǁ 2. = ǁ→ Processing math: 100% Example 1.2.2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/21 8/30/2020 1.2 Norm and angle Let → a ∈ ℝ n be a nonzero vector. Find the norm of the following vector (1.2.3) ǁ aǁ a. → ut io n 1 → Solution 1 Because → a ∈ ℝ n is a nonzero vector, ǁ→ aǁ ≠ 0. Let k = ǁ a ǁ . Then k > 0. By Theorem 1.2.1 → 1 ǁ→ aǁ aǁ = ǁk→ aǁ = kǁ→ aǁ = → 1 ǁ→ aǁ Definition 1.2.2. A vector with norm 1 is called a unit vector. , we see that every nonzero vector can be normalized to a unit vector by (1.2.3) Example 1.2.3. √2). Normalize → a to a unit vector. Solution Because ǁ→ aǁ = √ 1 2 + 2 2 + 3 3 + √2 2 1 ǁ→ aǁ Processing math: 100% = N ot Let → a = (1, 2, 3, . fo r By Example 1.2.2 ǁ→ aǁ = 1. D is tri b ǁ (ii) we have √16 = 4, a= → 1 4 the unit vector is (1, 2, 3, √2) = ( 1 2 , 1 2 , 3 4 , √2 4 ) . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/21 8/30/2020 1.2 Norm and angle Gram determinant Gram determinant can be used to obtain the Cauchy-Schwarz Inequality and the formula for the area of a parallelogram in ℝ n. ut io n A 2 × 2 matrix A is a rectangular array of four numbers a, b, c, d ∈ ℝ arranged in two rows and two columns: (1.2.4) ( ) a b c d . The determinant of A denoted by |A| is defined by (1.2.5) || | | Compute the following determinants. i. ii. | | | | 2 3 = ad − bc. fo r Example 1.2.4. a b c d N ot A = D is tri b A= 4 5 −1 2 −3 6 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/21 8/30/2020 1.2 Norm and angle iii. | | 1 0 −2 5 Solution iii. = (2)(5) − (3)(4) = 10 − 12 = − 2. 4 5 −1 2 −3 6 1 0 −2 5 ut io n ii. | | | | | | 2 3 = ( − 1)(6) − (2)( − 3) = − 6 + 6 = 0. = (1)(5) − (0)( − 2) = 5 − 0 = 5. From Example 1.2.4 D is tri b i. , we see that determinants can be negative, zero, or positive. Definition 1.2.3. → fo r Let → a, b ∈ ℝ n. The number N ot (1.2.6) → G(→ a, b) = | → a⋅→ a → a⋅b → → → → b⋅→ a b⋅b | → → = ǁ→ aǁ 2ǁ bǁ 2 − (→ a ⋅ b) 2 → is called the Gram determinant of → a and b. Lemma 1.2.1. → → Processing math: Let100% a, b ∈ ℝ n. Then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/21 8/30/2020 1.2 Norm and angle (1.2.7) → → → G(→ a, b)ǁ→ aǁ 2 = ǁǁ→ aǁ 2 b − (→ a ⋅ b)→ aǁ 2 ut io n Proof Let (1.2.8) → → → g = ǁ→ aǁ 2 b − (→ a ⋅ b)a. By Theorem 1.1.2 (2), (4) and Theorem 1.2.1 ǁ→ gǁ 2 = → D is tri b → (i), (iv), we have 2 ( → → ) → ǁǁ→ aǁ 2 bǁ − 2(ǁ→ aǁ 2 b) ⋅ (→ a ⋅ b)→ a + ǁ(→ a ⋅ b)→ aǁ → → 2 → 2 2 ǁ→ aǁ 4ǁ bǁ 2 − 2ǁ→ aǁ 2(→ a ⋅ b) + (→ a ⋅ b) ǁ→ aǁ 2 = → → [ 2 → → 2 ] . N ot This implies (1.2.7) fo r = ǁ→ aǁ 4ǁ bǁ 2 − ǁ→ aǁ 2(→ a ⋅ b) = ǁ→ aǁ 2 ǁ→ aǁ 2ǁ bǁ 2 − (→ a ⋅ b) . Using Lemma 1.2.1 , we prove the Cauchy-Schwarz inequality, which provides the relation between the dot product and the norm of the two vectors. Theorem 1.2.2 (Cauchy-Schwarz Inequality) Let → a, b ∈ ℝ n. Then → Processing math: 100% | a ⋅ b | ≤ ǁaǁǁbǁ. → → → → https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/21 8/30/2020 1.2 Norm and angle Proof → → If → a = 0, then we see that the Cauchy-Schwarz Inequality holds. If → a ≠ 0, then because → 2 → ǁǁ→ aǁ 2 b − (→ a ⋅ b)→ aǁ ≥ 0, by (1.2.7) → , G(→ a, b) ≥ 0. It follows that → 2 → and → |→ a ⋅ b| = √ → 2 (→ a ⋅ b) ≤ √ǁ aǁ 2ǁ bǁ 2 = ǁ aǁǁ bǁ. → → → → → ut io n (→ a ⋅ b) ≤ ǁ→ aǁ 2ǁ bǁ 2. |x 1y 1 + x 2y 2 + ⋯ + x ny n | ≤ √x 2 1 D is tri b In terms of components of → a = (x 1, x 2, ⋯, x n) and b = (y 1, y 2, ⋯, y n), the Cauchy-Schwarz inequality is equivalent to the following inequality. √ + x 22 + ⋯ + x 2n y 21 + y 22 + ⋯ + y 2n. → (1.2.9) If → a = 0 or b = 0, then the equality of the Cauchy-Schwarz inequality holds. The following result shows that if → Theorem 1.2.3. → N ot fo r a ≠ 0 and b ≠ 0, then the equality of the Cauchy-Schwarz inequality holds if and only if the two vectors are parallel. → Let → a, b ∈ ℝ n be nonzero vectors. Then the following assertions are equivalent. | → i. → a⋅b | = ǁaǁǁbǁ. → → → → ii. b = a→ ⋅ b ǁ a ǁ2 → → a. → iii. → a ǁ b. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/21 1.2 Norm and angle Proof i. implies (ii). Assume that (i) holds. Let → g be the same as in (1.2.8) [ [ → → → → ǁ→ gǁ 2 = ǁ→ aǁ 2 ǁ→ aǁ 2ǁ bǁ 2 − (→ a ⋅ b) 2 = ǁ→ aǁ 2 ǁ→ aǁ 2ǁ bǁ 2 − (ǁ→ aǁǁ bǁ) ] ] 2 . By (1.2.7) = 0. This implies that g = ǁ→ aǁ 2 b − (→ a ⋅ b)→ a = 0. → Hence, → a⋅b → → b= → ǁ aǁ 2 → D is tri b → → , we have ut io n 8/30/2020 → a. → ,→ a ǁ b. iii. implies (i). Assume that (iii) holds. By Definition 1.1.6 Then , there exists k ∈ ℝ such that → a = k b. N ot fo r ii. implies (iii). Assume that (ii) holds. By Definition 1.1.6 |a ⋅ b | = | (kb) ⋅ b | = | kǁ | bǁ → Hence, (i) holds By Theorem 1.2.2 → → → → 2 → → → → and ǁ→ aǁǁ bǁ = ǁk bǁǁ bǁ = | kǁ | bǁ . → , we prove the triangle inequality, which is illustrated by Figure 1.5 2 (I). Figure 1.5: (I) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/21 1.2 Norm and angle D is tri b ut io n 8/30/2020 Theorem 1.2.4. → → fo r Let → a, b ∈ ℝ n. Then → ǁ→ a + bǁ ≤ ǁ→ aǁ + ǁ bǁ. → → → → Figure 1.4: (I) N ot Moreover, equality holds if → a = 0 or b = 0 or → a ǁ b. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/21 1.2 Norm and angle (iii) and Theorem 1.2.2 → ǁ→ a + bǁ 2 → → , we have → → → → = ǁ→ aǁ 2 + 2→ a ⋅ b + ǁ bǁ 2 ≤ ǁ→ aǁ 2 + 2ǁ→ aǁǁ bǁ + ǁ bǁ 2 = (ǁ→ aǁ + ǁ bǁ) . → 2 fo r Proof By Theorem 1.2.1 D is tri b ut io n 8/30/2020 → N ot This implies that ǁ→ a + bǁ ≤ ǁ→ aǁ + ǁ bǁ. If → a ǁ b, by Theorem 1.2.3 → → ǁ→ a + bǁ 2 = (ǁ→ aǁ + ǁ bǁ) → , 2 → → and equality holds. It is obvious that equality holds if → a = 0 or b = 0. Let P(x 1, x 2, ⋯, x n) and Q(y 1, y 2, ⋯, y n) be two points in ℝ n. By (1.1.2) obtain the formula of the distance between P and Q : and Definition 1.2.1 , we (1.2.10) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/21 8/30/2020 1.2 Norm and angle → ǁPQǁ = √(y 1 − x 1) 2 + (y 2 − x 2) 2 + ⋯ + (y n − x n) 2. Example 1.2.5. Solution √(4 − 2) 2 + ( − 3 + 1) 2 + (1 − 5) 2 = 2√6. D is tri b → ǁP 1 P 2 ǁ = ut io n Find the distance between P 1(2, − 1, 5) and P 2(4, − 3, 1). Now, we derive the formula for points between two points in ℝ 2. In particular, we obtain the midpoint formula. Theorem 1.2.5. Let A(x 1, y 1) and B(x 2, y 2) be two different points in ℝ 2. Then the following assertions hold. fo r 1. If C(x, y) is a point on the line segment joining A and B, then there exists t ∈ [0, 1] such that (1.2.11) () ()() x1 N ot x y = (1 − t) 2. For each t ∈ [0, 1], C(x, y) given by (1.2.11) y1 +t x2 y2 . is a point on the line segment joining A and B. Proof → → 1. Because C is a point on the line segment joining A and B, then AC is parallel to AB. By Definition Processing math: 100% 1.1.6 , there exists t ∈ ℝ such that https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/21 8/30/2020 1.2 Norm and angle (1.2.12) → → AC = tAB. It follows that − ( ) [( ) ( )] =t y1 and () x y y2 − x1 y1 ( ) [( ) ( )] ( ) ( ) x1 y1 +t x2 − y2 x1 y1 , we have = (1 − t) x1 y1 +t x2 y2 . fo r By (1.2.12) (1.2.13) = x2 ut io n y x1 D is tri b () x N ot → → → ǁACǁ = ǁtABǁ = tǁABǁ. → → Because C is between A and B, ǁACǁ ≤ ǁABǁ. This, together with (1.2.13) 2. By (1.2.11) , we have Processing math: 100% () → x AC = − y ( ) [( ) ( )] x1 y1 = (1 − t) x2 y2 − x1 y1 , implies t ∈ [0, 1]. . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/21 8/30/2020 1.2 Norm and angle → → → → This implies that AC is parallel to AB. Because the starting point of AC and AB is A, C is a point on the line segment joining A and B. Note that Theorem 1.2.5 holds in ℝ n for n ≥ 3. 1 (2) with t = 2 , we obtain the following midpoint formula of the line segment joining A ut io n By Theorem 1.2.5 and B ( ) D is tri b (1.2.14) x1 + x2 y1 + y2 , . 2 2 Example 1.2.6. , the two midpoints of the segments AB and AC are ( By (1.2.10) N ot Solution By (1.2.14) fo r Let A(1, 2), B( − 2, 1), and C( − 1, 1) be three different points in ℝ 2. Find the distance between the midpoints of the segments AB and AC. 1−2 2+1 2 , 2 ) ( ) 1 3 2 2 = − , and ( 1−2 2+1 2 , 2 ) ( ) = 0, 3 2 . with n = 2, we obtain the distance between the two midpoints Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/21 8/30/2020 1.2 Norm and angle √( ) ( ) − 2 1 −0 + 2 3 3 2 1 − = . 2 2 2 Angle between two vectors → → → → a⋅b → ǁ→ aǁǁ bǁ ≤ 1. D is tri b −1 ≤ ut io n Let → a, b ∈ ℝ n be nonzero vectors. Then ǁ→ aǁ ≠ 0 and ǁ bǁ ≠ 0. By the Cauchy-Schwarz inequality given in Theorem 1.2.2 , we have Let θ be an angle whose radian measure varies from 0 to π. Then cos θ is strictly decreasing on [0, π] and − 1 ≤ cos θ ≤ 1 for θ ∈ [0, π]. → Hence, for any two nonzero vectors → a, b ∈ ℝ n, there exists a unique angle θ ∈ [0, π] such that fo r (1.2.15) → → → ǁ aǁ ǁ bǁ . N ot cos θ = → a⋅b If n = 2 or n = 3, then we can position the two vectors such that they have the same initial point. We denote → by θ the angle between the two vectors → a and b, which satisfies 0 ≤ θ ≤ π. By the cosine law, we have (1.2.16) Processing math: 100% → 2 → 2 → ǁ→ a − bǁ = ǁ→ aǁ 2 + ǁ bǁ − 2ǁ→ aǁǁ bǁ cos θ. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/21 8/30/2020 1.2 Norm and angle Thus, together with Theorem 1.2.1 (iv), we obtain → → ǁ→ aǁǁ bǁ cos θ = → a⋅b and (1.2.15) holds. Hence, it is natural to define the angle given in (1.2.15) as the angle between any → Definition 1.2.4. → ut io n two nonzero vectors → a, b in ℝ n. → Let → a, b be nonzero vectors in ℝ n. We define the angle, denoted by θ, between → a and b by (1.2.15) → only if → a ⋅ b < 0. Definition 1.2.5. → if and only if → a ⋅ b ≥ 0 and θ ∈ D is tri b [ ] π By the graph of y = cos θ for θ ∈ [0, π], we see that θ ∈ 0, 2 . ( ] π 2 , π if and π → fo r Two vectors → a, b ∈ ℝ n are said to be orthogonal (or perpendicular) if θ = 2 . We write → → By (1.2.15) , we obtain Theorem 1.2.6. → → N ot a ⊥ b. → Let → a, b ∈ ℝ n. Then → a ⊥ b if and only if → a ⋅ b = 0. → If the two vectors → a and b in Theorem 1.2.4 Theorem. are orthogonal, then we have the following Pythagorean Processing math: 100% Theorem 1.2.7 (Pythagorean Theorem) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/21 8/30/2020 1.2 Norm and angle → → Let → a, b ∈ ℝ n. Then → a ⊥ b if and only if (1.2.17) → 2 → 2 Proof By Theorem 1.2.1 ut io n ǁ→ a + bǁ = ǁ→ aǁ 2 + ǁ bǁ . (iii), we have (1.2.18) 2 2 ǁ→ a + bǁ = ǁ→ aǁ 2 + 2(→ a ⋅ b) + ǁ bǁ . → → → If a ⊥ b, then → a ⋅ b = 0. It follows from (1.2.18) → → → → then by (1.2.18) , a ⋅ b = 0 and a ⊥ b. → and (1.2.15) that (1.2.17) holds. Conversely, if (1.2.17) holds, , we have the following result. fo r By Theorem 1.2.3 → D is tri b → Theorem 1.2.8. → → N ot Let → a, b ∈ ℝ n be two nonzero vectors. Then → a ǁ b if and only if θ = 0 or θ = π. Proof By Theorem 1.2.3 → | → → ,→ a ǁ b if and only if → a⋅b | → | | = ǁaǁǁbǁ. Because a, b ∈ ℝn are two nonzero vectors, → → ǁ→ aǁ ≠ 0 and ǁ bǁ ≠ 0. This implies that → a ⋅ b ≠ 0 By (1.2.15) → → , Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/21 8/30/2020 1.2 Norm and angle → cos θ = | → → | → a⋅b → ǁ→ aǁǁ bǁ → = → a⋅b |a ⋅ b | → → → . → | → | → If → a ⋅ b > 0, then → a⋅b =→ a ⋅ b and cos θ = 1 and θ = 0. If → a ⋅ b < 0, then → a⋅b = −→ a ⋅ b, cos θ = − 1 ut io n and θ = π. Example 1.2.7. → Let a = ( − 2, 3, 1) and b = (1, 2, − 4). Show that a ⊥ b. → → Solution → Because → a ⋅ b = − 2 + 6 − 4 = 0, by (1.2.6) → , we have → a ⊥ b. Example 1.2.8. () x2 1 → and b = ( ) 2 −8 → . Find x ∈ ℝ such that → a ⊥ b. fo r Let → a= a ⋅ b = (x 2, 1) → ( ) 2 −8 N ot Solution → D is tri b → = 2x 2 − 8 = 0 if and only if x = 2 or x = − 2. Hence, when x = 2 or x = − 2, → a⋅b=0 → and → a ⊥ b. → Example 1.2.9. → Find the angle θ between → a and b. Processing math: 100% → i. → a = (2, − 1, 1) and b = (1, 1, 2). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/21 8/30/2020 1.2 Norm and angle → ii. → a = (1, − 1, 0, − 1) and b = (1, − 1, 1, − 1). → iii. → a = (a, a, a), where a > 0 is given and b = (a, 0, 0). → iv. → a = (1, − 1, 2) and b = (0, 1, − 1). Solution → a⋅b → cos θ = → ii. Because → a ⋅ b = 3, ǁ→ aǁ = → ǁ→ aǁǁ bǁ → √3 and ǁ bǁ = 2, → cos θ = √3a and → cos θ = → iv. Because → a ⋅ b = − 3, ǁ→ aǁ = → → ǁ aǁǁ bǁ = 3 = √6√6 by (1.2.15) → a⋅b → → ǁ aǁǁ bǁ = 3 2√3 = → 1 a2 √3a 2 √3 = 3 → √6 and ǁ bǁ = √2, → cos θ = and θ = 2 → a⋅b √3 and θ = 2 → ǁ→ aǁǁ bǁ = − √6√2 3 . and θ = arccos = − π 6 . , by (1.2.15) 3 π , ǁ bǁ = a, by (1.2.15) → a⋅b = , N ot → iii. Because → a ⋅ b = a 2 , ǁ→ aǁ = by (1.2.15) √6, ut io n ǁ bǁ = D is tri b → √6 and fo r → i. Because → a ⋅ b = 3, ǁ→ aǁ = √3 3 ≈ 54.74 ° . , √3 2 and θ = 5π . 6 Exercises Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/21 8/30/2020 1.2 Norm and angle 1. For each of the following vectors, find its norm and normalize it to a unit vector. i. → a = (1, 0, − 2); → ii. b = (1, 1, − 2); iii. → c = (2, 2, − 2). → 2. Let → a, b ∈ ℝ n. Show that the following identities hold. → 2 → 2 → 2 → 2 → 2 ut io n i. ǁ→ a + bǁ + ǁ→ a − bǁ = 2ǁ→ aǁ 2 + 2ǁ bǁ . → ii. ǁ→ a + bǁ − ǁ→ a − bǁ = 4(→ a ⋅ b). A= ( ) ( 1 1 4 −2 B= −2 2 1 −3 D is tri b 3. Evaluate the determinant of each of the following matrices. ) ( ) ( ) C= 2 −1 3 0 D= −2 1 −6 3 4. For each pair of vectors, verify that Cauchy-Schwarz and triangle inequalities hold. → i. → a = (1, 2), b = (2, − 1) → iii. → a = (1, − 1, 0, 1), b = (0, − 1, 2, 1) fo r → ii. → a = (1, 2, 1), b = (0, 2, − 2) → → 1. P 1(1, − 2), P 2( − 3, − 1) 2. P 1(3, − 1), P 2( − 2, 1) N ot 5. Let → a = (2, 1) and b = (1, x). Use Theorem 1.2.3 to find x ∈ ℝ such that → a ǁ b. 6. For each pair of points P 1 and P 2, find the distance between the two points. 3. P 1(1, − 1, 5), P 2(1, − 3, 1) 4. P 1(0, − 1, 2), P 2(1, − 3, 1) 5. P 1( − 1, − 1, 5, 3), P 2(0, − 2, − 3, − 1) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/21 8/30/2020 1.2 Norm and angle 7. Let A( − 1, − 2), B(2, 3), C( − 1, 1), and D(0, 1) be four different points in ℝ 2. Find the distance between the midpoints of the segments AB and CD. 8. For each pair of vectors, determine whether they are orthogonal. → i. → a = (1, 2), b = (2, − 1) → ii. a = (1, 1, 1), b = (0, 2, − 2) → → → → → iii. a = (1, − 1, − 1), b = (1, 2, − 2) iv. a = (1, − 1, 1), ut io n b = (0, 2, − 1) → v. a = (2, − 1, 0, 1), b = (0, − 1, 3, − 1) → 9. For each pair of vectors, find all values of x ∈ ℝ such that they are orthogonal. D is tri b → i. → a = ( − 1, 2), b = (2, x 2) → ii. → a = (1, 1, 1), b = (x, 2, − 2) → iii. → a = (5, x, 0, 1), b = (0, − 1, 3, − 1) 10. For each pair of vectors, find cos θ, where θ is the angle between them. → 1. → a = (1, − 1, 1) and b = (1, 1, − 2); → fo r → 2. → a = (1, 0, 1) and b = ( − 1, − 1, − 1). 11. Let → a = (6, 8) and b = (1, x). Find x ∈ ℝ such that → ii. → aǁb → N ot → i. → a⊥b π iii. the angle between → a and b is 4 . 12. Three vertices of a triangle are P 1(6, 6, 5, 8), P 2(6, 8, 6, 5), and P 3(5, 7, 4, 6). Processing math: 100% a. Show that the triangle is a right triangle. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/21 8/30/2020 1.2 Norm and angle b. Find the area of the triangle. 13. Suppose three vertices of a triangle in ℝ 5 are P(4, 5, 5, 6, 7), O(3, 6, 4, 8, 4), Q(5, 7, 7, 13, 6). ut io n a. Show that the triangle is a right triangle. b. Find the area of the triangle. 14. Four vertices of a quadrilateral are N ot fo r a. Show that the quadrilateral is a rectangle. b. Find the area of the quadrilateral. D is tri b A(4, 5, 5, 7), B(6, 3, 3, 9), C(5, 2, 2, 8), D(3, 4, 4, 6). Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 21/21 8/30/2020 1.3 Projection and areas of parallelograms 1.3 Projection and areas of parallelograms → In this section, we study the projection of a vector b on a vector → a in ℝ n. The projection has many ut io n applications. For example, it can be applied to find the height of a parallelogram in ℝ n later in this section, the volume of a parallelpiped in ℝ 3 (Section 6.1 ), and the distance from a point in ℝ 3 to a plane in ℝ 3 (Section 6.3 ). D is tri b Definition 1.3.1. → → Let → a, b be nonzero vectors in ℝ n. A vector → p in ℝ n is said to be a projection of b on → a if → p satisfies the following two conditions. 1. → p is parallel to → a. → N ot Figure 1.6: fo r 2. b − → p is perpendicular to → a. Processing math: 72% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/14 1.3 Projection and areas of parallelograms → N ot fo r D is tri b ut io n 8/30/2020 If a vector → p is a projection of b on → a, we write → → p = Proj a b. → Processing math: 72% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/14 8/30/2020 1.3 Projection and areas of parallelograms We refer to Figure 1.7 → → → → → → → → → → for the relations among the vectors → a, b, proj a b, and b − Proj a b. Moreover, if → → ( ) ( ) π a ⋅ b > 0, then the angle between → a and b belongs to 0, 2 → a ⋅ b < 0, then the angle between → a and b belongs to → → π 2 → and Proj a and → a have the same direction. If → , π and Proj a and → a have the opposite direction. If → → ut io n a ⋅ b = 0, then → a ⊥ b and Proj a b = 0. → N ot fo r D is tri b Figure 1.7: → The following result gives the formula for the projection of b on → a. Theorem 1.3.1. → Let → a, b be nonzero vectors in ℝ n. Then Processing math: 72% (1.3.1) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/14 8/30/2020 1.3 Projection and areas of parallelograms ( ) → → a⋅b → Proj a b = → ǁ→ aǁ 2 a. → By Definition 1.3.1 → → → , Proj a b ǁ → a and ( b − Proj a b) ⊥ → a. By Definition 1.1.6 → → there exists k ∈ ℝ such that → Proj a b = k→ a → and , fo r (1.3.3) and Theorem 1.2.6 D is tri b (1.3.2) ut io n Proof (b − Proja b) ⋅ a = 0. → → → , Theorem 1.1.2 → (2) and (3), (1.3.2) N ot Hence, by (1.3.3) → , and Theorem 1.2.1 (i), → a ⋅ b=→ a ⋅ Proj a b = → a ⋅ (k→ a) = k(→ a⋅→ a) = kǁ→ aǁ 2. → → → This imples k = a→ ⋅ b ǁ a→ ǁ 2 . This, together with (1.3.2) , implies (1.3.1) . Example 1.3.1. Processing math: 72% → → → Let a = (4, − 1, 2) and b = (2, − 1, 3). Find Proj a b. → https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/14 8/30/2020 1.3 Projection and areas of parallelograms Solution Because → → a ⋅ b = 4(2) + ( − 1)( − 1) + 2(3) = 15 we obtain 15 5 (4, − 1, 2) = (4, − 1, 2) = 21 7 → Proj a b = → ) 20 5 10 , − , . 7 7 7 D is tri b Theorem 1.3.2. ( ut io n by (1.3.1) and ǁ→ aǁ 2 = 16 + 1 + 4 = 21, → Let → a, b ∈ ℝ n be nonzero vectors. Then the following assertions hold. |a⋅b | → → → i. ǁProj a bǁ = ǁ a→ ǁ → ǁ Proj a b ǁ → ii. |cos θ| = → ǁbǁ . → , where θ is the angle between → a and b. fo r → ( ) → → i. Because Proj a b = → a→ ⋅ b ǁ a→ ǁ 2 → N ot Proof a, we have → ǁProj a bǁ = ǁ → Processing math: 72% ii. By (1.2.15) ( ) → ǁ aǁ 2 |a ⋅ b | → → a⋅b → → aǁ = → ǁ aǁ → 2 |a ⋅ b | → ǁ→ aǁ = → ǁ→ aǁ . and (i), we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/14 8/30/2020 1.3 Projection and areas of parallelograms |a ⋅ b | → → |cos θ | = → ǁ→ aǁǁ bǁ → ǁProj a bǁ → = → ǁ bǁ and the result (ii)holds. ( ) ( ) → π 1. If θ ∈ 0, 2 , then cos θ = 2. If θ ∈ π 2 ut io n , we see that the following assertions hold. ǁ Proj a b ǁ → → ǁbǁ . → , π , then cos θ = − ǁ Proj a b ǁ → → ǁbǁ . D is tri b By Theorem 1.3.2 Example 1.3.2. → Let → a = (1, − 1, 2) and b = (0, 1, − 1). → 1. Use Theorem 1.3.2 (i) to find ǁProj a bǁ 2. Use Theorem 1.3.2 (ii) to find the angle θ between → a and b. 1. Because → a ⋅ b = (1)(0) + ( − 1)(1) + (2)( − 1) = − 3 and ǁ→ aǁ = → by Theorem 1.3.2 √12 + ( − 1) 2 + 22 = √6, (i), |a ⋅ b | → → ǁProj a bǁ = → Processing math: 72% → N ot Solution fo r → → ǁ→ aǁ = |1 − 3| √6 = 3 √6 = √6 2 . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/14 1.3 Projection and areas of parallelograms → 2. Because ǁ bǁ = √02 + 12 + ( − 1) 2 = √2, by Theorem 1.3.2 √6 → ǁProj a bǁ 2 → |cos θ| = → ǁ bǁ (ii), = √2 = √3 2 . → Because → a ⋅ b = − 3 < 0, we have cos θ < 0. Hence, 2 and θ = π − π 5π = . 6 6 D is tri b cos θ = − |cos θ| = − √3 ut io n 8/30/2020 Areas of parallelograms → Let → a = (x 1, x 2, ⋯ , x n) and b = (y 1, y 2, ⋯ , y n) be nonzero vectors in ℝ n. By the condition (2) of Definition → → 1.3.1 , we see that the vector ( b − Proj a b) is always orthogonal to → a whenever the angle θ between → a and → ( ] ( ) π b belongs to 0, 2 or π → → , π . In ℝ 2, it is easy to see that the norm ǁ b − Proj a bǁ is the the height of the 2 → fo r → ( ] π → or θ ∈ N ot parallelogram determined by → a and b whenever θ ∈ 0, 2 ( ) π 2 , π . Hence, we define the norm of ( b − Proj a b) as the height of the parallelogram determined by → a and b in ℝ n → → → (see Figure 1.7 → → → ). We denote by A(→ a, b) the area of the parallelogram determined by → a and b. Then (1.3.4) → → → A(→ a, b) = ǁ→ aǁǁ b − Proj a bǁ. → Processing math: 72% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/14 8/30/2020 1.3 Projection and areas of parallelograms → → The formula for the area A(→ a, b) given in Definition 1.3.4 depends heavily on the projection Proj a b. By → Pythagorean Theorem 1.2.7 → , we derive a formula for A(→ a, b), which depends only and Theorem 1.3.2 → → on the Gram determinant of → a and b so it is easier to use it to compute A(→ a, b). Theorem 1.3.3. → ut io n Let → a and b be nonzero vectors in ℝ n. Then (1.3.5) √G( a, → → b) = → Proof → → → Because ( b − Proj a b) ⊥ → a and → a || Proj a b, we have → → → 2 √ǁaǁ2ǁbǁ2 − (a ⋅ b) . → → → D is tri b → A(→ a, b) = → → ( b − Proj a b) ⊥ Proj a b. → and (1.3.2) , → fo r By Theorems 1.2.7 → → → → → N ot ǁ b − Proj a bǁ 2=ǁ bǁ 2 − ǁProj a bǁ 2 = ǁ bǁ 2 − → and → 2 → → → A(→ a, b) = ǁ→ aǁ 2ǁ b − Proj a bǁ 2 = ǁ→ aǁ 2ǁ bǁ 2 − → Processing math: 72% This implies → A(→ a, b) = √G( a, → → | → |a ⋅ b| → → 2 ǁ→ aǁ 2 → a ⋅ b | 2 = G(→ a, b). → → b). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/14 8/30/2020 1.3 Projection and areas of parallelograms → → If → a ⊥ b, then the parallelogram determined by → a and b is called a rectangle. By Theorem 1.3.3 → → → , we see → that if a ⊥ b, then the area of the rectangle equals ǁ aǁǁ bǁ because a ⋅ b = 0. → → → → To give a formula of A( a, b) in terms of components of → a, b ∈ ℝ n, we prove the Lagrange’s Identity on the Gram determinant. Lemma 1.3.1 (Lagrange’s Identity) → Let → a = (x 1, x 2, ⋯ , x n) and b = (y 1, y 2, ⋯ , y n) be two vectors in ℝ n. Then n=1 → | | xi xj 2 y yj i=1j=i+1 i . N ot Proof By computation, we have ∑ ∑ fo r G(→ a, b) = n D is tri b (1.3.6) ut io n → Processing math: 72% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/14 1.3 Projection and areas of parallelograms ǁ→ aǁ 2 bǁ 2 = ( )( ) n ∑ x 2i ∑ y 2j i=1 ∑ (x iy i) ∑ (x iy i) 2 + ∑ ∑ i=1 j=i+1 ∑ (x iy i) 2 + ∑ ∑ i=1 j=i+1 and ∑ x iy i = 2 + ∑ (x iy i) i=1 n = ∑ (x iy i) i=1 ∑∑ j=1 ∑∑ i=1 j=i+1 2 2 + 2 2 n = ∑∑ i=1 j=i+1 xj yi n ∑ ∑ (x iy i)(x jy j) i = 1j = 1 n−1 n (x iy i)(x jy j) + n−1 n 2 2 2 i=1 j=i+1 (x i y j + x j y i ) N ot i=1 = 2 2 n−1 n ∑ (x iy i) j=1 i=j+1 xi yj + ∑ x iy i ∑ x jy j i=1 n n n 2 2 xi yj n−1 n ( ) n 2 i=1 = ∑∑ fo r ( ) n = + n−1 n i=1 (→ a ⋅ b) n−1 n 2 2 xi yj n−1 n n 2 i = 1j = 1 i=1 j=i+1 i=1 → ∑ ∑ x 2i y 2j ∑∑ + n = n n−1 n 2 i=1 = = j=1 n = n D is tri b | n → ut io n 8/30/2020 (x iy i)(x jy j) + ∑∑ j=1 i=j+1 (x iy i)(x jy j) n−1 n ∑∑ i=1 j=i+1 (x jy j)(x iy i) n−1 n 2 + 2∑ ∑ i=1 j=i+1 (x iy i)(x jy j). Processing math: 72% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/14 8/30/2020 1.3 Projection and areas of parallelograms − [ i=1 i=1 j=i+1 (x 2i y 2j + x 2j y 2i ) n−1 n ∑ (x iy i) 2 + 2 ∑ ∑ i=1 i=1 j=i+1 (x iy i)(x jy j) n ∑∑ i = 1j = i + 1 n (x 2i y 2j n−1 n = ∑ (x iy i) 2 + ∑ ∑ n n = [ n−1 n ∑∑ i=1 j=i+1 Hence, (1.3.6) [ x 2j y 2i ) + (x 2i y 2j + ] n − 2∑ ∑ x 2j y 2i ) ] ut io n G(→ a, b) = n i = 1j = i + 1 (x iy i)(x jy j) ] D is tri b → n−1 n − 2(x iy i)(x jy j) = ∑∑ i=1 j=i+1 (x iy j − x jy i) 2. holds. By Theorem 1.3.3 and Lemma 1.3.1 → , we obtain the following formula of A(→ a, b) in terms of → fo r components of → a, b ∈ ℝ n. N ot (1.3.7) √ | | n−1 → → A( a, b) = n ∑ ∑ i=1j=i+1 → xi xj 2 . yi yj → The formula for A(→ a, b) given in (1.3.7) only depends on the components of → a and b , where norms and dot products are not involved, but when n ≥ 4, it contains many terms of 2 × 2 determinants. So we only state the Processing math: 72% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/14 8/30/2020 1.3 Projection and areas of parallelograms simple cases when n = 2 or n = 3 as a corollary. For n ≥ 4, it would be simpler to use (1.3.5) to compute → A(→ a, b). In Corollary 1.3.1 → | | and → A( a, b) = ǁ → D is tri b x1 x2 2 G( a, b) = y1 y2 → → ut io n 1. Let → a = (x 1 , x 2) and b = (y 1, y 2). Then x1 x2 y1 y2 ǁ. fo r 2. Let a→=(x1,x2,x3) and b→=(y1,y2,y3). Then G(a→,b→)=| x1x2y1y2 |2+| x1x3y1y3 |2+| x2x3y2y3 |2 and N ot A(a→,b→)=| x1x2y1y2 |2+| x1x3y1y3 |2+| x2x3y2y3 |2. In Corollary 1.3.1 (2), the three determinants are obtained by deleting the third, second, and first columns of the following matrix, respectively: (x1x2x3y1y2y3). Example 1.3.3. Processing math: 72% 1. Find A(a→,b→), where a→=(1,2) and b→=(0,−1). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/14 8/30/2020 1.3 Projection and areas of parallelograms 2. Find A(a→, b→) and ǁb→−proja→ b→ǁ, where a→=(1, −2, 0) and b→=(−5,0,2). 3. Find A(a→,b→) and ǁb→−proja→ b→ǁ, where a→=(2, −1, 1, −1) and b→=(0, 1, −1, 2). ut io n Solution 1. By Corollary 1.3.1 (1), A(a→,b→)=| | 120−1 | |=| −1 |=1. 2. Because G(a,→b→)=| 1−2−50 |2+| 10−52 |2+| −2002 |2=120, by Theorem 1.3.3 , A(a→,b→)=G(a→,b→)=26. fo r Exercises D is tri b by Corollary 1.3.1 (2), A(a→,b→)=G(a→,b→)=120=230. By (1.3.4) , ǁb→−proja→ b→ǁ=A(a,→b→)ǁa→ǁ=2305=26. 3. Because G(a→, b→)=ǁa→ǁ2ǁb→ǁ2−(a→⋅b→)2=(7)(6)−(−4)2=26, N ot 1. For each pair of vectors a→ and b→ , find Proja→ b→ and verify that (b→−proja→ b→)⊥a→. 1. a→=(−1, 2), b→=(1, 1); 2. a→=(2, −1, 1), b→=(1, −1, 2); 3. a→=(−1, −2, 2), b→=(1, 0, −1); 4. a→=(0, −1, 1, −1), b→=(1, 1, −1, 2). 2. Let a→=(1,1,1) and b→=(1,−1,1). 1. Use Theorem 1.3.2 (i) to find ǁ Proja→ b→ ǁ. 2. Use Theorem 1.3.2 (ii) to find cos θ, where θ is the angle between a→ and b→. Processing math: 72% 3. Let a→=(1,0,1,2) and b→=(−12,12,12,−32). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/14 1.3 Projection and areas of parallelograms 1. Use Theorem 1.3.2 2. Use Theorem 1.3.2 (i) to find ǁ Proja→ b→ ǁ. (ii) to find θ, where θ is the angle between a→ and b→. 4. Find the area of the parallelogram determined by a→ and b→. 1. a→=(1,0), b→=(0,−1); 2. a→=(1,0), b→=(−1, 1); 3. a→=(0,2), b→=(1, 1); 4. a→=(1, −1), b→=(−1, 2). D is tri b 5. Find the area of the parallelogram determined by a→ and b→. 1. a→=(1, 0, 1), b→=(−1, 0, 1); 2. a→=(1, 0, 0), b→=(−1, 1, −2); 3. a→=(0, 2, −1), b→=(−5, 1, 1); 4. a→=(1, −1, −1), b→=(−1, 1, 2); 5. a→=(1, 0, 2, 1), b→=(2, 1, 0, 3); 6. a→=(−2, −1, 0, 3), b→=(3, 1, −1, 0). ut io n 8/30/2020 N ot fo r 6. Find G(a→, b→) and ǁb→ −Porja→ b→ǁ for each pair of vectors. a. a→=(−1, 2), b→=(1, −1); b. a→=(2, −1, 1), b→=(1, −1, 2); c. a→=(1, 2, 0, 1), b→ =(1, −1, 2, 3); d. a→ = (2, −1, 1, 0), b→ =(3, 0, 1, 3). Processing math: 72% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/14 8/30/2020 1.4 Linear combinations and spanning spaces 1.4 Linear combinations and spanning spaces ut io n In this section, we present the notions of linear combinations and spanning spaces, which will be used in Chapters 4–8. These notions employ the operations given in Definition 1.1.5 . Let m, n ∈ ℕ and let () () () a 11 → a1 = a 21 ⋮ a 12 → , a2 = a 22 ⋮ a 1n → , ⋯ , an = a m2 → N ot be vectors in ℝ m. Let b = (b 1, b 2⋯ , b n) T and S= a 2n ⋮ a mn fo r a m1 (1.4.2) D is tri b (1.4.1) { } → → → a 1, a 2, ⋯ , a n . Definition 1.4.1. Processing math: 100% If there exist real numbers x 1, x 2, ⋯ , x n ∈ ℝ such that https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/14 8/30/2020 1.4 Linear combinations and spanning spaces (1.4.3) → → → b = x 1 a 1 + x 2 a 2 + ⋯ + x na n , → → → → then b is said to be a linear combination of a 1, a 2 , ⋯, a nor S. ut io n → Example 1.4.1. Let () () () () () D is tri b → a1 = 1 2 0 → , a2 = 2 3 → , a3 = 1 1 → , a4 = 3 0 4 2 1 0 0 1 → , a5 = 1 0 0 0 0 . ( → → → a 1, a 2, a 3. iii. Let → a 1, N ot T fo r → → → → → → i. Let b = ( − 4, − 9, 1, 2) . Verify that 2a 1 − 3a 2 = b. Hence, b is a linear combination of a 1, a 2. → → → → → → T ii. Let b = 5, 7, 7, 9) . Verify that a 1 + 2a 2 + a 3 = b. Hence, b is a linear combination of → → → → → → → b = ( − 1, − 7, 6, 7) . Verify that 3a 1 − 2a 2 − a 3 + 5a 4 = b. Hence, b is a linear combination of → → → a 2, a 3, a 4. → T Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/14 8/30/2020 1.4 Linear combinations and spanning spaces → → → → → → → → iv. Let b = ( − 5, 1, 2, 4) T. Verify that a 1 + a 2 − 2a 3 + 5a 4 − 8a 5 = b. Hence, b is a linear combination → → → → → of a 1, a 2, a 3, a 4, a 5. () () ( ) ( ) () () () ( ) () 2−6 → → 0 3 i. 2a 1 − 3a 2 = 2 −3 2 1 4 2 1 → → → ii. a 1 + 2a 2 + a 3 = = 0 2 0−9 4 −9 = 4−3 3 1 2 0 1 + 3 2 1+4+0 = 1 0+6+1 2+2+3 → → → → iii. Let c = 3a 1 − 2a 2 − a 3 + 5a 4. Then () () () () ( ) 0 2 4 −2 3 1 2 0 − 1 3 1 = 0 −1 7 7 . 9 N ot c− =3 → 2 5 4+4+1 → 1 . 1 8−6 2 +2 −4 D is tri b 2 fo r 1 ut io n Solution +5 0 1 = −7 0 6 . 7 → → → → → iv. Let → c = a 1 + a 2 − 2a 3 + 5a 4 − 8a 5. Then Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/14 8/30/2020 1.4 Linear combinations and spanning spaces c = () () () () () ( ) 2 0 0 1 −5 0 3 1 0 0 1 2 4 + 1 −2 2 3 1 +5 1 0 −8 0 = 0 2 . 4 ut io n → 1 → → → b = 2a 1 − 3a 2. → D is tri b In Example 1.4.1 , we verify that the vector b is a linear combination of some vectors from → → → → → a 1, a 2, a 3, a 4, a 5. For example, in Example 1.4.1 (1), we verify that → → → → → A question is how to find the coefficients 2 and −3 of a 1 and a 2 if a 1, a 2, and b are given? In general, given → → → → → → → → vectors a 1, a 2, ⋯, a n and a vector b , how to show that b is a linear combination of a 1, a 2, ⋯, a n ? Or , how to find the numbers x 1, x 2, ⋯, x n ∈ ℝ such that (1.4.3) holds? fo r equivalently, by Definition 1.4.1 N ot In the following, we only give a simple example to show how to find these numbers, and give more detailed discussions in Section 4.6 after we study the approaches of solving systems of linear equations in Sections 4.2–4.4. Example 1.4.2. () () () → → → → 1 2 8 → Let a 1 = and a 2 = . Show that b = is a linear combination of a 1, a 2. 2 1 7 Solution Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/14 8/30/2020 1.4 Linear combinations and spanning spaces → → → Let x 1a 1 + x 2a 2 = b. Then () () () 1 + x2 2 2 8 = 1 7 . This implies = 2x 1 + x 2 By Definition 1.1.4 , we have = 2x 1 + x 2 = 7 . 8 7. fo r { x 1 + 2x 2 8 D is tri b ( ) () x 1 + 2x 2 ut io n x1 → → → → Solving the system, we get x 1 = 2 and x 2 = 3. Hence, b = 2a 1 + 3a 2 and b is a linear combination of a 1 → and a 2. N ot → Now, we discuss some special cases of Definition 1.4.1 . In Definition 1.4.1 → combination of a 1 if and only if there exists x 1 ∈ ℝ such that → , if n = 1, then b is a linear → b = x 1a 1. → Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/14 8/30/2020 1.4 Linear combinations and spanning spaces → → → → → → → → → → → If b ≠ 0 and a 1 ≠ 0 , then by Definition 1.1.6 , b is parallel to a 1. Hence, If b ≠ 0 and a 1 ≠ 0 , then b is a → → → linear combination of a 1. if and only if b is parallel to a 1. Theorem 1.4.1. Proof Let x 1 = x 2 = ⋯ = x n = 0. Then D is tri b → → → → → → 0 = 0a 1 + 0a 2 + ⋯ + 0a n = x 1a 1 + x 2a 2 + ⋯ + x na n. ut io n → The zero vector 0 ∈ ℝ m is a linear combination of S. → The result follows. → → fo r The following result shows that if b is a linear combination of S and we add finitely many vectors to S, then b is a linear combination of all these vectors. Theorem 1.4.2. N ot → → → → → Assume that b is a linear combination of S. Let b 1, b 2, ⋯, b k be vectors in ℝ m. Then b is a linear combination of T, where T= { } → → → → → → a 1, a 2, ⋯, a n, b 1, b 2, ⋯, b k . Proof Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/14 8/30/2020 1.4 Linear combinations and spanning spaces → To show that b is a linear combination of T, we need to find n + k numbers, say y 1, ⋯, y n, y n + 1, ⋯, y n + k such that (1.4.4) → → → → → b = y 1a 1 + y 2a 2 + ⋯ + y na n + y n + 1b 1 + ⋯ + y n + k b k . → Because b is a linear combination of S, by Definition 1.4.1 , there exist ut io n → {xi ∈ ℝ : i ∈ In } such that D is tri b (1.4.5) → → → b = x 1a 1 + x 2a 2 + ⋯ + x na n. → Comparing (1.4.4) with (1.4.5) , we can choose fo r y 1 = x 1, ⋯, y n = x n, y n + 1 = y n + 2 = ⋯ = y n + k = 0. Hence, we have By Definition 1.4.1 N ot → → → b = x 1a 1 + x 2a 2 + ⋯ + x na n → → → → → = x 1a 1 + x 2a 2 + ⋯ + x na n + 0b 1 + ⋯ + 0 b k . → → , b is a linear combination of T. Example 1.4.3. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/14 8/30/2020 1.4 Linear combinations and spanning spaces → Let a 1 = (1, 2, 3) T. Show that the following assertions hold. → → 1. 0 = (0, 0, 0) T is a linear combination of a 1. → 2. Let b = (2, 4, 6) . Then b is a linear combination of a 1. → → → → → T T T 3. Let a 2 = (1, 0, 1) and a 3 = (1, 1, 1) . Then b = (2, 4, 6) is a linear combination of a 1, a 2, and → a 3. T → Solution D is tri b ut io n → → The result (1) follows from Theorem 1.4.1 . Because b = 2a 1, the result (2) follows from Definition → → 1.4.1 . Because b is a linear combination of a 1, the result (3) follows from Theorem 1.4.2 . → , we obtain N ot (1.4.6) , and Definition 1.1.5 fo r By (1.4.3) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/14 8/30/2020 1.4 Linear combinations and spanning spaces b = ()() () ( )( ) ( ) ( ) = x1 a 12 a 1n a 21 a 22 a 2n ⋮ + x2 ⋮ + ⋯ + xn ⋮ a m1 a m2 a mn a 11x 1 a 12x 2 a 1mx n a 21x 1 a 22x 2 ⋮ a m1x 1 + ⋮ D is tri b = a 11 +⋯+ a m2x 2 ut io n → → → x 1a 1 + x 2a 2 + ⋯ + x na n → a 2mx n ⋮ a mnx n = fo r a 11x 1 + a 12x 2 + ⋯ + a 1nx n a 12x 1 + a 22x 2 + ⋯ + a 2nx n ⋮ . By Definition 1.1.4 , we have N ot a m1x 1 + a m2x 2 + ⋯ + a mnx n (1.4.7) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/14 8/30/2020 1.4 Linear combinations and spanning spaces ⋮ a m1x 1 + a m2x 2 + ⋯ + a mnx n = b m. Now we introduce spanning spaces, which will be used in Section 4.6 Definition 1.4.1 , the vector ut io n { a 11x 1 + a 12x 2 + ⋯ + a 1nx n = b 1 a 21x 1 + a 22x 2 + ⋯ + a 2nx n = b 2 and Chapters 5,7, and 8. By D is tri b → → → x 1a 1 + x 2a 2 + ⋯ + x na n is a linear combination of S for each set of x 1, ⋯, x n ∈ ℝ. We collect all these linear combinations together as a set called a spanning space of S. fo r Definition 1.4.2. (1.4.8) span S = By Definitions 1.4.1 Processing math: 100% Theorem and 1.4.2 N ot The spanning space of S, denoted by span S, is the set of all linear combinations of the vectors in S, that is, { } → → → x 1a 1 + x 2a 2 + ⋯ + x na n : x i ∈ ℝ, i ∈ I n . , we have 1.4.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/14 8/30/2020 1.4 Linear combinations and spanning spaces The following assertions are equivalent. → 1. A vector b ∈ span S. → → → 2. b is a linear combination of a 1, a 2, ⋯ , a n. → () ( ) D is tri b Example 1.4.4. → , b is a linear combination of ut io n → → → Let a 1, a 2, b be the same as in Example 1.4.2 . By Example 1.4.2 → → → → → a 1 and a 1 . By Theorem 1.4.3 , b ∈ span{a 1, a 2}. () → → → 6 → −4 2 → → Let a 1 = , a2 = , and b = . Show that b ∈ span{a 1, a 2}. 3 −2 1 → → → Let x 1a 1 + x 2a 2 = b. Then This implies () ( ) () 6 3 + x2 −4 −2 N ot x1 fo r Solution { = 2 1 . 6x 1 − 4x 2 = 2 3x 1 − 2x 2 = 1. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/14 8/30/2020 1.4 Linear combinations and spanning spaces The above two equations are the same, so the system is equivalent to the single equation 3x 1 − 2x 2 = 1. Let x 2 = 1. Then 3x 1 − 2x 2 = 1 implies x1 = 1 3 (1 + 2x 2) = 1 3 (1 + 2) = 1. → → → → → Hence, b = a 1 + a 2 and b ∈ span{a 1, a 2}. ut io n → () () → → → 1 → 0 2 Let e 1 = , e2 = . Show ℝ = span{ e 1 , e 2 }. 0 1 Solution { }{ = → → x 1 e 1 , x 2 e 2 : x 1, x 2 ∈ ℝ N ot Similarly, we can show (1.4.9) }{ = (x 1, x 2) : x 1, x 2 ∈ ℝ fo r → → span e 1 , e 2 D is tri b Example 1.4.5. ℝn { } = ℝ 2. } → → → = span e 1 , e 2 , ⋯, e n , → → → where e 1 , e 2 , ⋯, e n are the standard vectors in ℝ n given in (1.1.1). Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/14 8/30/2020 1.4 Linear combinations and spanning spaces Example 1.4.6. { } → → → → Let e 1 = (1, 0, 0) and e 2 = (0, 1, 0). Find span e 1 , e 2 . { }{ → → span e 1 , e 2 = → → x e 1 + y e 2 : x, y ∈ ℝ }{ = } { } (x, y, 0) : x, y ∈ ℝ . D is tri b → → The above example shows that the spanning space span e 1 , e 2 () () → 0 , a = 2 1 1 → 3 , and a = 3 2 → i. Let b = ( − 1, − 9, → ii. Let b = (3, 6, 6) T. → iii. Let b = (2, 7, 7) T. → Processing math: 100%a 2. Let () () () −1 1 3 . N ot 1 is the xy-plane in ℝ 3. fo r Exercises → 1. Let a 1 = ut io n Solution → → → − 4) . Verify whether 2a 1 − 3a 2 = b. → → → Verify whether a 1 − 2a 2 = b. → → → → Verify whether a 1 − 2a 2 + a 3 = b. T () → → 1 → 1 1 → → = , a = , and b = . Determine whether b is a linear combination of a 1 , a 2. 1 2 2 1 0 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/14 8/30/2020 1.4 Linear combinations and spanning spaces () () () → → → 4. Let a 1 = (1, − 1, 1) T, a 2 = (0, 1, 1) T, a 3 = (1, 3, 2) T, → → → T 1. Is 0 = (0, 0, 1) a linear combination of a 1, a 2, → → → T 2. Is b = (1, 0, 2) a linear combination of a 1, a 2, () ( ) → and a 4 = (0, 0, 1) T. → → a 3, a 4 ? → → a 3, a 4 ? { } () N ot fo r D is tri b → → → 1 → 2 4 → → 5. Let a 1 = , a2 = , and b = . Show that b ∈ span a 1, a 2 . 2 −1 3 ut io n → → → → 1 2 0 → 3. Let a 1 = and a 2 = . Show whether b = is a linear combination of a 1, a 2. 1 1 1 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/14 8/30/2020 Chapter 2 Matrices ut io n Chapter 2 Matrices 2.1 Matrices D is tri b Definition of a matrix Definition 2.1.1. An m × n matrix A is a rectangular array of mn numbers arranged in m rows and n columns: a 12 ⋯ a 1n a 21 a 22 ⋯ a 2n ⋱ ⋮ ⋯ a mn N ot ( a 11 fo r (2.1.1) A= ⋮ ⋮ a m1 a m2 ) . The ijth component of A, denoted a ij, is the number appearing in the ith row and jth column of A. a ij, is called an (i, j)-entry of A, or entry or element of A if there is no confusion. Sometimes, it is useful to write A = (a ij). Capital letters are usually applied to denote matrices. m × n is called the size of A. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/15 8/30/2020 Chapter 2 Matrices Let (2.1.2) () () () → a1 = a 21 ⋮ → , a= a 22 ⋮ a 1n → , ⋯, a n = a m2 a 2n ⋮ a mn . D is tri b a m1 a 12 ut io n a 11 → → → The vectors a 1, a 2, ⋯, a n are called the column vectors of the matrix A. We can rewrite the matrix A as (2.1.3) (2.1.4) N ot Let fo r → → → A = (a 1, a 2, ⋯, a n). → r 1 = (a 11, a 12, ⋯, a 1n), → r 2 = (a 21, a 22, ⋯, a 2n), ⋮ Processing math: 100% → r m = (a m1, a m2, ⋯, a mn). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/15 8/30/2020 Chapter 2 Matrices → → → The vectors r 1 , r 2 , ⋯, r m are called the row vectors of A. We can rewrite the matrix A as (2.1.5) () A= → r2 ut io n → r1 . → rm D is tri b ⋮ Example 2.1.1. ( ) 0 0 0 0 0 0 0 0 0 ( ) 1 2 3 4 is 2 × 2, the size of N ot The size of A = (2) is 1 × 1, the size of fo r → → → → → → m Note that a 1, a 2, ⋯, a n are in ℝ while r 1 , r 2 , ⋯, r m are in ℝ m. ( 1 2 0 3 4 −1 ) is 2 × 3, and the size of is 4 × 3. 0 0 0 Example Processing math: 100% 2.1.2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/15 8/30/2020 Chapter 2 Matrices Let A = ( 1 0 0 1 ) 2 2 1 1 . Use the column vectors and the row vectors of A to rewrite A. 0 1 2 4 Let () () () () 1 → 2 , a2 = 0 0 → 1 , a3 = 2 → 1 , and a 4 = 2 fo r →→→→ Then A can be rewritten as A = (a 1 a 2 a 3 a 4). 0 1 1 . 4 D is tri b → a1 = ut io n Solution N ot → → → Let r 1 = (1, 0, 0, 1), r 2 = (2, 2, 1, 1), and r 3 = (0, 1, 2, 4). Then A can be rewritten as A = () → r1 → r2 . → r3 Definition 2.1.2. 1. An m × n matrix is called a zero matrix if all entries of A are zero. 2. A 1 × n matrix is called a row matrix. 3. An m × 1 matrix is called a column matrix. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/15 8/30/2020 Chapter 2 Matrices Sometimes, we denote by 0 a zero matrix if there is no confusion. A 1 × n matrix and an m × 1 matrix can be treated as a row vector and a column vector, respectively. Example 2.1.3. ( ) 0 0 0 A = (0), B= ( 0 0 0 0 0 0 ) , and C = 0 0 0 0 0 0 . 0 0 0 () 0 0 () 2 is a 4 × 1 column matrix and 0 is a 3 × 1 column matrix. 3 N ot Transpose of a matrix 1 fo r 3. 0 D is tri b 2. (0 0 0) is a 1 × 3 row matrix and (1 3 5 7) is a 1 × 4 row matrix. ut io n 1. The following are zero matrices. Let A = (a ij) bean m × n matrix defined in (2.1.1) . The transpose of A, denoted by A T, is the n × m matrix obtained by interchanging the rows and columns of A. Hence, A T = (a ji) or (2.1.6) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/15 8/30/2020 Chapter 2 Matrices T From (2.1.6) T T ⋮ a 1n a 2n a m1 ⋯ a m2 ⋱ ⋮ ⋯ a mn ) . , we see that the ith column of A T and the ith row of A are the same. It is obvious that = A. Example 2.1.4. Find the transposes of the following matrices: D is tri b (A ) ( ⋮ ⋯ ut io n A = a 11 a 21 a 21 a 22 ( ) 7 C= 18 31 . 52 68 fo r A = (8) B = (1 3 5) 9 N ot Solution () 1 A T = (8), B T = ( ) 7 18 52 3 , CT = . 9 31 68 5 Operations on matrices Processing math: 100% Definition 2.1.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/15 8/30/2020 Chapter 2 Matrices Two matrices A and B are said to be equal if the following conditions hold: 1. A and B have the same size. 2. All the corresponding entries are the same. a ij = b ij for all i ∈ I m and j ∈ I n. Let A = ( ) 2 D is tri b Example 2.1.5. ( ) 1 ut io n If A and B are equal, then we write A = B. Let A = (a ij) and B = (b ij) be two m × n matrices. Then A = B if and only if 2 1 and B = . Find all x ∈ ℝ such that A = B. 3 4 3 x2 Solution fo r Note that A and B have the same size. Hence, if x 2 = 4, then A = B. This implies that when x = 2 or x = − 2, A = B. Let A = ( a 2x + y b 4x + 3y ) and B = ( ) 1 1 3 6 N ot Example 2.1.6. . Find all a, b, x, y ∈ ℝ such that A = B Solution Note that A and B have the same size. Hence, A = B if a = 1, b = 3, and x, y satisfy the following system. Processing math: 100% { 2x + y = 1, 4x + 3y = 6. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/15 8/30/2020 Chapter 2 Matrices Solving the above system, we obtain x = − 3 / 2 and y = 4. Hence, when a = 1, b = 3, x = − 3 / 2, and y = 4, A = B. Example 2.1.7. () 2 be two matrices. Determine if A = B. ut io n Let A = (1 2) and B = 1 Solution The sizes of A and B are 1 × 2 and 2 × 1 respectively. Because A and B have different sizes, A ≠ B. , if we treat A and B as vectors, then they are equal vectors. D is tri b Note that in Example 2.1.7 Definition 2.1.4. Let A = (a ij), B = (b ij) be m × n matrices and k a real number. We define ⋯ a 1n + b 1n a 21 + b 21 a 22 + b 22 ⋯ a 2n + b 2n ⋮ ⋮ ⋱ ⋮ a m1 + b m1 a m2 + b m2 Subtraction: A − B = Processing math: 100% ( ⋯ ) fo r ( a 12 + b 12 N ot Addition: A + B = a 11 + b 11 a mn + b mn . a 11 − b 11 a 12 − b 12 ⋯ a 1n − b 1n a 21 − b 21 a 22 − b 22 ⋯ a 2n − b 2n ⋮ ⋮ ⋱ ⋮ ⋯ a mn − b mn a m1 − b m1 a m2 − b m2 ) . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/15 Chapter 2 Matrices Scalar multiplication: kA = ( ka 11 ka 12 ⋯ ka 1n ka 21 ka 22 ⋯ ka 2n ⋮ ⋮ ⋱ ⋮ ⋯ ka mn ka m1 ka m2 ) . Example 2.1.8. Let ( 4 2 2 −3 0 ) and B = ( Compute ii. A − B = 0 5 −6 1 ) . ( ( 3−4 4+1 2+0 2 + 5 −3 − 6 0 + 1 3 − ( − 4) 2−5 4−1 N ot Solution i. A + B = 1 fo r i. A + B; ii. A − B; iii. − A iv. 3A. −4 D is tri b A= 3 ut io n 8/30/2020 ) ( = −1 7 2−0 − 3 − ( − 6) 0 − 1 5 2 −9 1 ) ( = 7 ) . 3 2 −3 3 −1 ) . Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/15 Chapter 2 Matrices iii. − A = iv. 3A = ( ( ( − 1)(3) ( − 1)(4) ( − 1)(2) ( − 1)(2) ( − 1)( − 3) ( − 1)(0) (3)(3) (3)(4) (3)(2) (3)(2) (3)( − 3) (3)(0) ) ( = ) ( = −3 −4 −2 −2 9 12 6 6 −9 0 3 0 ) . ) . The following result can be easily proved and its proof is left to the reader. Theorem 2.1.1. 10. (A − B) T = A T − B T, 11. (αA) T = αA T. Example 2.1.9. fo r N ot 1. A + 0 = A, 2. 0A = 0, 3. A + B = B + A, 4. (A + B) + C = A + (B + C), 5. α(A + B) = αA + αB, 6. (α + β)A = αA + βA, 7. (αβ)A = α(βA), 8. 1A = A, 9. (A + B) T = A T + B T, D is tri b Let A, B, and C be m × n matrices and let α, β ∈ ℝ. Then ut io n 8/30/2020 Let Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/15 8/30/2020 Chapter 2 Matrices ( ) ( ) ( ) 1 0 1 A= 1 2 3 2 1 1 B= 0 0 2 1 0 0 0 0 0 C= 1 4 2 0 1 1 0 0 2 Solution 2A − B + C = (2A − B) + C [ ( ) ( )] ( ) [( ) ( )] ( ) ( )( )( ) = 1 2 3 2 2 1 1 0 0 2 0 0 0 − 0 0 0 − 4 −1 −4 [2(A + B)] T = 2(A + B) T = 2(A T + B T) 2 0 0 2 1 0 0 + 0 1 1 2 0 0 2 2 = 4 −2 −1 3 −1 −4 3 . 4 [( ) ( )] ( ) ( ) 1 2 0 =2 0 1 1 1 4 2 −2 −1 2 + N ot = 1 0 0 fo r 0 0 4 0 1 1 0 0 2 1 2 3 4 2 2 1 + 1 4 2 2 0 2 = 1 0 0 D is tri b 1 0 1 ut io n Compute 2A − B + C and [2(A + B)] T. 0 1 0 Processing math: 100% 1 1 2 + 1 0 1 2 2 1 2 0 4 =2 2 1 4 4 1 4 3 0 2 4 4 2 = 4 2 8 . 8 2 8 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/15 8/30/2020 Chapter 2 Matrices Example 2.1.10. Find the matrix A if [ ( )] ( ) 2A − T T 0 0 = 2 −1 1 0 −1 . ut io n 1 T Solution Because 1 0 T T [( ) ] 1 T = (2A T) − 2 −1 ( ) 1 T 0 2 −1 = 2A − 0 = 2 −1 0 0 2 −1 , 1 0 −1 ( )( ) ( ) 0 ( ) 1 ( ) ( ) 1 N ot 2A = T T fo r we obtain and 0 2 −1 = 2(A T) − 2A − D is tri b [ ( )] 2A T − 1 0 −1 + 1 0 2 −1 = 1 1 2 −2 . Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/15 8/30/2020 Chapter 2 Matrices 1 ( ) 1 1 Hence, A = 2 = 2 −2 ( ) 1 1 2 2 . 1 −1 ut io n Exercises ( ) ) ( ) 1 A = (2) B = ( ) ( 5 8 1 4 C= 10 3 8 10 3 4 ) D= 0 0 0 10 2 0 0 1 6 9 10 D is tri b 1. Find the sizes of the following matrices: 2. Use column vectors and row vectors to rewrite each of the following matrices. ( ) ( 2 1 8 1 B= 5 4 10 1 9 7 2 3 10 8 7 4 10 4 3 C= 9 8 7 6 5 4 2 1 fo r A= 3 3 3 0 −1 −2 () ( 33 C = (4) B = ( 3 11 2 ) C= 8 12 4. Let A = Processing math: 100% ( ) 9 3 N ot 3. Find the transposes of the following matrices: D= 3 9 − 19 −2 8 5 3 −7 −9 ) ( ) 9 3 and B = . Find all x ∈ ℝ such that A = B. 6 4 6 x2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/15 8/30/2020 Chapter 2 Matrices ( 6. Let E = ( 7. Let A = ( 19 4 120 6 25 ) and D = 122 123 124 125 126 127 128 a b 3x + 2y − x + y ( 12 5 x 2 19 4 ) ( and F = ) 6 ) 120 x2 122 123 124 x3 ( ) 3 6 ) . Find all x ∈ ℝ such that E = F. 26x − 4 127 128 1 1 and B = . Find all x ∈ ℝ such that C = D. ut io n 5. Let C = 12 5 25 . Find all a, b, x, y ∈ ℝ such that A = B. C= ( a+b 2b − a x − 2y 5x + 3y ) and D = Find all a, b, x, y ∈ ℝ such that C = D. 6 3 4 −1 −8 ) and B = ( 7 −8 9 0 −1 0 A + B; − A; 4A − 2B; 100A + B. 10. Let A = ) 0 8 14 . . Compute N ot i. ii. iii. iv. ( −2 ( ) 6 fo r 9. Let A = D is tri b 8. Let ( ) ( ) ( ) 9 5 1 2 2 2 4 7 0 8 0 0 , B= 0 3 2 0 0 0 , and C = 4 6 8 0 3 1 . 0 0 2 Processing math: Compute 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/15 8/30/2020 Chapter 2 Matrices 1. 3A − 2B + C, 2. [3(A + B)] T, 11. Find the matrix A if ) T . [( ) ( ) ] ( 3A T − −7 −2 T T −6 9 = − 5 − 10 33 12 ) ut io n ( 1 3. 4A + 2 B − C . [ ( )] ( ) 1 B+ 8 3 1 4 T −3 −5 6 8 −9 −4 2 T = ( 23 − 16 17 − 16 26.5 2 ) . N ot fo r 2 6 3 D is tri b 12. Find the matrix B if Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/15 8/30/2020 2.2 Product of two matrices Product of a matrix and a vector Definition 2.2.1. a 12 ⋯ a 1n a 21 a 22 ⋯ a 2n ⋮ ⋮ a m1 a m2 ⋱ ⋮ ⋯ a mn )( ) ( x1 x2 ⋮ xn = → → a 11x 1 + a 12x 2 + ⋯ + a 1na n a 21x 1 + a 22x 2 + ⋯ + a 2nx n ⋮ a m1x 1 + a m2x 2 + ⋯ + a mnx n → → D is tri b ( a 11 and X = (x 1, ⋯, x n) T. The product AX of A and X is defined by ) . fo r Let A be the same as in (2.1.1) ut io n 2.2 Product of two matrices → Note that AX is a vector in ℝ m, where m is the number of rows of A while X is a vector in ℝ n, where n is the → → N ot number of columns of A. AX is called the image of X under A. → → We denote by A(ℝ n) the set of all the vectors AX as X varies in ℝ n, that is, (2.2.1) A(ℝ n) = { AX ∈ ℝ → → m: X } ∈ ℝn . Loading [MathJax]/jax/output/HTML-CSS/jax.js By Definition 2.2.1 , we see https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/22 8/30/2020 2.2 Product of two matrices ( a 21x 1 + a 22x 2 + ⋯ + a 2na x ⋮ a m1x 1 + a m2x 2 + ⋯ + a mnx n → → → where r 1 , r 2 , ⋯, r m are the row vectors of A given in (2.1.4) Example 2.2.1. 0 3 −1 4 ) → , X= ( ) () () Solution → x2 , X = 1 x3 1 → → rm ⋅ X ) . 0 fo r ( 2 ⋮ , → 2 , and X = 2 1 → → → 2 . Compute AX, AX 1, and AX 2. 2 N ot Let A = 1 x1 )( = → → r2 ⋅ X D is tri b → AX = a 11x 1 + a 12x 2 + ⋯ + a 1nx n → → r1 ⋅ X ut io n (2.2.2) → → Let r 1 = (1, 2, 0) and r 2 = (1, − 3, 4) be the row vectors of A. Then Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/22 8/30/2020 2.2 Product of two matrices = (1)(x 1) + (2)(x 2) + (0)(x 3) = x 1 + 2x 2 and = (3)(x 1) + ( − 1)(x 2) + (4)(x 3) = 3x 1 − x 2 + 4x 3. x3 Hence, → ( 2 0 3 −1 4 We can rewrite AX in a simple way. ) ()( ) x2 x3 = → → r1 ⋅ X = ( fo r AX = 1 x1 → → r2 ⋅ X x 1 + 2x 2 3x 1 − x 2 + 4x 3 ) . N ot → D is tri b () x1 → → r 2 ⋅ X = (1, − 3, 4) x 2 ut io n () x1 → → r 1 ⋅ X = (1, 2, 0) x 2 x3 Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/22 2.2 Product of two matrices AX = 0 ) ()( x2 x3 x 1 + 2x 2 3x 1 − x 2 + 4x 3 ( 1 ( 1 2 0 3 −1 4 2 0 3 −1 4 = ) )( ) ( = 1 )( ) ( (1)(1) + (2)(2) + (0)(1) (3)(1) + ( − 1)(2) + (4)(1) 0 2 2 (3)(x 1) + ( − 1)(x 2) + (4)(x 3) . 1 2 (1)(x 1) + (2)(x 2) + (0)(x 3) = ) ) () 5 (1)(0) + (2)(2) + (0)(2) (3)(0) + ( − 1)(2) + (4)(2) = 5 . ) () = 4 6 . fo r → AX 2 = 2 3 −1 4 ( = → AX 1 = 1 ut io n ( → x1 D is tri b 8/30/2020 Example 2.2.2. N ot Let A= ( ) −4 1 5 −1 3 0 2 4 () () → 0 , X= and X 1 = . 1 4 → 1 Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/22 8/30/2020 2.2 Product of two matrices → → Compute AX and AX 1. Solution → AX 1 = 5 −1 3 0 2 4 (2)(1) + (4)(1) −4 1 − 4(0) + (1)(4) 5 −1 0 3 0 4 2 4 () 1 1 = = (5)(1) + ( − 1)(1) (3)(1) + (0)(1) )() )() −3 = 4 3 . 6 ut io n ( − 4)(1) + (1)(1) (5)(0) + ( − 1)(4) (3)(0) + (0)(4) (2)(0) + (4)(4) 4 = −4 0 . 16 , a matrix and a vector can be multiplied together only when the number of columns of the fo r From (2.2.2) ( ) ( ( )( ) ( 1 D is tri b → AX = −4 → Example 2.2.3. Let N ot matrix equals the number of components of the vector. Hence, if the vectors →r i and X have a different → → number of components, then the scalar product r i ⋅ X is not defined. Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/22 8/30/2020 2.2 Product of two matrices ( ) () A= 2 3 1 → −1 , X = 1 1 1 1 −1 () 1 () → → 0 , X2 = , and X 3 = 0 1 2 4 → Determine whether A X i is defined for i = 1, 2, 3. → → → AX 2 is defined, but AX 1 and AX 3 are not defined. Theorem 2.2.1. , X, Y ∈ ℝ n and α, β ∈ ℝ. → → → → fo r Then D is tri b Solution Let A be the same as in (2.1.1) . ut io n −1 → ˉ + β(AY). A(αX + βY) = α(AX) N ot Example 2.2.4. Let ( ) () () 1 1 A= 0 → 2 0 −1 , X = 1 0 1 1 2 −1 0 → , and Y = 1 . −1 Loading [MathJax]/jax/output/HTML-CSS/jax.js → → Compute A(2X + 3Y). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/22 8/30/2020 2.2 Product of two matrices Solution → → → → A(2X + 3Y) = 2(AX) + 3(AY) ( )( ) ( )( ) () ( ) () ( ) ( ) = 2 2 0 −1 1 0 1 3 = 2 3 0 1 1 1 2 + 3 2 0 −1 1 0 1 −1 6 9 −1 1 +3 1 −1 = 6 3 3 + 0 → → → Let a 1, a 2, ⋯, a n be the column vectors of A given in (2.1.2) expressed by a linear combination of these column vectors: = −3 0 1 9 . −3 . By (1.4.6) and (2.2.2) → , AX can be fo r (2.2.3) 0 ut io n 0 D is tri b 1 1 → → → AX = x 1 a 1 + x 2 a 2 + ⋯ + x n a n . Example 2.2.5. ( ) () 1 2 3 Let A = 4 5 6 2 → and X = 7 8 9 N ot → → 3 . Write AX as a linear combination of the column vectors of A. 4 Loading [MathJax]/jax/output/HTML-CSS/jax.js Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/22 8/30/2020 2.2 Product of two matrices () () () 1 2 AX = 2 4 7 +3 5 8 → 3 +4 6 . 9 Theorem 2.2.2. { → → → a 1, a 2, ⋯, a n } be the set of the column vectors of A given in (2.1.2) (2.2.4) A(ℝ n) = span S. Proof → → → , there exists a vector X ∈ ℝ n such that TX = b. By fo r → Let b = (b 1, b 2, ⋯, b m) T ∈ A(ℝ n). By (2.2.1) (2.2.3) and (1.4.8) , . Then D is tri b Let S = ut io n The following result gives the relation between A(ℝ n) and the spanning space. N ot → → → b = AX = x 1a 1 + x 2a 2 + ⋯ + x na n ∈ span S. → → → This shows that A(ℝ n) is a subset of span S. Conversely, if b ∈ span S, then by Theorem 1.4.3 , there → exists a vector X ∈ ℝ n such that → → → b = x 1a 1 + x 2a 2 + ⋯ + x na n. → Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/22 8/30/2020 2.2 Product of two matrices This, together with (2.2.3) → → , implies b = AX ∈ A(ℝ n). This shows that span S is a subset and (2.2.1) of A(ℝ n). We have shown that A(ℝ n). is a subset of span S and span S is a subset of A(ℝ n). Hence, (2.2.4) holds. Product of two matrices Bn × r = ⋯ b 21 b 22 ⋯ ⋯ ⋯ ⋱ b n1 b n2 ⋯ ⋯ b nr ) () () () b 12 N ot b 11 → b1 = b 2r fo r and let b 1r ut io n ( b 11 b 12 . Let D is tri b Let A be the same as in (2.1.1) → → → and r 1 , r 2 , ⋯, r m be the row vectors of A given in (2.1.4) b 21 ⋮ b n1 → , b2 = b 22 ⋮ b n2 → , ⋯, b r = b 1r b 2r ⋮ c nr be the column vectors of B. Definition 2.2.2. Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/22 8/30/2020 2.2 Product of two matrices The product AB of A and B is defined by ( → → → → r2 ⋅ b1 r2 ⋅ b2 ⋮ ⋮ → → → → rm ⋅ b1 rm ⋅ b2 From Definition 2.2.5, we see that ( → → r 1 ⋅ br ⋯ → → r2 ⋅ br ⋱ ⋮ ⋯ → → rm ⋅ br ) N ot → → → AB = Ab 1 Ab 2 ⋯ A b r and the size of AB is m × r ) . fo r (2.2.6) ⋯ D is tri b AB = → → → → r1 ⋅ b1 r1 ⋅ b2 ut io n (2.2.5) The order for computing AB is to compute the first row → → → → → → r 1 ⋅ b 1, r 1 ⋅ b 2, ⋯, r 1 ⋅ b r Loading [MathJax]/jax/output/HTML-CSS/jax.js then the second row https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/22 8/30/2020 2.2 Product of two matrices → → → → → → r 2 ⋅ b 1, r 2 ⋅ b 2, ⋯, r 2 ⋅ b r and so on until the last row Example 2.2.6. Let ( ) 1 3 −2 4 and B = Find AB and the sizes of A, B, and AB. ( 3 −2 0 ) D is tri b A= ut io n → → → → → → r m ⋅ b 1, r m ⋅ b 2, ⋯, r m ⋅ b r . 5 6 0 . ( 3 −2 0 5 6 0 ) = ((1)(3) + (3)(5), (1)( − 2) + (3)(6), (1)(0) + (3)(0)) N ot (1 3) fo r The order of computing AB is to compute the first row of AB, = (18, 16, 0) and then compute the second row of AB, (−2 4) ( 3 −2 0 5 6 0 ) = (( − 2)(3) + (4)(5), ( − 2)( − 2) + (4)(6), ( − 2)(0) + (4)(0)) = (14, 28, 0). Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/22 8/30/2020 2.2 Product of two matrices Solution = 3 3 −2 0 −2 4 5 6 0 (1)(3) + (3)(5) (1)( − 2) + (3)(6) (1)(0) + (3)(0) ( − 2)(3) + (4)(5) ( − 2)( − 2) + (4)(6) ( − 2)(0) + (4)(0) 18 16 0 14 28 0 ) ut io n = ( )( ) ( ( ) 1 . D is tri b AB = The sizes of A, B, and AB are 2 × 2, 2 × 3, 2 × 3, respectively. Example 2.2.7. Let 2 6 0 ) ( ) 1 4 fo r ( 4 and B = 0 −1 3 . 2 7 5 N ot A= 1 2 4 Compute AB and find the sizes of A, B, and AB. The order of computing AB is to compute the first row of AB, Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/22 8/30/2020 2.2 Product of two matrices ( ) 4 1 4 (1 2 4) 0 − 1 3 2 7 5 = ((1)(4) + (2)(0) + (4)(2), (1)(1) + (2)( − 1) + (4)(7), (1)(4) + (2)(3) + (4)(5) = (12, 27, 30) ( ) (2 6 1 4 0) 0 − 1 3 2 7 5 = ((2)(4) + (6)(0) + (0)(2), (2)(1) + (6)( − 1) + (0)(7), (2)(4) + (6)(4) + (6)(3) + (0)(5) D is tri b 4 ut io n and then compute the second row of AB, = (8, − 4, 26). Solution = = )( 1 4 ) ( 1 2 4 ( ( (1)(4) + (2)(0) + (4)(2) (1)(1) + (2)( − 1) + (4)(7) (1)(4) + (2)(3) + (4)(5) 2 6 0 0 −1 3 2 7 5 N ot AB = 4 fo r By computation, we have (2)(4) + (6)(0) + (0)(2) (2)(1) + (6)( − 1) + (0)(7) (2)(4) + (6)(3) + (0)(5) 12 27 30 8 − 4 26 ) ) . Loading [MathJax]/jax/output/HTML-CSS/jax.js The sizes of A, B, and AB are 2 × 3, 3 × 3, and 2 × 3, respectively. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/22 8/30/2020 2.2 Product of two matrices Example 2.2.8. Let ( 4 1 5 ) and B = ( 7 −1 4 2 5 −3 1 7 ) 0 −4 . 2 3 ut io n A= 2 0 −3 Compute AB and find the sizes of A, B, and AB. Solution ( 4 1 5 )( 7 −1 4 7 2 5 0 −4 −3 1 2 3 )( = 23 − 5 2 15 26 39 fo r AB = 2 0 −3 D is tri b We always follow the order like the above two examples to compute AB, but we do not need to mention the order every time when we compute AB. 6 5 ) . N ot The sizes of A, B, and AB are 2 × 3, 3 × 4, , and 2 × 4, respectively. → In (2.2.6) , for each i = 1, 2, ⋯, r, the product A b i requires that the number of columns of the matrix A be → equal to the number of components of the vector b i . Hence, the product of two matrices A and B requires that the number of columns of A be equal to the number of rows of B. Symbolically, (2.2.7) Loading [MathJax]/jax/output/HTML-CSS/jax.js (m × n)(n × r) = m × r. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/22 8/30/2020 2.2 Product of two matrices The inner numbers must be the same and the outer numbers give the size of the product AB. Hence, if the inner numbers are not the same, the product AB is not defined. Example 2.2.9. A= 2 1 and B = 3 0 ( ) 1 2 3 4 . D is tri b ( ) 1 0 ut io n 1. Let A, B, and C be matrices such that AB = C. Assume that the sizes of A and C are 5 × 3 and 5 × 4. respectively. Find the size of B. 2. Let Are AB and BA defined? If so, compute them. If not, explain why. 3. Let A = ( ) 1 3 −2 4 and B = ( ) 3 −2 5 6 . Are AB and BA defined? If so, compute them. If not, explain why. Is AB equal to BA? fo r Solution N ot 1. Because AB = C and the sizes of A and C are 5 × 3 and 5 × 4, respectively, by (2.2.7) , the size of B is 3 × 4, 2. The size of A is 3 × 2 and the size of B is 2 × 2. Hence, AB is defined because the number of columns of A and the number of rows of B are the same and equal 2. ( )( ) ( ) 1 0 AB = 2 1 3 0 1 2 3 4 1 2 = 5 8 . 3 6 Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/22 8/30/2020 2.2 Product of two matrices BA is not defined because the number of columns of B is 2 and the number of rows of A is 3 and they are not the same. 3. AB and BA are defined because their sizes are 2 × 2. ( )( ) ( ( )( ) ( 7 3 −2 4 3 −2 5 6 = 14 28 ) and 3 −2 5 6 1 −2 4 Hence, AB ≠ BA. Following (2.2.6) 3 = 1 − 7 39 ) . D is tri b BA = ut io n AB = 18 16 1 , the following properties can be easily proved. The proofs are left to the reader. The following assertions hold: fo r Theorem 2.2.3. N ot 1. (Associative law for matrix multiplication) Let A : = A m × n, B : = B n × p and C : = C p × q. Then A(BC) = (AB)C. 2. (Distributive laws for matrix multiplication) i. Let A : = A m × n, B : = B n × p and C : = C n × q. Then A(B + C) = AB + AC. Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/22 8/30/2020 2.2 Product of two matrices ii. Let A : = A m × n, B : = B m × n and C : = C n × q. Then (A + B)C = AC + BC. 3. (AB) T = B TA T. Let ( ) ( 0 2 , B= 2 −1 4 3 1 5 ) ( ) 0 , and C = Show that A(BC) = (AB)C and (AB) T = B TA T. 4 3 −5 0 2 . 6 N ot fo r Solution −2 1 D is tri b A= 1 −3 ut io n Example 2.2.10. Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/22 8/30/2020 2.2 Product of two matrices ( 3 1 )( −2 1 4 3 2 −5 0 6 )( = − 24 − 7 24 − 21 − 3 35 − 24 − 7 24 0 − 21 − 3 35 2 2 −1 4 0 3 2 ( 1 − 7 − 4 − 11 6 2 10 )( = 6 0 2 −2 1 4 3 2 −5 0 6 −1 1 5 0 −3 2 = 2−9 (AB) T = ( −1 − 3 39 2 10 ) − 81 70 ) )( ) ( ) 0+2 −7 = 2 − 42 − 6 −7 = 4 − 15 0 + 10 − 7 − 4 − 11 T 6 )( = 0+6 N ot 4 1 70 10 fo r ( )( ) ( 3 − 81 − 42 − 6 Hence, A(BC) = (AB)C. 2 2 − 7 − 4 − 11 = 5 39 ut io n 1 −3 1 −3 (AB)C = B TA T = ) ( )( ) ( ) ( )( ) ( ) A(BC) = AB = 5 0 D is tri b BC = 2 −1 4 6 −4 2 . − 11 10 6 −4 2 . − 11 10 Hence, (AB) T = B TA T. Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/22 8/30/2020 2.2 Product of two matrices Exercises 1 1 −1 2 ) → , X= ( ) () ( ) () −2 −1 2. 2. A = ( ) () () 3. Let A = 3 1 0 1 0 1 → , X= 0 → x2 , X = 1 x3 2 → 1 , X = 2 2 → → 1 . Compute AX, AX , and AX . 1 2 1 () → → 1 → , X1 = . Compute AX and AX 1. 2a 2 −a → 1 − 1 , X1 = 2 1 1 () → 1 → , X2 = , X3 = 0 0 −1 () 1 1 1 . 2 ( ) () () ( ) () 1 → 1 0 −1 , X = 1 1 2 0 1 0 2 2 1 −1 , Y= −1 1 1 −1 5. Let A = 3 → → → → . Compute A(X − 3Y). −1 1 and X = 0 N ot 4. Let A = 0 fo r → Determine whether A X i . is defined for each i = 1, 2, 3. 0 1 → ut io n ( 1 D is tri b 1. Let A = 2 x1 → − 1 . Write AX as a linear combination of the column vectors of A. 0 6. We are interested in predicting the population changes among three cities in a country. It is known that in the current year, 20% and 30% of the populations of City 1 will move to Cities 2 and 3, respectively, Loading [MathJax]/jax/output/HTML-CSS/jax.js 10% and 20% of the populations of City 2 will move to Cities 1 and 3, respectively, and 25% and 10% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/22 8/30/2020 2.2 Product of two matrices of the populations of City 3 will move to Cities 1 and 2, respectively. Assume that 200,000, 600,000, and 500,000 people live in Cities, 1, 2, and 3, respectively, in the currently year. Find the populations in the three cities in the following year. ( ( 1 0 −2 9. Let A = −2 3 2 3 and B = 1 ) ( 1 −2 1 3 and B = and B = ( ) and B = 1 1 10. 10. Let A = 1 3 1 3 1 ) . Find AB and the sizes of A, B, and AB. ( ) ( ) 2 1 −1 1 0 1 −1 2 4 0 ) 3 2 −1 4 −1 2 −1 2 2 0 ( ) 2 1 −2 3 . Compute AB and find the sizes of A, B, and AB. 1 1 − 2 . Compute AB and find the sizes of A, B, and AB. 0 1 . Are AB and BA defined? If so, compute them. If not, explain why. ( 2 −1 −2 3 ) and B = ( ) 0 −2 4 2 . Are AB and BA defined? N ot 11. Let A = ut io n 8. Let A = 0 1 −1 1 D is tri b ( ) 2 fo r 7. Let A = If so, compute them. If not, explain why. Is AB equal to BA? 12. Let A= ( 1 −1 −1 2 ) ( , B= 0 −1 2 1 1 3 ) , and C = ( ) 1 −1 1 0 3 −2 0 2 . 3 Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/22 8/30/2020 2.2 Product of two matrices ( ) 1 1 1 1 . 1 1 0 1 D is tri b A= 0 1 1 1 ut io n Show that A(BC) = (AB)C and (AB) T = B TA T. 13. Two students buy three items in a bookstore. Students 1 and 2 buy 3,2,4 and 6,2,3, respectively. The unit prices and unit taxes are 5, 4, 6 (dollars) and 0.08, 0.07, 0.05, respectively. Use a matrix to show the total prices and taxes paid by each student. 14. Assume that three individuals have contracted a contagious disease and have had direct contact with four people in a second group. The direct contacts can be expressed by a matrix Now, assume that the second group has had a variety of direct contacts with five people in a third group. The direct contacts between groups 2 and 3 are expressed by a matrix ( ) 1 0 0 1 0 0 0 0 1 0 . fo r B= 1 1 0 1 1 N ot 1 0 0 1 0 i. Find the total number of indirect contacts between groups 1 and 3. ii. How many indirect contacts are between the second individual in group 1 and the fourth individual in group 3? 15. 15. There are three product items, P1, P2, and P3, for sale from a large company. In the first day, 5, 10, and 100 are sought out. The corresponding unit profits are 500, 400, and 20 (in hundreds of dollars) and the corresponding unit taxes are 3, 2, and 1. Find the total profits and taxes in the first Loading [MathJax]/jax/output/HTML-CSS/jax.js day. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 21/22 2.2 Product of two matrices N ot fo r D is tri b ut io n 8/30/2020 Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 22/22 8/30/2020 2.3 Square matrices 2.3 Square matrices (2.3.1) ( ⋮ ⋮ a n1 a n2 ⋯ a 1n ⋯ a 2n ⋱ ⋮ ⋯ a nn ) D is tri b A= a 11 a 12 a 21 a 22 ut io n An n × n matrix is called a square matrix of order n, that is, . fo r The entries a 11, a 22, ⋯, a nn are said to be on the main diagonal of A. The sum of these entries, denoted by tr(A), is called the trace of A, that is, N ot (2.3.2) n tr(A) = ∑ a ii = a 11 + a 22 + ⋯ + a nn. l=1 Example 2.3.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/11 8/30/2020 2.3 Square matrices Let A = ( 15 21 44 25 ) . Then A is a square matrix of order 2, the numbers 15, 25 are on the main diagonal of A, and tr(A) = 15 + 25 = 40. ut io n Symmetric matrices A square matrix A is said to be symmetric if A T = A. By (2.3.3) we see that A is symmetric if and only if D is tri b (2.3.3) a ij = a ji for i, j ∈ I n. a 11 a 12 ⋯ N ot ( fo r We can write a symmetric matrix in an explicit form A= a 12 a 22 ⋮ ⋮ a 1n a 2n a 1n ⋯ a 2n ⋱ ⋮ ⋯ a nn ) . From the above matrix, we see that A is symmetric if the ith row and the ith column are the same for each i ∈ I n. Example 2.3.2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/11 8/30/2020 2.3 Square matrices 1. The following matrices are symmetric. −3 −3 0 )( 4 5 4 −3 0 5 0 7 )( ) 1 x2 2 x2 0 x 2 x 3 2. The following matrices are not symmetric. 2 0 ( )( 1 4 5 2 −3 0 5 0 7 Definition 2.3.1. 0 1 2 2 2 2 ) fo r A square matrix is said to be 2 x2 + 1 2 D is tri b ( ) 7 −3 ut io n ( 7 1 N ot i. lower triangular if all the entries above the main diagonal are zero; ii. upper triangular if all the entries below the main diagonal are zero; iii. triangular if it is lower or upper triangular; iv. diagonal if it is lower and upper triangular; v. identity matrix if it is a diagonal matrix whose entries on the main diagonal are 1. We denote by I or I n an n × n identity matrix. Hence, (2.3.4) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/11 2.3 Square matrices ( In = 1 0 ⋯ 0 0 1 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ 1 ) . Example 2.3.3. 1. The following matrices are lower triangular. 2 0 ( ) ( ) ( ) 1 0 0 0 0 0 1 0 0 2 0 0 0 0 0 1 0 0 0 1 1 0 0 0 x 0 0 D is tri b ( ) 0 0 ut io n 8/30/2020 2. The following matrices are upper triangular. 0 0 0 0 0 1 N ot 0 0 1 0 1 )( ) 0 0 0 fo r ( )( 0 1 0 0 0 0 0 0 3. The following matrices are diagonal. ( )( 1 0 0 2 1 0 0 0 0 0 0 0 1 )( ) 0 0 0 0 0 0 0 0 0 4. The following matrices are not triangular matrices. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/11 8/30/2020 2.3 Square matrices ( )( ) ( 1 1 0 1 0 1 0 2 0 2 0 0 1 1 1 0 0 1 0 0 0 0 0 0 ) ( ) ( ) 0 1 0 , I4 = 0 0 1 ( ) 1 0 0 0 0 1 0 0 0 0 1 0 . 0 0 0 1 D is tri b 1 0 I 1 = (1), I 2 = , I3 = 0 1 1 0 0 ut io n 5. The following are identity matrices. The following theorem gives some properties of triangular matrices. The proofs are left to the reader. Theorem 2.3.1. Example 2.3.4. Let A = ( ) 1 0 2 1 and B = ( ) 3 0 4 1 N ot fo r i. If A is an upper triangular (lower triangular) matrix, then AT is a lower triangular (upper triangular) matrix. ii. If A and B are lower triangular matrices, then AB is a lower triangular matrix. iii. If A and B are upper triangular matrices, then AB is an upper triangular matrix. . Compute AB. Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/11 8/30/2020 2.3 Square matrices AB = ( )( ) ( ) 1 0 3 0 2 1 4 1 = 3 0 10 1 . The above example shows that the product of two lower triangular matrices is a lower triangular matrix. (2.3.5) D is tri b A 0 = I n, A 2 = A ⋅ A, A 3 = A 2 ⋅ A, ⋯ , A n = A n − 1 ⋅ A. Example 2.3.5. Let A = ( ) 1 2 1 3 . Find A 0, A 2, and A 3. ( ) ( ) ( )( ) ( ) ( )( ) ( 1 2 0 1 3 = 1 0 0 1 , N ot A2 = fo r Solution A0 = ut io n For a square matrix, we can define its power as follows. 1 2 1 2 1 3 1 3 A 3 = A 2A = 3 = 3 8 4 11 8 1 2 4 11 1 3 = , 11 30 15 41 ) . We can write a diagonal matrix as https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/11 8/30/2020 2.3 Square matrices (2.3.6) ( ) 0 ⋯ 0 0 a2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ an . It is easy to verify that for k ∈ ℕ, D is tri b diag(a 1, a 2, ⋯ , a n) k = diag(a 1k , a 2k , ⋯ , a nk ), ut io n diag(a 1,a 2,⋯ , a n) = a1 that is, (2.3.7) ⋯ 0 0 a2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ an k = a 1k 0 ⋯ 0 0 a 2k ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ a nk N ot 0 fo r ( )( ) a1 . Example 2.3.6. Find A 5 if https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/11 8/30/2020 2.3 Square matrices ( ) 1 A= 0 0 0 −2 0 . 0 0 2 Because A is a diagonal matrix, it follows from (2.3.7) ( 0 0 0 ( − 2) 5 0 0 0 25 )( ) 1 = It is easy to verify that for r, s ∈ ℕ, 0 − 32 0 0 . 32 0 0 fo r (2.3.8) 0 D is tri b A5 = 15 that ut io n Solution s A r ⋅ A s = A r + s and (A r) = A rs. N ot Using the power of a square matrix, we introduce the notion of a matrix polynomial, which is a polynomial with square matrices as variables. Let P(x) = a 0 + a 1x + ⋯ + a mx m be a polynomial of x. Then the polynomial evaluated at a matrix A is (2.3.9) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/11 8/30/2020 2.3 Square matrices p(A) = a 0I n + a 1A + a 2A 2 + ⋯ + a mA m, which is called a matrix polynomial of degree m. Let P(x) = 4 − 3x + 2x 2 and A = ( ) −1 2 0 3 . Compute the matrix polynomial P(A). Solution ( ) ( ) ( ) ( )( )( ) ( ) −1 2 P(A) = 4I 2 − 3A + 2A = 4 −3 0 1 0 4 0 0 4 − −3 6 0 9 2 8 0 18 = 3 9 +2 2 0 13 0 3 . fo r Exercises + −1 2 2 D is tri b 1 0 2 = ut io n Example 2.3.7. A1 = N ot 1. For each of the following matrices, find its trace and determine whether it is symmetric. ( ) 0 −1 −1 0 ( ) ( ) 5 y x3 1 4 − 3 − 1 , A3 = 5 0 7 x3 y x . 1 x z 1 , A2 = 4 2. Identify which of the following matrices are upper triangular, diagonal, triangular, or an identity matrix. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/11 2.3 Square matrices ( ) ( ( ) ( ) ( ( ) ( ) ( ( ) ( ) ( ( ) 0 0 1 0 0 A2 = 2 0 0 1 0 0 x 0 0 A7 = 2 0 0 2 A5 = 0 0 0 2 0 1 ( ) 4. Let A = ( ) 0 −1 1 2 0 . Find A 0, A 2, A 3, and A 4. 0 0 − 2 0 . Find A 6. 0 0 3 5. Let P(x) = 1 − 2x − x 2 and A = 0 0 0 0 0 1 ) ) 1 0 0 0 2 0 0 0 0 0 A 12 = −1 1 1 ) ) 0 0 1 0 1 0 1 0 0 N ot 1 A9 = 0 0 A 11 = 0 0 0 fo r 2 0 0 1 0 1 3. Let A = A6 = 1 0 1 0 0 A8 = 0 0 0 1 0 1 0 1 1 0 1 A 10 = A3 = 0 0 1 1 0 x A4 = 0 0 0 D is tri b A1 = 0 2 ut io n 8/30/2020 ( ) −1 1 1 2 . Compute P(A). 6. Let P(x) = x 2 − x − 4. Compute P(A) and P(B), where https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/11 8/30/2020 2.3 Square matrices ( ) ( 3 A= 1 3 −2 4 5 0 3 and B = −1 6 −2 −1 3 0 2 −1 3 4 ) . D is tri b ut io n 7. Assume that the population of a species has a life span of less than 3 years. The females are divided into three age groups: group 1 contains those of ages less than 1, group 2 contains those of age between 1 to 2, and group 3 contains those of age 2. Suppose that the females in group 1 do not give birth and the average numbers of newborn females produced by one female in group 2 and 3, respectively, are 3 and 2. Suppose that 50% of the females in group 1 survive to age 1 and 80% of the females in group 2 live to age 2. Assume that the numbers of females in groups 1, 2, and 3 are X 1, X 2, and X 3, respectively. i. Find the number of females in the three groups in the following year. ii. Find the number of females in the three groups in n years. iii. If the current population distribution (x 1, x 2, x 3) = (1000,500,100), then use result 2 to predict the N ot fo r population distribution in 2 years. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/11 8/30/2020 2.4 Row echelon and reduced row echelon matrices 2.4 Row echelon and reduced row echelon matrices Arowinan m × n matrix is said to be a zero row if all the entries in that row are zero. If one of the entries in a row is nonzero, then it is called a nonzero row. Example 2.4.1. ( ) 1 0 0 In the matrix 2 1 0 , the first two rows are nonzero rows and the last row is a zero row. 0 0 0 ut io n The first nonzero entry (starting from the left) in a nonzero row is called the leading entry of the row. If a leading entry is 1, it is called a leading one. Example 2.4.2. A= ( ) 1 6 0 4 0 2 0 5 3 0 0 10 The numbers 1, 2, and 3 are the leading entries of A. Definition 2.4.1. N ot Solution 0 fo r 0 0 0 . D is tri b Find the leading entries of A, where An m × n matrix A is said to be a row echelon matrix if it satisfies the following conditions (1) and (2); or to be a reduced row echelon matrix if it satisfies the following conditions (1), (2), (3), and (4). 1. Each nonzero row (if any) lies above every zero row, that is, all zero rows appear at the bottom of the matrix. 2. For any two successive nonzero rows, the leading entry in the lower row occurs further to the right than the leading entry in the higher row. 3. Each leading entry is 1. 4. For each column containing a leading 1, all entries above the leading 1 are zero. Remark 2.4.1. i. For a row echelon matrix, if a column contains a leading entry, then by condition (2), all entries below the leading entry of that column must be zero. ii. A reduced row echelon matrix must be a row echelon matrix. iii. For a reduced row echelon matrix, if a column contains a leading 1, then by conditions (2) and (4), all entries except the leading 1 of that column must be zero. Processing math: 26% Example 2.4.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/9 8/30/2020 2.4 Row echelon and reduced row echelon matrices 1. The following matrices are row echelon matrices. ( )( 1 0 0 1 , 2 0 3 5 )( 2 0 1 )( ) 1 0 0 6 0 0 0 2 3 0 , 0 1 0 , . 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 ( )( )( 1 0 0 0 0 1 0 0 1 0 )( 1 0 0 0 ) 0 0 1 , 0 0 0 , 1 0 0 0 , 0 0 0 0 . 0 1 0 0 0 1 0 0 0 0 0 0 0 1 ( )( ) 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 6 0 , 0 0 0 0 0 0 1 3 0 0 0 0 0 1 , ( 0 1 0 0 −1 0 0 0 1 0 0 0 0 0 ( )( 2 ) . fo r 4. The following matrices are row echelon matrices, but not reduced row echelon matrices: 2 0 1 D is tri b 3. The following matrices are reduced row echelon matrices. ut io n 2. The following matrices are not row echelon matrices. 1 0 2 6 0 )( ) N ot 0 0 1 3 0 1 0 0 2 0 0 1 0 , , . 0 0 0 0 1 0 0 2 1 0 0 0 0 0 0 0 0 0 Solution We only consider (4). These matrices are row echelon matrices. However, the first matrix contains a leading entry 2 that is not a leading 1; column 3 of the second matrix has the leading 1 of the second row but also contains a nonzero entry 2; and the third matrix has a leading entry that is not 1. Hence, they are not reduced row echelon matrices. Row operations. In many cases, we need to change a matrix to a row echelon matrix or a reduced row echelon matrix and wish that the new matrix and the original matrix share some common properties. Thus, one can obtain properties of the original matrix by studying the row echelon matrix or the reduced row echelon matrix. The powerful tools used to change a matrix to a row echelon matrix or reduced row echelon matrix are the following (elementary) row operations. 1. First row operation R j(c) : Multiply theith row of a matrix by a nonzero number c; 2. Second row operation R i(c) + R j : Add the ith row multiplied by a nonzero number c to the jth row; Processing math: 26% 3. Third row operation R i , j: Interchange the ith row and the jth row. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/9 8/30/2020 2.4 Row echelon and reduced row echelon matrices Remark 2.4.2. The first row operation R i(c) is used to change only the ith row, so other rows remain unchanged. The second row operation R i(c) + R j is used to change only the jth row, so other rows including the ith row remain unchanged. The third row operation R i , j is used to interchange only the ith row and the jth row, so other rows remain unchanged. Now, we give some examples to show how to use the above row operations to change one matrix to another. These row operations in these examples are used to show how to perform the row operations and are chosen for practice. There are no reasons why we choose these row operations here. You can choose other row operations for practice as well. Later, when we want to change a matrix to a row echelon matrix or reduced row echelon matrix, we must choose suitable row operations to obtain its row echelon matrices. Example 2.4.4. ( ) 1 ut io n 0 1 Let A = 1 2 −3 . 2 1 1 ( ) ( ) 0 1 1 0 1 2 − 3 R ( − 2)→ 2 2 1 1 ii. We apply R 2( − 2) + R 3 to change row 3 of A, so rows 1 and 2 remain unchanged. ( ) 0 1 1 2 − 3 R ( − 2) + R → 2 3 2 1 1 1 ( ) N ot A= 1 1 −2 −4 6 . 2 1 1 fo r A= D is tri b i. We apply R 2( − 2) to change row 2 of A, so rows 1 and 3 remain unchanged. 0 1 1 2 1 −3 . 7 0 −3 iii. We apply R 1 , 2 to change rows 1 and 2 of A, so row 3 remains unchanged. ( ) ( ) 0 1 A= 1 1 2 −3 1 2 −3 R → 1,2 2 1 1 0 1 1 2 1 1 . Example 2.4.5. Let Processing math: 26% A= ( 1 2 1 1 ) −1 −1 . −3 4 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/9 8/30/2020 2.4 Row echelon and reduced row echelon matrices Use the second row operation to change the numbers 2 and − 3 in A to 0. Solution To change the number 2 in the first column and second row of A, we use the row operation R 1( − 2) + R 2 to change the entire second row. A= ( 1 1 1 2 ) − 1 − 1 R ( − 2) + R → 1 2 −3 4 1 ( 1 1 1 0 ) −3 −3 . −3 4 1 ( 1 0 1 1 ) ( ) 1 − 3 − 3 R (3) + R → 1 3 −3 4 1 1 1 0 −3 −3 . 0 7 4 ut io n Similarly, we use R 1(3) + R 3 to change the entire third row in order to change the number − 3 to 0. D is tri b From the above, we see that row 1 is always kept unchanged when we changed rows 2 and 3. Hence, we can combine the process as follows: A=(1112−1−1−341)→R1(3)+R3R1(−2)+R2(1110−3−3074). Example 2.4.6. fo r Let B=(1110−3−3074). Use the second row operation to change the number 7 in B to 0. Solution Combining Examples 2.4.5 and 2.4.6 , we get N ot B=(1110−3−3074)R2(73)+R3→(1110−3−300−3). A=(1112−1−1−341)→R1(3)+R3R1(−2)+R2(1110−3−3074)R2(73)+R3→(1110−3−300−3):=C. Note that the last matrix C is a row echelon matrix of A and is obtained by performing second row operations three times. Methods of finding row echelon matrices After we intuitively get some ideas on how to change a matrix to its echelon matrix from Examples 2.4.5 and 2.4.6 , we learn how to apply row operations to find a row echelon matrix for a given matrix. To find a row echelon matrix, we need to do several rounds of the same steps, depending on the number of leading entries. The first round is to find the first leading entry. Here are some key steps to find the first leading entry. Step 1. Move all zero rows, if any, to the bottom of the matrix. Look at each row to see whether there is a common factor in the row. If there is a common factor in a row, then we would use the first row operation to eliminate the common factor. After eliminating all the common factors in the rows by using the first row operation, the resulting matrix has entries that are smaller than or equal to the corresponding entries of the given matrix. Smaller entries make later computations easier. Processing math: 26% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/9 8/30/2020 2.4 Row echelon and reduced row echelon matrices Step 2. After eliminating all the common factors, we would start to seek the first leading entry in the first nonzero column of the resulting matrix. Case I. The first nonzero column of the resulting matrix contains number 1or −1. i. If the number 1 or −1 is on the first row, then the number 1 or −1 can be used as the first leading entry. ii. If the number 1 or −1 is not on the first row, then we use the third row operation to interchange one of the rows that contains the number 1 or −1 with row one. So the number 1 or −1 is moved to the first row and can be used as the first leading entry. Case II. The first nonzero column of the resulting matrix does not contain number 1 or −1. ut io n a. We choose the smallest number of the first nonzero column of the resulting matrix and use the methods in (i)and (ii) to move the row containing the smallest number to the first row. The smallest number in the first row, in general, can be used as the first leading entry. b. In some cases, we can continue to change the smallest number to 1 or −1 by using the second row operation. Some examples will be provided later to exhibit the smart method, for example see Example 2.4.9 . D is tri b Step 3. Note that the row one contains the first leading entry 1 or −1 or the smallest number, and there is a column that contains the first leading entry as well. If all the entries below the first leading entry in the column are zero, then the first round is completed. If there are nonzero entries below the first leading entry in the column, then we would use the second row operation and the row one to change each of these nonzero entries to 0. Note that all other entries in the row that contain such a nonzero number must be changed following the second row operation. After all these nonzero entries are changed to 0, the first round is completed. The second round is to find the second leading entry from the last matrix obtained above. Example 2.4.7. Find a row echelon matrix of the matrix fo r Step 4. Repeat the above Steps 1-3 to find the second leading entry below row one of the last matrix obtained in the above Step 3, and then continue the process to find the third, fourth, etc., leading entry, if any, until you get a row echelon matrix. Solution N ot A=(06−32−26201). (1) Step 1. There are no zero rows in A. Look at each row to see whether there is a common factor in the row. Row 1 contains a common factor 3 and row 2 contains a common factor 2. So, we first use the first row operation to eliminate the two common factor. A=(06−32−26201)→R2(12)R1(13)(02−11−13201):=B. Step 2. Check the first nonzero column of B to see if there is number 1 or −1. The column one of B is the first nonzero column, which contains the number 1 as well. The row that contains the number 1 is not row one, so we need to move it to row one by using the third row operation. Therefore, we interchange rows 1 and 2 of B and the number 1 is the first leading entry. We use the symbol a to denote a leading entry a. B=(02−11−13201)R1,2→(1−1302−1201):=C. Step 3. If there are nonzero entries below the first leading entry in the column of C, then use the second row operation and row one to change each of these nonzero entries to 0. Processing math: 26% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 5/9 8/30/2020 2.4 Row echelon and reduced row echelon matrices We use the symbol ↓ to show the direction of changing all nonzero entries below a leading entry to zero. C=(1−1302−1201)↓R1(−2)+R3→(1−1302−102−5):=D. Step 4. Repeat the above steps 1-3 to find the second leading entry below row one of the last matrix D. In order to find the second leading entry of D, we consider the following matrix below row one of D: D1:=(02−102−5) We repeat Steps 1-3. Step 1. Look at each row of the matrix D1 to see whether there is a common factor in the row. The answer is that there are no common factors in each row of D1 ut io n Step 2. Check the first nonzero column of D1 to see if there are nonzero numbers. The first nonzero column is the second column of D1 and the first and second rows contains a nonzero number 2. So, the number 2 in the first row of D1 is the second leading entry for D. D=(1−1302−102−5). D is tri b Step 3. If there are nonzero entries below the second leading entry 2 in the column, then use the second row operation and the row 2 of D to make the nonzero entries in the column zero. In D, there is a nonzero number 2 in the third row below the second leading entry 2. We change the nonzero number 2 to 0 by using row 2 that contains the second leading entry 2. The last matrix is a row echelon matrix and the number −4 is the third leading entry. We rewrite the process as follows: fo r D=(1−1302−102−5)↓R2(−1)+R3→(1−1302−100−4). N ot A=(06−32−26201)→R2(12)R1(13)(02−11−13201)R1,2→(1−1302−1201)↓R1(−2)+R3→(1−1302−102−5)↓R2(−1)+R3→(1−1302−100−4). In the following, we always follow Steps 1-3 in each round without further statement. Example 2.4.8. Find a row echelon matrix of the matrix A=(1302011100241−1000031). Solution A= (1302011100241−1000031)↓→R1(−2)+R3R1(−1)+R2(130200−21−200−21−5000031)↓R2(−1)+R3→(130200−21−20000−3000031)R3(13)→(130200−21−20000−1000031)↓R3(3)+R4 The last matrix is a row echelon matrix of A. Remark Processing math: 26% 2.4.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 6/9 8/30/2020 2.4 Row echelon and reduced row echelon matrices Note that the row echelon matrix of a matrix is not unique. For example, in Example 2.4.8 , we can continue to use row operations to change the above row echelon matrix to another row echelon matrix such as (130200−21−20000−1000001)R3(−1)→(130200−21−200001000001). The last matrix is another row echelon matrix of A. The following example provides a smart method, which is mentioned in Step 2 of the first round. Example 2.4.9. A=(2−102030−1102−41−1051030) ut io n Find a row echelon matrix of the matrix D is tri b Analysis: The first number 2 in the first row of A would be the first leading entry. But if we use it as the first leading entry to change all the nonzero numbers 3, 2, 5 in column one to 0 by using the second row operations and row one, then the problem we encounter is that there are computations of fractions in rows 2 and 4. To avoid the computation of fractions, we can try to use suitable second row operations to get 1 or −1 in the first nonzero column. For the above matrix A, we have several choices to get 1 or −1. For example, we can use one of the following: i. R2(−1)+R1 ii. R1(−1)+R2 and then R1, 2. iii. R1(−2)+R4 and then R1, 4. fo r In the following, we use (i)toget −1 and take the number −1 as the first leading entry. We also use the smart method to choose the second and third leading entries. Solution N ot A=(2−102030−111201−10510−3−1)R2(−1)+R1→(−1−111−130−111201−10510−3−1)↓ →R1(5)+R4R1(3)+R2R1(2)+R3(−1−111−10−324−20−231−20−452−6)R3(−2)+R2→ (−1−111−101−4220−231−20−452−6)↓→R2(2)+R3R2(4)+R4(−1−111−101−42200−55200−11102)R2(−2)+R4→(−1−111−101−42200−55200−10−2)R3, 4→(−1−111−101−42200−10−2 R3(−5)+R4→(−1−111−101−42200−10−2000512). The last matrix is a row echelon matrix of A. Methods of finding reduced row echelon matrices The basic idea is to use row operations to change a given matrix A to a row echelon matrix B, where we change A from its top to bottom, and then continue using row operations to change all the nonzero numbers above the leading entries in the columns that contain the leading entries to 0, where we change B from its bottom to top. We denote by l1, l2, ⋯, lr, Processing math: 26% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 7/9 8/30/2020 2.4 Row echelon and reduced row echelon matrices the leading entries of the row echelon matrix B in order. To find the reduced row echelon matrix of A, we first use the second row operation and the row that contains the last leading entry lr to change all the nonzero numbers above the leading entry lr from the lower one to the top one to 0. Then we do the same thing to lr−1, ⋯, l2 in order. We have studied how to change a matrix to a row echelon matrix, so we first give examples demonstrating how to change a row echelon matrix to a reduced row echelon matrix and then provide examples to show how to change a matrix to a reduced row echelon matrix. Example 2.4.10. Find the reduced row echelon matrix of the following row echelon matrix B=(1−1302−100−4). Solution ut io n (1) Note that the leading entries in order are l1=1, l2=2, and l3=−4. To find the reduced row echelon matrix for the row echelon matrix, the first step is to change all the nonzero numbers above l3=−4 to be zero. The nonzero numbers above l3=−4 are −1 and 3, which are in the second row and the first row of B, respectively. We use the second row operations and the row 3 that contains the leading entry l3=−4 to change −1 and 3 to 0 in order. We use the symbol ↑ to show the direction of changing all nonzero entries above a leading entry to zero. D is tri b B=(1−1302−100−4)R3(−14)→(1−1302−1001)↑→R3(−3)+R1R3(1)+R2(1−10020001). (2) The second step is to change the nonzero number above l2=2 to 0 by using row 2 that contains l2=2. (1−10020001)R2(12)→(1−10010001)↑R2(1)+R1→(100010001). Find the reduced row echelon matrix of N ot Example 2.4.11. fo r The last matrix is the reduced row echelon matrix of B. A=(1302011100241−1000031). Solution The row echelon matrix B of A is given in Example 2.4.8 . We change the echelon matrix B to the reduced echelon matrix. B= (130200−21−20000−1000001)R3(−1)→(130200−21−200001000001)↑→R3(−2)+R1R3(2)+R2(130000−21000001000001)R2(−12)→(1300001−12000001000001)↑R2(−3)+R1→(1032 The last matrix is the reduced row echelon matrix of A. Example 2.4.12. Find the reduced row echelon matrix of A=(210130314213). Processing math: 26% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 8/9 8/30/2020 2.4 Row echelon and reduced row echelon matrices Solution A= (210130314213)R2(−1)+R1→(−11−3030314213)↓→R1(4)+R3R1(3)+R2(−11−3003−6106−113)↓R2(−2)+R3→(−11−3003−610011)↑→R3(3)+R1R3(6)+R2(−110303070011)R2(13)→( The last matrix is the reduced row echelon matrix of A. In Example 2.4.12 , we employ the smart method used in Example 2.4.9 entry in row one to −1. By Remark 2.4.3 . We use the row operation R2(−1)+R1 to change row one of A in order to change the leading , we know that the row echelon matrices of a matrix are not unique. But its row echelon matrix is unique. We give this result below without proof. Theorem 2.4.1. ut io n Any m×n matrix can be changed using row operations to a unique reduced row echelon matrix. Exercises D is tri b 1. Which of the following matrices are row echelon matrices? A=(1000)B=(9435900074)C=(250090008)D=(30160008860000100000)E=(100001010)F=(141000066)G=(081729800000)H=(000040580961) 2. Which of the following matrices are reduced row echelon matrices? A=(1001010100100000)B=(15060001300000100000)C=(1111100011)D=(201010001)E=(10040011300000100000)F=(1002000210) N ot fo r 3. Let A=(10121−1−32−2). Use the second row operation to change the numbers 2 and −3 in the first column of A to zero. 4. Let A=(10101−3031). Use the second row operation to change the number 3 in A to 0. 5. For each of the following matrices, find its row echelon matrix. Clearly mark every leading entry a by using the symbol a and use the symbol ↓ to show the direction of eliminating nonzero entries below leading entries. A=(10121−1−32−2)B=(01−21−11−210)C=(210−130325−2−13) 6. For each of the following matrices, find its reduced row echelon matrix. Clearly mark every leading entry a by using the symbol a and use the symbols ↓ and ↑ to show the directions of eliminating nonzero entries below or above leading entries. A=(2−4201−300−6)B=(11010036−3−6000−2200002)C=(06−32−26201)D=(210130325−2−13)E=(100−11−2112−212)F=(1121111−22211)G= (10111−312−2−5−2−148031−3−70) Processing math: 26% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 9/9 8/30/2020 2.5 Ranks, nullities, and pivot columns 2.5 Ranks, nullities, and pivot columns ut io n Let A be an m × n matrix. From Remark 2.4.3 , we see that the row echelon matrices of A are not unique, but the total number of leading entries for each row echelon matrix of A is the same. Hence, no matter which row operations are used to change A to a row echelon matrix, the total number of the leading entries of each row echelon matrix of A is unchanged. This leads us to define the total number of the leading entries as the rank of A. D is tri b Definition 2.5.1. Let A be an m × n matrix and let B be one of its row echelon matrices. fo r 1. The total number of leading entries of the row echelon matrix B is called the rank of A, denoted by r(A). 2. The number n − r(A) is called the nullity of A, denoted by null(A), that is, (2.5.1) N ot null(A) = n − r(A). 3. The columns of A corresponding to the columns of B containing the leading entries are called the pivot columns (or pivot column vectors) of A. According to Definition 2.5.1 , the rank of A is equal to the rank of each of its row echelon matrices. The pivot columns of A will be used in Section 7.3 to find the basis of column space of A (see Theorem 7.3.1 ). By Definitions 2.4.1 and 2.5.1 , we obtain the following result. Theorem 2.5.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/9 8/30/2020 2.5 Ranks, nullities, and pivot columns 1. The rank of A. 2. The number of leading entries of B. 3. The number of rows of B containing the leading entries. 4. The number of columns of B containing the leading entries. Remark 2.5.1. ut io n Let A be an m × n matrix and let B be one of its row echelon matrices. Then the following numbers are equal. D is tri b By Theorem 2.5.1 (3), we see that the number of zero rows of B equals m − r(A), and if A contains q zero rows, then r(A) ≤ m − q. By Theorem 2.5.1 (4), if A contains k zero columns, then r(A) ≤ n − k. . The following result gives the relation between the rank of a matrix and its size. Theorem 2.5.2. Proof fo r Let A be an m × n matrix. Then r(A) ≤ min{m, n}, where min{m, n} represents the minimum of m and n. Example 2.5.1. N ot By Theorem 2.5.1 (3), the number of rows of B containing the leading entries is smaller than or equal to m. By Theorem 2.5.1 (1)-(3), r(A) ≤ m. Similarly, by Theorem 2.5.1 (4), the number of columns of B containing the leading entries is smaller than or equal to n, and thus r(A) ≤ n. This implies that r(A) is smaller than or equal to the minimum of m and n. Let https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/9 8/30/2020 2.5 Ranks, nullities, and pivot columns ( ) ( ) 1 2 −1 2 1 B= C= −2 −4 2 2 −2 6 4 0 1 Show r(A) ≤ 3, r(B) ≤ 2, and r(C) ≤ 3. Solution , we have r(A) ≤ min{3, 4} = 3. Similarly, r(B) ≤ min{2, 3} = 2 and r(C) ≤ min{4, 3} = 3. Theorem 2.5.3. 0 2 1 1 1 1 2 0 1 D is tri b Because A is a 3 × 4 matrix, by Theorem 2.5.2 ( ) 0 4 2 ut io n A= 1 −1 3 2 Proof fo r Let A be an m × n matrix. Assume that the last k rows of A are zero rows. Then r(A) ≤ m − k. N ot Let C be the (m − k) × n matrix consisting of rows 1 to (m − k) of A. We use row operations to change C to a reduced row echelon matrix denoted by F. Let B be the m × n matrix whose first m − k rows are the same as F and other rows are zero rows. Then B is the row echelon matrix of A and by Theorem 2.5.2 , r(B) ≤ m − k. . It follows that r(A) ≤ m − k. Now, we give the definition of row equivalent matrices. Definition 2.5.2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/9 8/30/2020 2.5 Ranks, nullities, and pivot columns Two m × n matrices A and B are said to be row equivalent if B can be obtained from A by a finite sequence of row operations. The following result shows that the ranks of two row equivalent matrices are the same. Theorem 2.5.4. ut io n Let A and B be row equivalent m × n matrices. Then r(A) = r(B). The method of finding the rank of a matrix is to follow Steps 1-3 given in Section 2.4 for each round until a row echelon matrix is obtained, and then to count the leading entries of the row echelon matrix. D is tri b Example 2.5.2. For each of the following matrices, find its rank, nullity, and pivot columns. A= 0 1 0 0 Solution ( 1 2 −1 2 −1 3 ) ( ) 0 2 1 1 C= 1 1 1 4 2 0 1 3 N ot 0 0 0 0 B= fo r ( ) 0 4 2 1 1 0 0 0 Because A itself is a reduced row echelon matrix, we do not need to use any row operations. Note that A has 2 leading entries, r(A) = 2. Because A is a 3 × 4 matrix, null(B) = 3 − 2 = 1. Because the columns 1 and 2 of A contain the leading entries, the column vectors ()() 1 0 0 , 0 1 are the pivot columns of A. 0 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/9 8/30/2020 2.5 Ranks, nullities, and pivot columns Because B is not a row echelon matrix, we need to use row operations to find its row echelon matrix. B= ( 1 2 2 −1 −1 3 )↓ R 1( − 2) + R 2→ ( 1 2 0 −5 −1 5 ) : = B *. ut io n The last matrix B* is a row echelon matrix of B and there are two leading entries. Hence, r(B) = 2. Because B is a 2 × 3 matrix, null(B) = 3 − 2 = 1. D is tri b Because the columns 1 and 2 of B* contain the leading entries, the columns 1 and 2 of B : the pivot columns of B. ()( ) 1 2 , 2 −1 are N ot fo r We need to use row operations to change C to its row echelon matrix. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 5/9 8/30/2020 2.5 Ranks, nullities, and pivot columns ↓ ( ) ( ) ↓ ↓ ( ) ( ) ( ) 0 4 2 1 1 1 1 4 R 1, 3→ 2 0 1 3 0 2 1 1 0 4 2 1 R 1( − 2) + R 4→ 2 0 1 3 1 1 1 4 0 2 1 1 0 4 2 1 R2 ( − 2 ) + R3 → 4 0 2 1 1 0 2 1 1 0 0 0 −4 : = C *. 0 0 0 −1 fo r R 3( − 4) + R 4→ 1 1 1 4 0 0 0 −1 R2 ( 1 ) + R4 0 −2 −1 −5 1 1 1 ut io n 0 2 1 1 D is tri b C = 1 1 1 4 0 0 0 0 N ot The last matrix is a row echelon matrix of C and there are three leading entries. Hence, r(C) = 3 and null(C) = 4 − 3 = 1. Because the columns 1, 2, 4 of C* contain the leading entries, the columns 1, 2, 4 of C: ()()() 0 0 1 2 4 , 2 1 0 1 , 1 4 3 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 6/9 8/30/2020 2.5 Ranks, nullities, and pivot columns are the pivot columns of C. We state the following result without proof. The result shows that the rank of A T is equal to the rank of A. Theorem 2.5.5. 1. r(A T) = r(A); 2. null(A T) = m − r(A). Example 2.5.3. Find the rank of ( ) 1 0 0 0 1 2 0 0 1 1 −1 1 fo r A= . 0 N ot 2 −1 0 Solution to get the rank of A. D is tri b For some matrices, you can find the rank of A T and then use Theorem 2.5.5 ut io n Let A be an m × n matrix. Then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 7/9 8/30/2020 2.5 Ranks, nullities, and pivot columns AT = ↓ ( ) ( ) 1 1 1 2 0 2 1 −1 0 0 1 0 R 3(1) + R 4→ 0 2 0 2 1 −1 0 0 1 0 0 0 0 0 . ut io n 0 0 −1 1 1 1 The last matrix is a row echelon matrix and contains three leading entries. Hence, r(A T) = 3. By Theorem , r(A T) = r(A T) = 3. In Example 2.5.3 , if we use row operations on A to find the rank of A, we would have needed to use five D is tri b 2.5.5 row operations while we only need to use one row operation on A T to get the rank of A. Exercises N ot fo r 1. For each of the following matrices, find its rank, nullity, and pivot columns. Clearly mark every leading entry a by using the symbol a and use the symbol ↓ to show the direction of eliminating nonzero entries below leading entries. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 8/9 2.5 Ranks, nullities, and pivot columns ( 0 0 −1 0 0 0 0 1 1 ( B= 2 2 −1 −1 3 ) C= ( ) 1 0 2 1 1 1 0 1 4 2 0 1 −3 0 −2 0 2 0 1 1 1 E= −2 1 1 0 0 0 0 1 −1 0 1 −1 0 1 0 0 1 0 1 −2 1 0 0 2 −2 0 0 −1 −1 −2 −1 0 2 −1 0 −1 G= 5 0 3 2 7 3 0 1 2 5 −1 1 1 N ot fo r F = 0 0 ) 0 1 4 2 ( )( ) ( )( ) −1 D = 0 1 D is tri b A = 1 ut io n 8/30/2020 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 9/9 8/30/2020 2.6 Elementary matrices 2.6 Elementary matrices ut io n An elementary matrix is a special type of square matrix that is obtained by performing a single row operation to the identity matrix. Such matrices can be used to derive the algorithm for finding the inverse of an invertible matrix (see Section 2.7 ). Definition 2.6.1. D is tri b An m × m matrix is called an elementary matrix if it can be obtained by performing a single row operation to the m × m identity matrix. Example 2.6.1. The following matrices are elementary matrices. ( ) ( ) ( ) 0 1 0 0 0 1 E2 = 0 0 −2 1 0 0 0 1 E3 = N ot E1 = 1 fo r 2 0 0 0 0 1 0 1 0 ( )( )( ) 1 0 0 0 E4 = 1 0 0 0 1 0 0 0 0 3 0 0 0 0 1 1 0 4 0 E5 = 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 E6 = 0 0 0 1 0 0 1 0 0 1 0 0 Processing math: 100% Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/14 2.6 Elementary matrices I3 = ( ) ( ) 1 0 0 2 0 0 0 1 0 R (2)→ 1 0 0 1 0 1 0 0 0 1 ( ) ( ) ( ) ( ) 1 0 0 1 0 1 0 R 1( − 2) + R 2→ 0 0 1 I3 = 0 0 −2 1 0 0 1 0 0 1 0 0 0 1 0 R → 2, 3 0 0 1 0 0 1 = E 2. 0 1 D is tri b I3 = = E 1. ut io n 8/30/2020 0 1 0 = E 3. matrices. The elementary matrices E 1 − E 6 in Example 2.6.1 fo r The matrices E 1, E 2, and E 3 are 3 × 3 elementary matrices. Similarly, E 4, E 5, and E 6 are obtained by using the row operations: R 3(3), R 3(4) + R 1, and R 2 , 4 to I 4, , respectively, and so they are elementary can be changed back to the corresponding identity N ot matrices by performing a suitable single row operation. For example, ( ) ( ) 2 0 0 E1 = 1 0 1 0 R ( )→ 1 2 0 0 1 1 0 0 0 1 0 0 0 1 = I 3. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/14 8/30/2020 2.6 Elementary matrices ( ) E2 = 0 0 − 2 1 0 R (2) + R → 1 2 0 0 1 E3 = ( ) 1 0 0 0 1 0 0 0 1 ( ) ( ) 1 0 0 1 0 0 0 0 1 R 2 , 3→ 0 1 0 0 1 0 = I 3. = I 3. 0 0 1 ut io n 1 D is tri b From the above, we see that the row operations are invertible. Hence, we introduce the definition of inverse row operation. Definition 2.6.2. () 1 i. R i c is the inverse row operation of R i(c). iii. R i , j is the inverse row operation of R i , j. fo r ii. R i( − c) + R j is the inverse row operation of R i(c) + R j. i i. I m R i(c)→ N ot We use the symbols E i(c), E i ( c ) + j, and E i , j to denote the elementary matrix obtained by performing the row operation R i(c) + R (c) + R , and R i , j, respectively, to the m × m identity matrix I. Then we have j () 1 E i(c) R i c → I m. ii. I m R i(c) + R j→ E i ( c ) + j R 1( − c) + R j→ iii. I m R i , j→ E i , j R i , j→ I m. I m. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/14 8/30/2020 2.6 Elementary matrices Example 2.6.2. R 2(5)→ ( ) () ( ) 0 5 1 0 1 R2 5 → . 0 1 ( ) 1 0 R 1(2) + R 2→ 0 1 ( ) 1 0 0 1 ( ) R 1 , 2→ 1 0 R 1( − 2) + R 2→ 2 1 ( ) 0 1 1 0 R 1 , 2→ It is easy to show (2.6.1) () 1 c = I m; E i ( c ) + j E i () 1 c ( ) 1 0 0 1 0 1 . . +j = I m, E i , j E i , j = I m. fo r E i(c)E i ( ) 1 0 ut io n 0 1 1 0 D is tri b ( ) 1 0 Theorem 2.6.1. N ot The following result shows that if a single row operation is used on an m × n matrix A, then the resulting matrix is equal to the product of the elementary matrix E and A, where E is the matrix obtained by performing the same row operation on the n × n identity matrix In. Let A be an m × n matrix. Then the following assertions hold. i. A R i(c)→ B if and only if E i(c)A = B. ii. A R i(c) + R j→ B if and only if E i ( c ) + jA = B. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/14 8/30/2020 2.6 Elementary matrices iii. A R i , j→ B if and only if E i , jA = B. a i1 ⋯ a ii⋯ a in ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 ⋯ 0⋯ ⋮ ⋮ 0 ⋮ ⋮ 1 ⋮ a m1 ⋯ a mi⋯ a mn )( ) a 11 a 12 ⋯ a 1n ⋮ ⋮ ⋮ ⋮ ca i1 ca i2 ⋯ ca in . ⋮ ⋮ ⋮ ⋮ = a m1 a m2 ⋯ a mn D is tri b 0 0 ( )( ) a 1j⋯ a 1n ⋮ a 11 ⋯ a i1⋯ ⋮ ⋮ ⋮ ⋮ ⋮ ⋯ 1… 0⋯ 0 a i1 a ij⋯ a in ⋮ ⋮ ⋮ ⋮ 0 ⋯ c⋯ 1⋯ 0 a j1 ⋯ a ji⋯ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 ⋯ 0⋯ 0⋯ 1 ⋯ 0⋯ 0⋯ 0 ⋮ ⋮ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 1 ⋯ a ii… ⋮ ⋮ fo r ii. ( )( ⋯ c⋯ ⋯ 0⋯ ⋮ ⋮ a jj⋯ a jn ⋮ ⋮ N ot i. a 1n ⋮ a 11 ⋯ a 1i⋯ ⋮ ⋮ ⋮ 1 ut io n Proof a m1 ⋯ a mi⋯ a mj⋯ a mn Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/14 8/30/2020 2.6 Elementary matrices ⋯ a 1i⋯ a 1j⋯ a 1n ⋮ ⋮ ⋮ ⋮ ⋮ a i1 ⋯ a ii⋯ a ij⋯ a in ⋮ ⋮ ⋮ ⋮ ⋮ = . ut io n ( ) ( )( ) a 11 ca i1 + a j1 ⋯ ca ii + a ji⋯ ca ij + a jj⋯ ca in + a jn ⋮ ⋮ ⋮ a m1 ⋯ a mi⋯ a mj⋯ a mn D is tri b ⋮ a 1n ⋮ ⋮ ⋯ 0⋯ 1⋯ 0 a i1 ⋯ a ii⋯ a ij⋯ a in ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 ⋯ 1⋯ 0⋯ 0 a j1 ⋯ a ji⋯ a jj⋯ a jn ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 ⋯ 0⋯ 0⋯ ⋯ 0⋯ 0⋯ 0 ⋮ ⋮ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 1 fo r a 1j⋯ ⋮ a 11 ⋯ a 1i⋯ ⋮ ⋮ ⋮ 1 N ot iii. ⋮ a m1 ⋯ a mi⋯ a mj⋯ a mn Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/14 2.6 Elementary matrices ( ) a 11 ⋯ a 1i⋯ = a 1j⋯ a 1n ⋮ ⋮ ⋮ ⋮ ⋮ a j1 ⋯ a ji⋯ a jj⋯ a jn ⋮ ⋮ ⋮ ⋮ ⋮ a i1 ⋯ a ii⋯ a ij⋯ a in ⋮ ⋮ ⋮ ⋮ ⋮ D is tri b a m1 ⋯ a mi⋯ a mj⋯ a mn Example 2.6.3. Let ( ) ( ) 1 0 0 0 1 0 i. Apply R 2( − 2) to I and A: and A = 1 1 2 −3 . 2 1 1 N ot 0 0 1 0 1 fo r I= . ut io n 8/30/2020 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/14 8/30/2020 2.6 Elementary matrices ( ) ( ) ( ) ( ) 1 0 1 0 R ( − 2)→ 2 0 0 1 0 −2 0 0 1 A = 1 2 − 3 R ( − 2)→ 2 2 1 1 iii. Apply R 1 , 2 to I and A: 1 1 2 1 = B 1. 1 ( ) ( ) 1 0 0 0 1 0 0 −2 1 0 1 1 1 2 −3 fo r 1 2 − 3 R 2( − 2) + R 3→ 2 1 1 = E2 0 −3 = B 2. 7 N ot Show that E 2A = B 2. 1 1 −2 −4 6 0 1 0 R 2( − 2) + R 3→ 0 0 1 0 1 = E1 D is tri b ( ) ( ) 0 0 0 1 0 0 A = 0 1 Show that E 1A = B 1. ii. Apply R 2( − 2) + R 3 to I and A: I = 0 ut io n I = 1 0 0 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/14 2.6 Elementary matrices I = ( ) ( ) ( ) ( ) 1 0 0 0 1 0 0 1 0 R → 1, 2 0 0 1 1 0 0 0 1 A = = E3 0 0 1 1 1 2 −3 1 2 −3 R → 1, 2 2 1 1 0 1 1 2 1 1 = B 3. Solution ( )( ) ( ) 1 E 2A = 0 0 0 1 1 0 −2 0 1 2 −3 0 2 1 1 0 = 1 1 1 −2 −4 6 2 1 ( )( ) ( ) ( )( ) ( ) E 3A = 1 0 0 0 1 1 0 1 0 1 2 −3 = 0 1 1 1 2 −3 0 −2 1 2 1 1 0 −3 0 1 0 0 1 1 1 2 −3 1 0 0 1 2 −3 0 0 1 2 1 1 = = B 1. 1 fo r 0 N ot E 1A = D is tri b Show that E 3A = B 3. ut io n 8/30/2020 7 0 1 1 2 1 1 = B 2. = B 3. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/14 8/30/2020 2.6 Elementary matrices By using Theorem 2.6.1 , we can prove the following result, which gives the relation between a matrix and the resulting matrix obtained by using multiple row operations on the matrix. Theorem 2.6.2. If we use k row operations on an m × n matrix A and denote the resulting matrix by B,then ut io n B = E kE k − 1⋯E 2E 1A, where E 1, E 2, ⋯ , E k are m × m elementary matrices corresponding to the row operations used on A. D is tri b Proof We denote the row operations by r 1, r 2, ⋯ , r k in order. Then I r1 , we have → E 1A To understand Theorem 2.6.2 B as follows: r2 2 ) , we give the following example. We use two row operations to change A to −1 1 −1 3 2 1 0 ( → E 2(E 1A) = E 2E 1A ⋯ r k→ E k E k − 1⋯E 2E 1A = B. ( ) ( ) 0 A= fo r A → E i for each i ∈ {1, 2, ⋯ , k} N ot and by Theorem 2.6.1 ri 1 −1 R 1 , 2→ 0 2 2 0 3 − 1 R ( − 2) + R → 1 3 1 ( ) 1 −1 3 0 2 −1 0 2 −5 = B. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/14 2.6 Elementary matrices I= ( ) ( ) 1 0 0 0 1 0 0 1 0 R → 1, 2 0 0 1 1 0 0 0 0 1 ( ) 1 0 0 I= ( ) 1 0 0 0 1 0 = E 2. −2 0 1 , we obtain B = E 2E 1A. In the following, we directly verify that the result B = E 2E 1A holds. 0 0 0 1 0 0 0 1 0 1 0 0 1 −1 3 −2 0 1 0 0 1 2 1 2 −1 0 0 2 −1 0 0 2 −5 1 0 0 1 −1 3 0 1 0 0 2 −1 −2 0 1 2 1 N ot = ( )( )( ) ( )( ) ( ) 1 fo r E 1 E 2A = D is tri b By Theorem 2.6.2 0 1 0 R 1( − 2) + R 3→ 0 0 1 = E 1. ut io n 8/30/2020 1 −1 = 3 We end this section with the ranks of elementary matrices. By Theorem 2.5.4 result. = B. , we have the following Corollary 2.6.1. Let E be an m × m elementary matrix. Then r(E) = m. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/14 8/30/2020 2.6 Elementary matrices Proof Because E and I are row equivalent and r(I) = m, it follows from Theorem 2.5.4 that r(E) = m. By Theorem 2.6.2 , we obtain the following result, which shows that the rank of the product of finitely many elementary matrices is n. ut io n Theorem 2.6.3. Let E 1, ⋯ , E k be m × m elementary matrices and E = E kE k − 1⋯E 2E 1. Then r(E) = m. By Theorem 2.6.2 r(E) = r(E 1) = m. D is tri b Proof , E and E1 are row equivalent. By Theorem 2.5.4 Exercises and Corollary 2.6.1 , 2. 3. 4. ( ( ( ( 2 0 0 1 0 1 1 0 1 5 0 1 2 0 0 2 ) ) ) ) N ot 1. fo r 1. Determine which of the following matrices are elementary matrices. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/14 8/30/2020 2.6 Elementary matrices 5. 6. ( ) ( ) 0 2 2 0 1 0 −2 1 ( ) ( ) ( ) 7. ut io n 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 −1 1 0 1 0 1 ( ) ( ) 1 0 0 0 10. D is tri b 0 0 1 0 0 0 0 5 0 0 0 0 1 fo r 9. −1 1 0 N ot 8. 0 0 1 0 0 0 11. 0 1 0 0 0 0 0 0 0 0 0 1 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/14 8/30/2020 2.6 Elementary matrices ( ) 1 0 0 0 12. 0 0 0 1 0 0 1 0 ut io n 0 1 0 0 2. Use row operations to change the above matrices (1), (2), (3), (6), (7), (8), (10), and (12) to an identity matrix. ( ) () 1 0 0 1 . Verify the following assertions. 1 1. E 2(5)E 2 5 2. E 1 ( 2 ) + 2E 1 D is tri b 3. Let I = = I; () 1 2 +2 = I; fo r 3. E 1 , 2E 1 , 2 = I. 4. Find a matrix E such that B = EA, where ( ) ( ) A= 2 −1 N ot 0 1 −1 3 2 1 0 and B = 1 −1 0 2 0 0 3 −1 . −4 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/14 8/30/2020 2.7 The inverse of a square matrix 2.7 The inverse of a square matrix In this section, we study how to determine a square matrix to be invertible by using its rank, and if it is invertible, how to find its inverse. After we study Section 3.4 shall see that determinants of matrices also can be used to determine whether a square matrix is invertible. , we Definition 2.7.1. Let A be an n × n matrix. If there exists an n × n matrix B such that AB = I n, ut io n then A is said to be invertible (or nonsingular) and B is called the inverse of A. We write B = A − 1. If A is not invertible, then A is said to be singular. We only introduce the notion of inverse matrices for square matrices. Therefore, if a matrix is not a square matrix, then it has no inverses. , if A is invertible, then AA − 1 = I n. Moreover, A is not invertible if and only if for each n × n matrix B, AB ≠ I n. D is tri b By Definition 2.7.1 We state the following theorem without proof, which provides some equivalent results for inverse matrices. Theorem 2.7.1. Let A and B be n × n square matrices. Then the following are equivalent. fo r i. AB = I n. ii. BA = I n. By Theorem 2.7.1 N ot iii. AB = BA = I n. , we can show that if A is invertible, then its inverse is unique. Indeed, assume that there are two n × n matrices B 1 and B 2 satisfying AB 1 = I n and (ii), B 1A = I n. Hence, AB 2 = I n. Then, by Theorem 2.7.1 B 1 = B 1I n = B 1(AB 2) = (B 1A)B 2 = I nB 2 = B 2. This shows that the inverse of A is unique. Example 2.7.1. (1) Show that B is an inverse of A if A= (2) 100% Show Processing math: that ( ) ( ) 0 0 0 0 and 1 1 1 1 ( 2 −5 −1 3 ) and B = ( ) 3 5 1 2 . are not invertible. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/12 8/30/2020 2.7 The inverse of a square matrix Solution ( ) b 11 b 12 b 21 b 22 2 −5 −1 3 )( ) ( ) 3 5 1 2 = 1 0 0 1 = I 2, by Definition 2.7.1 , B is the inverse of A. be a 2 × 2 matrix. Because ( )( by Definition 2.7.1 , ( ) 1 0 0 1 , ( ) 1 1 1 1 b 11 b 12 0 0 b 21 b 22 )( )( ) = 0 0 1 0 ≠ 0 0 , 0 1 is not invertible. Because ( )( by Definition 2.7.1 0 0 1 1 b 11 b 12 1 1 b 21 b 22 )( = b 11 + b 21 b 12 + b 22 b 11 + b 21 b 12 + b 22 is not invertible. ≠ 1 0 0 1 , fo r Example 2.7.2. )( ) ut io n (2) Let ( D is tri b (1) Because AB = N ot Assume that a i ≠ 0 for each i ∈ I n. Then ( ) a1 0 ⋯ 0 0 a2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ an ( ) 1 a1 −1 = 0 0 1 a2 ⋯ 0 ⋯ 0 ⋮ ⋮ ⋮ ⋱ 0 0 ⋯ . 1 an Solution ( −1 −1 −1 Because a i ≠ 0 for each i ∈ I n, diag a 1 , a 2 , ⋯ , a n ) exists. It is easy to verity that Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/12 8/30/2020 2.7 The inverse of a square matrix ( ) ( −1 −1 −1 diag a 1, a 2, ⋯ , a n diag a 1 , a 2 , ⋯ , a n The result follows from Definition 2.7.1 )=I . n . In Example 2.7.2 , we use Definition 2.7.1 to verify whether a 2 × 2 matrix is invertible. But it is not a good method to use Definition 2.7.1 to verify whether a square matrix with a large size is invertible. Later, we shall study some other useful methods such as using ranks and determinants to determine whether square matrices are invertible. By Definition 2.7.1 and (2.6.1) , we obtain the following result, which shows that every elementary matrix is invertible. Theorem 2.7.2. ut io n Every elementary matrix is invertible and its inverse is an invertible elementary matrix. Theorem 2.7.3. Proof By Theorem 2.6.2 , there exists an n × n matrix E = E kE k − 1⋯E 2E 1 such that B = EA. By Theorem 2.7.2 invertible. By Definition 2.7.1 D is tri b Assume that B is a row echelon matrix of an n × n matrix A. Then there exists an invertible n × n matrix E = E kE k − 1⋯E 2E 1 such that B = EA, where E i is an n × n elementary matrix for i = 1, ⋯ , k. , we prove the following result, which will be used to prove Theorem 2.7.3 , E is fo r Theorem 2.7.4. . , E i is an n × n invertible matrix and by Theorem 2.7.4 N ot If A and B are n × n invertible matrices, then AB is invertible and (AB) − 1 = B − 1A − 1. Proof Because (B − 1A − 1)(AB) = B − 1(A − 1A)B = B − 1IB = B − 1B = I, (AB) − 1 = B − 1A − 1 and AB is invertible. Now, we show how to find the inverse matrix of a 2 × 2 matrix A given in (1.2.4) . Theorem 2.7.5. Let A = ( ) a b c d be the same as in (1.2.4) . Then the following assertions hold. Processing math: 100% 1. A is invertible if and only if |A| ≠ 0. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/12 8/30/2020 2.7 The inverse of a square matrix 2. If A is invertible, then (2.7.1) A −1 = 1 ( d |A| − c −b a ) . Proof Let B= ( ) x1 x2 y1 y2 ut io n (2.7.2) . AB = ( )( a b x1 x2 c d y1 y2 )( = D is tri b We find x 1, x 2, y 1, y 2 such that AB = I. Because ax 1 + by 1 ax 2 + by 2 cx 1 + dy 1 cx 2 + dy 2 1 0 = 0 1 , fo r we obtain )( ) (2.7.3) N ot ax 1 + by 1 = 1, cx 1 + dy 1 = 0. (2.7.4) ax 2 + by 2 = 0, cx 2 + dy 2 = 1. By (2.7.3) , we obtain adx 1 + bdy 1 = d, bcx 1 + bdy 1 = 0. Subtracting the above two equations implies (ad − bc)x 1 = d and |A|x 1 = d. Also, by (2.7.3) , we obtain acx 1 + bcy 1 = c, and acx 1 + ady 1 = 0. Subtracting the above two equations implies |A|y 1 = − c. Similarly, applying the above method to (2.7.4) , we obtain |A|x 2 = − b and |A|y 2 = a. Hence, if AB = I, then Processing math: 100% (2.7.5) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/12 8/30/2020 2.7 The inverse of a square matrix |A | x1 = d, | A | y1 = that is, the condition (2.7.5) − c, | A | x2 = | | − b, and A y 2 = a, is a necessary condition for AB = I. (1) Assume that A is invertible. By Example 2.7.1 assume that |A| ≠ 0. By (2.7.5) , (2), A is not a zero matrix. So one of a, b, c, d is not zero. This, together with (2.7.5) , implies |A| ≠ 0. Conversely, (2.7.6) x1 = d −c −b a , y = , x = and y 2 = , |A| 1 |A| 2 |A| |A| ut io n that is, (2.7.7) and AB = I. Hence, by Definition 2.7.1 ( ) ( ) x1 x2 y1 y2 = d 1 −b |A| − c a , A is invertible. (2) By the above proof, we see that if A is invertible, then A − 1 = B. fo r Example 2.7.3. D is tri b B= For each of the following matrices, determine whether it is invertible. If so, calculate its inverse. ( ) ( 2 −4 1 ) N ot A= 3 B= 1 2 −2 −4 . Solution Because |A | = | | 2 −4 1 3 = (2)(3) − ( − 4)(1) = 10 ≠ 0, by Theorem 2.7.5 A −1 = Processing math: 100% Because |B | = | 1 2 −2 −4 | 1 , A is invertible and ( ) ( ) 3 4 |A| − 1 2 = 1 3 4 10 − 1 2 = ( ) 3 2 10 5 1 1 − 10 . 5 = (1)( − 4) − (2)( − 2) = 0, B is not invertible. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/12 8/30/2020 2.7 The inverse of a square matrix Example 2.7.4. For each of the following matrices, find all x ∈ ℝ such that it is invertible. A= ( ) ( ) 1 x3 1 1 B= x 4 1 x Solution Because if |A | = | | x 4 1 x 1 1 = 1 − (x 3)(1) = 1 − x 3 = 0, then x = 1. Hence, when x ≠ 1, |A| ≠ 0 and by Theorem 2.7.5 , A is invertible. Because if = (x)(x) − (4)(1) = x 2 − 4 = 0, then x = − 2 or x = 2. Hence, when x ≠ − 2 and x ≠ 2, |B| ≠ 0 and by Theorem 2.7.5 , B is invertible. ut io n |B | = | | 1 x3 . D is tri b In Theorem 2.7.5 (1), we use the determinants of 2 × 2 matrices to determine whether they are invertible. The result remains true for n × n matrices, see Theorem 3.4.1 In Corollary 8.1.3 , we shall show how to use the eigenvalues of a matrix to determine whether it is invertible. Here we show that we can show how to use the rank of an n × n matrix to determine whether it is invertible. Theorem 2.7.6. Let A be an n × n matrix. Then A is invertible if and only if r(A) = n. Proof N ot fo r (1) Assume that A is invertible. Then there exists an n × n matrix B such that AB = I. If r(A) < n, then by Theorem 2.7.3 , there exists an invertible matrix E such that C = EA, where C is the reduced row echelon matrix of A. By Theorem 2.6.3 , r(E) = n and by Theorem 2.5.4 , r(C) = r(A) < n. This implies that the rows from r(A) + 1 to n of the reduced row echelon matrix C are zero rows and thus, the rows from r(A) + 1 to n of CB are zero rows. By Theorem 2.5.3 , r(CB) ≤ r(A) < n. On the other hand, because AB = I, we have CB = (EA)B = E(AB) = EI = E and n = r(E) = r(CB) < n, a contradiction. (2) Assume that r(A) = n and B is the reduced row echelon matrix of A. Then r(B) = n and B = I n. By Theorem 2.7.3 I n = EA and A is invertible. By Theorem 2.7.5 3.4.1 (i). , there exists an invertible matrix E such that (1), we see that when n = 2, the determinant of A can be used to justify whether A is invertible. The result remains true when n ≥ 3, see Theorem Example 2.7.5. For each of the following matrices, determine whether it is invertible. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/12 8/30/2020 2.7 The inverse of a square matrix A= ( ) ( ) 1 1 1 1 0 1 1 0 −1 0 B= 1 0 1 1 1 1 0 1 Solution 0 1 1 1 0 1 1 1 0 1 1 0 −1 0 ( ) 1 1 1 R 2(1) + R 3→ 0 1 1 . 0 0 1 , A is invertible. ( )↓ 1 B= 1 1 0 −1 0 1 Then r(B) = 2. By Theorem 2.7.6 ( )↓ 1 0 1 ( )↓ 1 R 1( − 1) + R 3→ 1 1 0 −1 1 ( ) 1 R 2( − 1) + R 3→ 0 −1 0 , B is not invertible. Corollary 2.7.1. 1 1 0 −1 0 . 0 0 0 D is tri b Then r(A) = 3. By Theorem 2.7.6 R 1( − 1) + R 3→ ut io n ( )↓ 1 1 1 A= Proof N ot fo r 1. An n × n matrix is invertible if and only if its reduced row echelon matrix is the identity matrix In. 2. An n × n matrix is not invertible if and only if its row echelon matrix contains at least one zero row. We only prove (1). Assume that B is the reduced row echelon matrix of A. By Theorem 2.5.4 , r(A) = r(B). If A is invertible, it follows from Theorem 2.7.6 that r(A) = n. Hence, r(B) = n and B = I n. Conversely, if the reduced row echelon matrix of A is the identity matrix I n, then by Theorem 2.7.3 , there exists an invertible matrix E such that I n = EA and A is invertible. Corollary 2.7.2. An n × n matrix is invertible if and only if there exists a finite sequence of elementary matrices F 1, F 2, ⋯ , F k − 1, F k such that A = F 1F 2⋯F k − 1F k. Proof By Theorem 2.7.3 and Corollary 2.7.1 , A is invertible if and only if there exists an invertible matrix E = E kE k − 1⋯E 2E 1 such that I n = EA, where E i is an elementary matrix for i = 1, ⋯ , k if and only if Processing math: 100% A = E − 1 = (E kE k − 1⋯E 2E 1) − 1 = E 1− 1E 2− 1⋯E k−−11E k− 1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/12 8/30/2020 2.7 The inverse of a square matrix Let F i = −1 Ei for i ∈ {1, 2, ⋯ , k}. By Theorem 2.7.2 , Fi is an invertible elementary matrix for i ∈ {1, 2, ⋯ , k}. Corollary 2.7.3. Let A and B be m × n matrices. Assume that P is an m × m invertible matrix such that B = PA. Thenr(A) = r(B). Proof Because P is an m × m invertible matrix, by Corollary 2.7.2 , there exists a finite sequence of m × m elementary matrices F 1, F 2, ⋯ , F k − 1, F k such that P = F 1F 2⋯F k − 1F k and thus, B = F 1F 2⋯F k − 1F kA. This shows that we use k row operations to change A to B and A and B are row equivalent. By Theorem 2.5.4 r(A) = r(B). The following result shows that the converse of Theorem 2.7.4 holds. ut io n Theorem 2.7.7. Assume that A and B are n × n matrices and AB is invertible. Then both A and B are invertible. Proof , r(C) = n. Let D be the reduced row echelon matrix of A. By Theorem 2.7.3 (2.7.8) EC = EAB = DB. D is tri b Let C = AB. Because AB is invertible, by Theorem 2.7.6 matrix E such that D = EA. Hence, , , there exists an invertible N ot fo r By Theorem 2.5.4 , r(A) = r(D) and r(DB) = r(C) = n. This implies r(D) = n. In fact, if r(D) < n, then because D is a row echelon matrix, by Corollary 2.7.1 (2), D contains at least one zero row, so the last row of D must be a zero row. By (2.2.5) , we see that the last row of DB is a zero row. By Theorem 2.5.3 , we have r(DB) < n, a contradiction. Hence, r(A) = r(D) = n. By Theorem 2.7.6 , A is invertible. Now, because A is invertible, by Corollary 2.7.1 (1), D = I and by (2.7.8) , EC = B. By Theorem 2.5.4 , r(B) = r(C) = n and by Theorem 2.7.6 , B is invertible. Methods of finding the inverses By Theorem 2.7.3 and Corollary 2.7.1 Because EA = I n and EI n = E = A −1 , we see that if an n × n matrix A is invertible, then there exists a matrix E = E kE k − 1⋯E 2E 1 such that EA = I n. Hence, E = A − 1. , so | | | E(A I n) = (EA EI n) = (I n A − 1). This provides a method to find the inverse of A. Theorem 2.7.8. i. If we can use row operations to change (A / I n) to a matrix (I n / B), then A − 1 = B. ii. If we can use row operations to change (A|In) to a row echelon matrix (C|D), where C contains at least one zero row, then A is not invertible. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/12 8/30/2020 2.7 The inverse of a square matrix Proof | | 1. Assume that we use row operations to change (A / I n) to a matrix (I n / B). From the left sides of (A I n) and (I n B), we see that A and I n are row equivalent. By Theorems (2.5.4), r(A) = r(I n) = n. By Theorem 2.7.6 , A is invertible. By Theorem 2.7.3 | , there exists an invertible matrix E = E kE k − 1⋯E 2E 1 such that | EA = I n. This implies E = A − 1. From the right sides of (A I n) and (I n B), we see that EI n = B. This implies B = EI n = E = A − 1. | 2. Assume that we use row operations to change (A I n) to the reduced row echelon matrix (C | D), where C contains at least one zero row. From the left sides of | (A I n) and (C | D), we see that A and C are row equivalent. By Theorems (2.5.4) and 2.5.3 , r(A) = r(C) < n. It follows from Theorem 2.7.6 that A is not invertible. By Theorem 2.7.8 ut io n Remark 2.7.1 , the method to find the inverse of an n × n matrix A is given below. | i. Change the matrix (A I n) to a row echelon matrix (C | D). If C contains at least one zero row, then A is not invertible. D is tri b ii. If C does not contain any zero rows, then A is invertible and we continue changing the row echelon matrix (C | D) to the reduced row echelon matrix, which | definitely is of the form (I n A − 1). fo r Example 2.7.6. For each of the following matrices, determine whether it is invertible. If so, find its inverse. ( −3 −4 5 ) ( ) ( ) 1 1 1 N ot A= 2 B= 0 1 1 1 0 1 C= 1 −3 4 2 −5 7 0 −1 1 Solution (A | I 2) = ( | ) 2 −3 1 0 −4 5 0 1 R 1(2) + R 2→ Hence, A is invertible and A = 0 −1 2 1 R 2( − 1)→ ( | 2 −3 1 0 0 1 −2 −1 ) R 2(3) + R 1→ ( | 2 0 −5 −3 0 1 −2 −1 )() R1 1 2 ( | ) | 5 → 3 1 0 − −2 2 0 1 −2 −1 = (I 2 A − 1). ( ) 5 −1 ( | ) 2 −3 1 0 3 −2 −2 . −2 −1 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/12 8/30/2020 2.7 The inverse of a square matrix ( | ) 1 1 1 1 0 0 Hence B is invertible and B − 1 = ( 1 −1 1 0 −1 1 ( | ) 1 1 1 1 0 1 1 0 0 0 1 0 R (1) + R → 2 3 0 −1 0 −1 0 1 0 1 0 1 1 0 ( | ) ( | ) 1 −3 2 − 5 7 0 1 0 R 1( − 2) + R 2→ 0 −1 1 0 0 1 0 4 1 0 0 0 ) − 1 R ( − 1) + R → 2 1 1 1 ( | 1 0 0 1 −1 0 0 1 0 1 0 −1 1 1 0 0 1 −1 ) | = (I 3 B ( | ) 1 −3 1 − 1 − 2 1 0 R 2(1) + R 3→ 0 −1 1 0 0 1 , C is not invertible. 0 1 0 0 4 1 0 0 −1 −2 1 0 . 0 −2 1 1 D is tri b , it is easy to prove the following result, which provides some properties of inverse matrices. We omit these proofs. Let A be an invertible n × n matrix. Then the following assertions hold. −1 −1 −1 1 ) Theorem 2.7.9. 1. (A − 1) 2 0 0 1 − 1 1 1 R3 ( − 1 ) + R2 0 0 1 − 1 Because the left side of the last matrix contains a row of zeros, by Theorem 2.7.8 By Definition 2.7.1 0 0 R ( −1) +R 1 1 0 3 1 1 0 0 1 0 → −1 . 1 1 −3 4 1 0 0 (C | I 3) = ( | ) ( | 1 1 1 ut io n (B | I 3) = 0 1 1 0 1 0 R ( − 1) + R → 1 3 1 0 1 0 0 1 = A. 3. A n is invertible for n ∈ ℕ and (A n) 4. A T is invertible and −1 (A T) = −1 n n = (A − 1) . In this case, we write (A − n) = (A − 1) . T (A − 1) . N ot 5. If A is symmetric, then A − 1 is symmetric. fo r 1 2. If k ≠ 0, then kA is invertible and (kA) − 1 = k A − 1. 6. AA T and A T A are invertible. 7. If A is triangular, then A is invertible if and only if its diagonal entries are all nonzero. 8. If A is lower triangular, then A − 1 is lower triangular. If A is upper triangular, then A − 1 is upper triangular. Exercises 1. Show that B is an inverse of A if A= ( ) 1 1 −1 1 ( ) 1 and B = 1 2 −2 1 1 2 2 . Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/12 8/30/2020 2.7 The inverse of a square matrix 2. Use Definition 2.7.1 to show that ( 1 −1 −2 2 ) is not invertible. 3. For each of the following matrices, determine whether it is invertible. If so, find its inverse. ( )( ) 3 A = 1 0 0 0 0 0 1 0 0 2 0 0 B= 0 4 0 0 3 0 0 0 0 4 0 0 4 C = 4. Let A = ( ) 5. Let A = ( ) 1 1 1 2 1 3 3 −4 2 3 D= 2 4 −1 −2 ) . Find all x ∈ ℝ such that A is invertible. and B = ( ) 3 2 2 2 D is tri b 1 x2 ( ) ( ut io n 1 . Find A − 1, B − 1, and (AB) − 1 and verify (AB) − 1 = B − 1A − 1. 6. For each of the following matrices, use its rank to determine whether it is invertible. A= 1 5 −5 8 ( 0 0 0 2 1 2 0 −2 −3 1 −2 1 −1 0 0 ) C= ( N ot 1 0 B= 1 2 1 4 1 −2 4 fo r ( ) 0 2 −5 −2 −1 −4 ) 7. For each of the following matrices, use row operations to determine if it is invertible. If so, find its inverse. A = ( ) ( D = ( ) ( ) ( ) −1 2 4 7 1 −1 8. Let A = 9. Let A = Processing math: 100% ( ) ( ) −1 2 1 3 −4 1 3 1 3 . Find A − 1, (A T) −1 and verify that (A T) 1 −1 2 0 −1 = (A − 1) . −1 = (A − 1) . 1 −3 −4 5 1 0 −1 . Find A 3, A − 1, (A − 1) , and verify (A 3) B= ) ( 1 1 E= C= −2 4 2 2 0 −5 3 2 1 −2 1 ) 0 −8 −9 F= 2 4 −1 1 −2 −5 3 T https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/12 8/30/2020 2.7 The inverse of a square matrix ( ) 1 3 3 2 . Find A − 1. Is A − 1 symmetric? N ot fo r D is tri b ut io n 10. Let A = Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/12 8/30/2020 Chapter 3 Determinants 3.1 Determinants of 3 × 3 matrices ut io n Chapter 3 Determinants D is tri b The determinant of a 2 × 2 matrix is defined in (1.2.4) . It is applied to introduce the Gram determinant in Section 1.2 and compute the areas of parallelograms in Section 1.3 . In this section, we introduce the determinants of 3 × 3 matrices. Let ( a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 N ot A= ) fo r (3.1.1) be a 3 × 3 matrix. The determinant of A, denoted by |A | , is defined by the number (3.1.2) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/13 8/30/2020 Chapter 3 Determinants |A| = | | a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 (a 11a 22a 33 + a 21a 32a 13 + a 31a 12a 23) − (a 13a 22a 31 + a 23a 32a 11 + a 33a 21a 12). ut io n = Note that the three numbers in each of the terms are from different rows and different columns of A. Example 3.1.1. D is tri b Calculate the determinant | | 1 −2 −1 3 . |A | = 0 2 4 −5 −3 , we obtain N ot By (3.1.3) fo r Solution |A| = [(1)(2)( − 3) + (0)( − 5)( − 1) + (4)( − 2)(3)] − [( − 1)(2)(4) + (3)( − 5)(1) + ( − 3)(0)( − 2)] = − 7. By (3.1.3) , we rewrite |A| as follows. (3.1.3) Processing math: 100% |A| = (a 11a 22a 33 + a 12a 23a 31 + a 13a 21a 32) − (a 13a 22a 31 + a 11a 23a 32 + a 12a 21a 33). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/13 8/30/2020 Chapter 3 Determinants This gives us another way to compute |A | . Indeed, if we write |A| and adjoin it to its first two columns, we get | | a 11 a 12 a 13 a 11 a 12 a 21 a 22 a 23 a 21 a 22 a 31 a 32 a 33 a 31 a 32 : D is tri b The first three terms in (3.1.3) ut io n (3.1.4) a 11a 22a 33, a 12a 23a 31, a 13a 21a 32, can be obtained from (3.1.4) in the following way: Drawing an arrow from a 11 to a 33 through a 22 (from the top left to the bottom right), we see that the product of three terms fo r the three numbers a 11, a 22, a 33 on the arrow is the first product in (3.1.3) . Drawing an arrow from a 12 to a 31 through a 23 and from a 13 to a 32 through a 21, we get the other two products. Similarly, we can get the N ot a 13a 22a 31, a 11a 23a 32, a 12a 21a 33 by drawing an arrow from a 13 to a 31 through a 22 (from the top right to the bottom left), from a 11 to a 32 through a 23 and from a 12 to a 33 through a 21. Example 3.1.2. Calculate the determinant Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/13 8/30/2020 Chapter 3 Determinants | | 2 4 6 |A | = − 4 5 6 . 7 −8 9 By (3.1.4) , we obtain | | 2 ut io n Solution 4 6 2 4 D is tri b |A | = − 4 5 6 − 4 5 = [(2)(5)(9) + (4)(6)(7) + (6)( − 4)( − 8)] − [(6)(5)(7) + (2)(6)( − 8) + (4)( − 4)(9)] = 480. 7 −8 9 7 −8 Before we introduce the determinants for matrices with orders greater than 3, we reorganize the six terms in (3.1.3) . The basic idea is to use the determinants of 2 × 2 matrices to compute 3 × 3 matrices. N ot fo r (3.1.5) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/13 8/30/2020 Chapter 3 Determinants |A| = | | a 11 a 12 a 13 a 21 a 22 a 23 = (a a a + a a a + a a a ) 11 22 33 12 23 31 13 21 32 a 31 a 32 a 33 − (a 13a 22a 31 + a 11a 23a 32 + a 12a 21a 33) | | | | | | D is tri b a 22 a 23 a 21 a 23 a 21 a 22 = a 11 − a 12 + a 13 a 32 a 33 a 31 a 33 a 31 a 32 = a 11M 11 − a 12M 12 + a 13M 13, where (3.1.6) Remark 3.1.1. | | | | fo r | | a 22 a 23 a 21 a 23 a 21 a 22 , M 12 = , M 13 = . a 32 a 33 a 31 a 33 a 31 a 32 N ot M 11 = ut io n = a 11(a 22a 33 − a 23a 32) − a 12(a 21a 33 − a 23a 31) + a 13(a 21a 32 − a 22a 31) M 11 is the determinant of the matrix obtained from A by deleting row 1 and column 1 of A, M 12 is the determinant of the matrix obtained from A by deleting row 1 and column 2, and M 13 is the determinant of the matrix obtained from A by deleting row 1 and column 3. The expression on the right side of (3.1.5) corresponding to the first row of A. Processing math: 100% is called an expansion of minors of the determinant |A| https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/13 8/30/2020 Chapter 3 Determinants Note that the signs in (3.1.5) change and follow the following rule: ( − 1) 1 + 1, ( − 1) 1 + 2, ( − 1) 1 + 3. These powers are obtained from subscripts of a 11, a 12, and a 13, respectively. ut io n We define (3.1.7) By (3.1.5) , we obtain (3.1.8) |A| = a 11M 11 − a 12M 12 + a 13M 13 D is tri b A 11 = ( − 1) 1 + 1M 11, A 12 = ( − 1) 1 + 2M 12, A 13 = ( − 1) 1 + 3M 13. fo r = a 11[( − 1) 1 + 1M 11] + a 12[( − 1) 1 + 2M 12] + a 13[( − 1) 1 + 3M 13] Definition 3.1.1. N ot = a 11A 11 + a 12A 12 + a 13A 13. Let M ij be the determinant of the 3 × 3 matrix obtained from A by deleting the ith row and the jth column of A. M ij is called the minor of a ij or the ijth minor of A. (3.1.9) A ij = ( − 1) i + jM ij Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/13 8/30/2020 Chapter 3 Determinants is called the cofactor of a ij or the ijth factor of A. The expression on the right side of (3.1.8) corresponding to the first row of A. is called an expansion of cofactors of the determinant |A| ut io n Similarly, by reorganizing the six terms in (3.1.3) , we obtain the expansion of minors or cofactors of the determinant |A| corresponding to the second row or the third row or each column of A. Each of these expressions equals |A | . Theorem 3.1.1. D is tri b (3.1.10) |A| = a 21A 21 + a 22A 22 + a 23A 23 = a 31A 31 + a 32A 32 + a 33A 33 = a 11A 11 + a 21A 21 + a 31A 31 = a 12A 12 + a 22A 22 + a 32A 32 = a 13A 13 + a 23A 23 + a 33A 33, fo r Example 3.1.3. ( ) 2 4 −4 5 N ot Let A= 7 6 6 . −8 9 1. Compute |A| by using the minors corresponding to the first row of A. 2. Compute |A| by using the cofactors corresponding to the first row of A. Processing math: 100% Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/13 8/30/2020 Chapter 3 Determinants , we have M 11 = By (3.1.5) | | 5 6 = 93, M 12 = −8 9 | | −4 6 7 9 = − 78, M = | −4 5 7 −8 | = − 3. , we get ut io n 1. By (3.1.6) |A | = 2M11 − 4M12 + 6M13 = 2(93) − 4( − 78) + 6( − 3) = 480. , we obtain D is tri b 2. By (3.1.7) A 11 = ( − 1) 1 + 1M 11 = M 11 = 93. A 12 = ( − 1) 1 + 2M 12 = − M 12 = − ( − 78) = 78. A 13 = ( − 1) 1 + 3M 11 = M 13 = − 3. , |A | = 2A 11 + 4A 12 + 6A 13 = 2(93) + 4(78) + 6( − 3) = 480. fo r By (3.1.8) Let N ot Example 3.1.4. A= ( ) 2 4 −4 5 7 6 6 . −8 9 1. Find M 21, M 22, M 23, A 21, A 22, A 23 and compute |A| using the cofactors. Processing math: 100% 2. Find M 31, M 32, M 33, A 31, A 32, A 33 and compute |A| using the cofactors. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/13 8/30/2020 Chapter 3 Determinants Solution = 1. M 22 = M 23 = | | | | | | 4 6 −8 9 2 6 7 9 2 4 7 8 = 36 + 48 = 84, A 21 = ( − 1) 2 + 1M 21 = − M 21 = − 84. = 18 − 42 = − 24, A 22 = ( − 1) 2 + 2M 22 = M 22 = − 24. = − 16 − 28 = − 44, A 23 = ( − 1) 2 + 3M 23 = − M 23 = 44. = M 32 = M 33 By Theorem 3.1.1 = | | | | | | 4 6 = 24 − 30 = − 6, A 31 = ( − 1) 3 + 1M 31 = M 31 = − 6. 5 6 2 6 −4 6 2 4 −4 5 = 12 + 24 = 36, A 32 = ( − 1) 3 + 2M 32 = − M 32 = − 36. fo r M 31 = 10 + 16 = 26, A 33 = ( − 1) 3 + 3M 33 = M 33 = 26. N ot 2. D is tri b By Theorem 3.1.1 , we have |A | = a 21A 21 + a 22A 22 + a 23A 23 = ( − 4)( − 84) + (5)( − 24) + (6)(44) = 480. ut io n M 21 , we have |A | = a31A31 + a32A32 + a33A33 = (7)( − 6) + ( − 8)( − 36) + (9)(26) = 480. It is easy to show that the determinant of a 2 × 2 matrix is equal to the determinant of its transpose. (3.1.11) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/13 8/30/2020 Chapter 3 Determinants | | | | | | a b c d = a b T c d = a c b d = ad − bc. The following result shows that it is true for 3 × 3 matrices. Let A be the same in (3.1.1) . Show |A | = ut io n Theorem 3.1.2. |A |. T Using the method stated in (3.1.3) | | , we compute A T as follows. (3.1.12) | | D is tri b Proof N ot Comparing the right sides of (3.1.3) fo r a 11 a 21 a 31 a 11 a 21 |A T | = a 12 a 22 a 32 a 12 a 22 = (a 11a 22a 33 + a 21a 32a 13 + a 31a 12a 23) − (a 31a 22a 13 + a 11a 32a 23 + a 21a 12a 33). a 13 a 23 a 33 a 13 a 23 and (3.1.12) | | = |A |. shows A T It is easy to see that the expansion of cofactors of the determinant |A| corresponding to the ith row of A is | | the same as the expansion of minors or cofactors of the determinant A T corresponding to the ith column of A T. For example, the expansion of cofactors of the determinant |A| corresponding to the first row of A Processing math: is 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/13 8/30/2020 Chapter 3 Determinants | | | | | | a 22 a 23 a 21 a 23 a 21 a 22 |A | = a 11 − a 12 + a 13 . a 32 a 33 a 31 a 33 a 31 a 32 | | | | | | | | a 22 a 23 a 21 a 23 a 21 a 22 |A T | = a 11 − a 12 + a 13 a 32 a 33 a 31 a 33 a 31 a 32 , we see that the the corresponding minors equal. D is tri b By (3.1.11) ut io n and the expansion of cofactors of the determinant A T corresponding to the first row of A T is The following result shows that the determinants of upper or lower triangular matrices equal the products of entries on the main diagonals. Theorem 3.1.3. Proof 0 || | fo r 0 a 11 0 a 22 a 23 = a 21 a 22 0 0 0 = a 11a 22a 33 = a 11a 22a 33. N ot | a 11 a 12 a 13 a 33 a 31 a 32 a 33 We only prove the second equality because the first equality follows from Theorem 3.1.2 expansion of cofactors, we have . By the Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/13 8/30/2020 Chapter 3 Determinants | | a 11 0 0 | | Example 3.1.5. Compute each of the following determinants. | | | | 3 −8 9 6 0 0 D is tri b |A | = 0 − 2 4 , | B | = 3 − 1 0 0 0 4 1 0 5 Solution ( ) a b c d N ot Exercises 1. Let A = , |A | = 3( − 2)(4) = − 24; | B | = 6( − 1)(5) = − 30. fo r By Theorem 3.1.3 ut io n a 22 0 a 21 a 22 0 = a A + 0A + 0A = a = a 11a 22a 33. 11 11 12 13 11 a 23 a 33 a 31 a 32 a 33 , where a, b, c, d ∈ ℝ. Show that |λI − A | = λ 2 || − tr(A)λ + A for each λ ∈ ℝ. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/13 8/30/2020 Chapter 3 Determinants ( ) ( ) ( ) 2 4 1 2 −1 1 2. Let A 1 = , A2 = , A3 = For each i = 1, 2, 3, find all λ ∈ ℝ such that 3 −2 −1 4 −1 −3 |λI − Ai | = 0. 3. Use (3.1.3) and (3.1.4) to calculate each of the following determinants. | | | | | | 2 3 |A | = − 2 1 2 3 −1 4 2 −2 4 |B| = 4 0 3 1 1 2 0 −1 3 |C| = 4 0 1 0 ut io n 1 2 1 D is tri b 4. Compute the determinant |A| by using its expansion of cofactors corresponding to each of its rows, where | | 1 2 3 5. Compute each of the following determinants. fo r |A | = − 2 1 2 . 3 −1 4 N ot | | | | 2 3 −4 |A | = 0 6 0 0 0 5 |B| = −2 0 0 1 1 0 0 3 8 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/13 8/30/2020 3.2 Determinants of Math Equation matrices 3.2 Determinants of n × n matrices , that is, (3.2.1) a11 a12 ⋯ a1n ⎜ a21 ⎜ A = ⎜ ⎜ ⎜ ⋮ a22 ⋯ ⋮ ⋱ a2n ⎟ ⎟ ⎟. ⎟ ⋮ ⎟ an2 ⋯ ann ⎛ ⎝ Following Definition 3.1.1 fo rD is tri bu tio n Let A be an n × n square matrix given in (2.3.1) an1 ⎞ ⎠ , we introduce the following definition. Definition 3.2.1. (3.2.2) N ot Let Mij be the determinant of the (n − 1) × (n − 1) matrix obtained from A by deleting the ith row and the jth column of A. Mij is called the minor of aij or the ijth minor of A. Aij = (−1) i+j Mij is called the cofactor of aij or the ijth factor of A. Example 3.2.1. Let https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/8 8/30/2020 3.2 Determinants of Math Equation matrices ⎛ 1 ⎜2 A = ⎜ ⎜1 ⎝ M24 , A32 , 5 6 4 0 3 5 9 0 2 ⎞ ⎟ ⎟. −2 ⎟ 7 ⎠ and A24 . Solution ∣1 M32 = M24 = 5 6∣ 2 0 3 ∣4 2 7∣ ∣1 −3 ∣ ∣ ∣ ∣ ∣ = 8, 5∣ ∣ 1 5 9 ∣4 0 2∣ ∣ Definition 3.2.2. The determinant of A is defined by A32 = (−1) = −192, 3+2 M32 = −M32 = −8, A24 = (−1) 2+4 M24 = M24 = −192. N ot (3.2.3) ∣ fo rD is tri bu tio n Find M32 , 4 −3 n |A| = ∑ a1k A1k = a11 A11 + a12 A12 + ⋯ + a1n A1n . k=1 The expression on the right side of (3.2.3) first row of A. is called an expansion of cofactors of the determinant |A| via the https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/8 8/30/2020 3.2 Determinants of Math Equation matrices In Definition 3.2.2 , we use the cofactors of the first row of A to define the determinant |A|. Like Theorems 3.1.1 and 3.1.12, we can use the expansion of cofactors of each row or each column of A to compute the determinant |A|, but the proof of the result is difficult for n ≥ 4. Hence, we state the result without proof. Theorem 3.2.1. j ∈ In , n fo rD is tri bu tio n For each i, n T |A| = ∑ aik Aik = ∑ akj Akj = |A k=1 |. k=1 By Theorem 3.2.1 , we can choose any row of A together with its cofactors to compute the determinant of A. Hence, the smart choice would be to choose the row or column that contains the most zero entries. Theorem 3.2.2. If A has a row of zeros or a column of zeros, then |A| = 0. Proof Assume that all entries of the ith row are zero, that is, N ot ai1 = ai2 = ⋯ = ain = 0. Using the expansion of cofactors of the ith row and Theorem 3.2.1 , we have |A| = ai1 Ai1 + ai2 Ai2 + ⋯ + ain Ain = 0Ai1 + 0Ai2 + ⋯ + 0Ain = 0. If A has a column of zeros, then by Theorem 3.2.1 , |A| = 0. Example 3.2.2. Evaluate each of the following determinants. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/8 8/30/2020 3.2 Determinants of Math Equation matrices ∣0 ∣2 |A| = ∣ ∣ 0 ∣1 5∣ 3 0 0 ∣ ∣ |B| = ∣0 3 4∣ 0 1∣ 0 0 1 0 ∣3 1 2 0∣ ∣ ∣1 ∣ 1 ∣ 4∣ −2 2 ∣ ∣ 3 5 2∣ −1 3 ∣ 4∣ 2 1 9 6 ∣3 2 4 8∣ ∣ |C| = ∣ ∣0 ∣ ∣ ∣ ∣ fo rD is tri bu tio n Solution Because the matrix A contains a zero row, |A| = 0. We use the expansion of cofactors of row 3 of B to compute |B|. ∣0 |B| = 0A31 + 0A32 + A33 + 0A34 = A33 = 2 4∣ 0 1 1 ∣3 1 0∣ ∣ ∣ ∣ ∣ = −6. We use the expansion of cofactors of row 2 of C to compute |C|. |C| = 0 + (−1)A22 + 3A23 + 4A24 = M22 − 3M23 + 4M24 ∣1 2∣ 2 9 6 ∣3 4 8∣ ∣ ∣ ∣ ∣ ∣1 −3 3 2∣ 2 1 6 ∣3 2 8∣ ∣ ∣ N ot = 5 Similar to Theorem 3.1.3 , by Theorems 3.2.3 determinant of a lower or upper triangular matrix. and 3.2.1 ∣ ∣ ∣1 +4 3 5∣ 2 1 9 ∣3 2 4∣ ∣ ∣ ∣ ∣ = 160. , we obtain the following result on the Theorem 3.2.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/8 3.2 Determinants of Math Equation matrices ∣ a11 ∣ 0 ∣ ∣ 0 a12 a13 a22 a23 0 ∣ ∣ ∣ ∣ ⋯ ⋯ a33 ⋯ ⋮ ⋮ ⋮ ⋱ 0 0 0 ⋯ a1n ∣ 0 ∣ a11 0 0 ⋯ 0 a21 a22 0 ⋯ 0 ∣a 31 = ∣ ∣ a32 a33 ⋯ 0 ⋮ ⋮ ⋱ an2 an3 ⋯ ∣ ∣ ∣ ∣ ∣ 0 ⋮ ∣ ∣ ∣ ∣ ann ∣ ⋮ ∣ an1 Proof Let A be the lower triangular matrix. Using Theorem 3.2.1 ∣ a22 ∣ ∣ a32 0 ⋯ 0 a33 ⋯ 0 |A| = a11 ∣ ∣ ∣ ∣ ∣ ∣ a33 ∣ ∣ ∣ = a11 a22 ∣ ⋮ ∣ an2 ⋮ an3 ⋱ ⋯ ⋮ By Theorem 3.2.3 ⋮ ∣ ∣ ∣ = a11 a22 ⋯ ann . ∣ ⋮ ∣ ann ∣ repeatedly, we get ⋯ ∣ 0 ⋱ ∣ ∣ ∣ an3 The second result follows from Theorem 3.2.1 ∣ ∣ ∣ ann ∣ ∣ fo rD is tri bu tio n 8/30/2020 ∣ = a11 a22 ⋯ ann . ⋮ ∣ ann ∣ ⋯ and the above result. , the determinant of the identity matrix In is 1. Evaluate ∣5 ∣ |A| = ∣2 ∣ ∣ 1 ∣0 0 N ot Example 3.2.3. 0 0 3 0 0 6 7 0 0 2 ∣ ∣ ∣2 ∣ ∣ ∣ ∣ −1 ∣ |B| = 1 7 0 2 −5 ∣0 0 1 ∣ ∣ −2 ∣ ∣ ∣ ∣ ∣ |C| = ∣ ∣ ∣ ∣ 3 0 1∣ 0 0 2 ∣ 4∣ 0 0 1 3 0 0 0 2∣ ∣ ∣ Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 5/8 8/30/2020 3.2 Determinants of Math Equation matrices , we have ∣5 0 0 0∣ 3 0 ∣ 0∣ 1 6 7 0 ∣0 0 2 1∣ ∣2 1 7 0 2 −5 ∣0 0 1 ∣ |A| = ∣2 ∣ ∣ |B| = ∣ ∣ ∣ −2 ∣ |C| = ∣ ∣ ∣ ∣ 3 0 0 0 2 0 0 1 0 0 0 Exercises 1. For each of the following matrices, find M32 , ⎜ A = ⎜ ⎜ ⎝ 1 −1 = 5 × 3 × 7 × (−1)1 = 105. ∣ ∣ ∣ ∣ = 2 × 2 × 1 = 4. ∣ 1∣ ∣ 4∣ 3 ∣ = (−2) × 0 × 1 × 2 = 0. ∣ 2∣ M24 , A32 , A24 , M41 , M43 , A41 , A43 2 3 ⎞ N ot ⎛ ∣ fo rD is tri bu tio n By Theorem 3.2.3 2 2 0 1 2 3 −3 0 1 3 ⎛ ⎟ ⎟ −1 ⎟ ⎜ B = ⎜ ⎜ ⎠ ⎝ 6 1 0 2 −1 2 0 0 1 1 0 1 −1 0 1 ⎞ ⎟ ⎟. −1 ⎟ 2 ⎠ 2. Compute each of the following determinants by using its expansion of cofactors of each row. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 6/8 8/30/2020 3.2 Determinants of Math Equation matrices ∣ ∣ |A| = ∣ ∣ ∣ 1 0 2 2 0 1 ∣ −1 −1 ∣ ∣ ∣ ∣ ∣ ∣ 0 1 0 1 −1 0 1 3 |B| = ∣ ∣ ∣ 1 −1 2 3 2 2 0 3 1 2 3 −1 0 1 6 ∣ ∣ ∣ ∣ −3 ∣ ∣ ∣ . ∣ ∣ fo rD is tri bu tio n 3. Evaluate |A| by using the expansion of cofactors of a suitable row, where ∣0 0 0 1∣ 1 0 ∣ 1∣ 0 0 1 0 ∣1 1 −1 ∣ |A| = ∣0 ∣ ∣ ∣ . ∣ 0∣ 4. Evaluate each of the following determinants by inspection. 0 0∣ 0 2 0 ∣0 0 6∣ ∣1 −1 3 5 0∣ 0 0 0 0 0 |D| = ∣ 0 0 1 2 7∣ 6 3 −4 −2 6 10 |A| = ∣ ∣ ∣ ∣ ∣ ∣ 5 ∣2 ∣ 1 ∣ |F | = ∣ −2 ∣ ∣ ∣ ∣2 ∣ |B| = ∣ ∣ ∣ 2 ∣0 0 0 0 2 0 0 −1 6 −6 0 4 0 2 0 −2 1 0∣ 0 ∣ ∣ ∣1 |C| = 2∣ 3 0 1 2 ∣0 0 ∣ ∣ ∣ ∣ ∣ −1 ∣ 3 5 0∣ 0 3 0 0 0 |E| = ∣ 0 0 −1 2 7∣ 0 0 0 −4 ∣0 0 0 0 ∣ ∣ −2 0 0 0∣ ∣ ∣ ∣ ∣ 0 3 0 ∣ 0∣ 0 0 −1 0 0 0 6 ∣1 2 −1 ∣ ∣ ∣ N ot ∣1 ∣ ∣ ∣ ∣ ∣ 0∣ ∣ |G| = ∣ ∣ ∣ −1 ∣ ∣ 0 6 ∣ ∣ ∣ ∣ 2∣ ∣ ∣ 2∣ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 7/8 8/30/2020 3.2 Determinants of Math Equation matrices https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 8/8 8/30/2020 3.3 Evaluating determinants by row operations 3.3 Evaluating determinants by row operations fo rD is tri bu tio n By Theorem 3.2.3 , we see that the determinant of an upper triangular matrix can be easily calculated. Hence, we can use row operations to change an n × n matrix A to an upper triangular matrix B and then evaluate |A| by calculating |B|. One can show that there exists a constant k ∈ R R such that (3.3.1) |A| = k|B|. How can we find the number k? To do that, we need to know the relations between |A| and |B| if B is obtained from A by a single row operation. Theorem 3.3.1. Let A be an n × n matrix and c ∈ R with c ≠ 0. Then the following assertions hold. Ri ( 1 c ) Ri,j 2. If A − → B, then |A| = −|B|. Ri (c)+Rj N ot 1. If A − − − → B, then |A| = c|B|. 3. If A −−−−→ B, them |A| = |B|. Proof 1. Without loss of generalization, we assume that A and B are of the following forms: https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/26 8/30/2020 3.3 Evaluating determinants by row operations a11 ⎜ ⋮ ⎜ ⎜ A = ⎜ cai1 ⎜ ⎜ ⎜ ⎜ ⋮ ⎝ an1 Ri ( 1 c a12 ⋯ a1n ⋮ ⋮ ⋮ cai2 ⋯ cain ⋮ ⋮ ⋮ an2 ⋯ ann ⎛ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎜ ⎜ ⋮ ⎜ ⎜ and B = ⎜ ai1 ⎜ ⎜ ⎜ ⋮ ⎜ ⎠ ⎝ ) Then A − − − → B and Aik j ≠ i. By Theorem 3.2.1 (3.3.2) a11 an1 a12 ⋯ a1n ⋮ ⋮ ⋮ ai2 ⋯ ain ⋮ ⋮ ⋮ an2 ⋮ ann ⎞ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ ⎠ fo rD is tri bu tio n ⎛ for each k ∈ In because the jth rows of A and B are the same for , we have = Bik n n n |A| = ∑(caik )Aik = c ∑(aik )Aik = c ∑(aik )Bik = c|B|. k=1 k=1 k=1 2. We first prove that the result holds when we interchange two successive rows. Let a11 ⎜ ⋮ ⎜ ⎜ ⎜ ai1 A = ⎜ ⎜ ⎜ a(i + 1)1 ⎜ ⎜ ⎜ ⋮ ⎝ Ri, an1 a12 ⋯ ⋮ ⋮ a12 ⋯ a(i + 1)2 ⋯ ⋮ ⋮ an2 a1n ⎞ ⋯ ⎛ a11 ⎟ ⎜ ⋮ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ a(i + 1)1 ain ⎟, B = ⎜ ⎟ ⎜ a(i + 1)n ⎟ ai1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⋮ ⋮ ⋮ N ot ⎛ ann ⎠ ⎝ an1 a12 ⋯ ⋮ ⋮ a(i + 1)2 ⋯ ai2 ⋯ ⋮ ⋮ an2 ⋯ a1n ⎞ ⎟ ⎟ ⎟ a(i + 1)n ⎟ ⎟. ⎟ ain ⎟ ⎟ ⎟ ⎟ ⋮ ⋮ ann ⎠ (i+1) Then A −−−→ B, that is, we get B by interchanging the ith row and the (i + 1)th row of A. By Theorem 3.2.1 , we expand by cofactors via the ith row of A and the (i + 1)th row of B and get https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/26 8/30/2020 3.3 Evaluating determinants by row operations (3.3.3) n n |A| = ∑ aik Aik and |B| = ∑ k=1 aik B(i+1)k , k=1 where Aik and B(i+1)k are the cofactors of A and B, respectively. We note that the (i + 1)th row of ⋯ ain ). easy to see that Mik We denote by Mik the minors of A and by U(i+1)k the minors of B. Then it is = U (i+1)k for each k ∈ In . Then by (3.2.2) B(i+1)k = (−1) This, together with (3.3.3) , implies n fo rD is tri bu tio n B is (ai1 ai2 i+1+k , we have U (i+1)k = −(−1) n i+k Mik = −Aik . n |A| = ∑ aik Aik = − ∑ aik (−Aik ) = − ∑ aik B(i+1)k = −|B|. k=1 k=1 N ot Now we assume k=1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/26 8/30/2020 3.3 Evaluating determinants by row operations ⎛ a11 ⎜ ⎜ ⋮ ⎜ ⎜ a i1 ⎜ ⎜ A = ⎜ ⋮ ⎜ ⎜ ⎜ a j1 ⎜ ⎜ ⎜ ⎜ ⋮ ⎝ an1 a12 ⋯ a1n ⋮ ⋮ ⋮ ai2 ⋯ ain ⋮ ⋮ ⋮ aj2 ⋯ ajn ⋮ ⋮ ⋮ an2 ⋯ ann ⎝ Ri,j A− → B. a12 ⋯ a1n ⋮ ⋮ ⋮ ⋮ a(i − 1)1 a(i − 1)2 ⋯ a(i − 1)n aj1 aj2 ⋯ ajn a(i + 1)1 a(i + 1)2 … a(i + 1)n ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ fo rD is tri bu tio n ⎛ ⎜ ⎜ ⎜ ⎜ ⎞ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ and B = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎠ ⎜ ⎜ ⎜ ⎜ a11 ⋮ ⋮ ⋮ ⋮ a(j − 1)1 a(j − 1)2 ⋯ a(j − 1)n ai1 a12 ⋯ ain a(j + 1)1 a(j + 1)2 ⋯ a(j + 1)n ⋮ ⋮ ⋮ ⋮ an1 an2 ⋯ ann ⎠ We can get B from A by interchanging two successive rows several times by the following ways: i. Move the ith row vector: (ai1 ai2 ⋯ ain ) to the (i + 1)th row by interchanging the ith row and the (i + 1)th row of A, then move the vector (ai1 ai2 ⋯ ain ) from the i + 1th row to the (i + 2)th row by interchanging (ai1 ai2 ⋯ ain ) and (a(i+2)1 a(i+2)2 ⋯ a(i+2)n ). By times, we move (ai1 a12 ⋯ ain ) to the jth row. The resulting matrix is N ot j−i https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/26 8/30/2020 3.3 Evaluating determinants by row operations ⎛ ⎝ and (3.3.4) a12 ⋯ a1n ⋮ ⋮ ⋮ ⋮ a(i+1)1 a(i−1)2 ⋯ a(i−1)n a(i+1)1 a(i+1)2 ⋯ a(i+1)n ⋮ ⋮ ⋮ ⋮ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ fo rD is tri bu tio n ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ C := ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ a11 a(j−i)1 a(j−i)2 ⋯ a(j−i)n aj1 aj2 ⋯ ajn ai1 ai2 ⋯ ain a(j+1)1 a(j+1)2 ⋯ a(j+i)n ⋮ ⋮ ⋮ ⋮ an1 an2 ⋯ ann j−i ∣C ∣. ∣ ∣ N ot ∣A∣ = (−1) ∣ ∣ ⎠ Note that the vector (aj1 aj2 ⋯ ajn ) is in the (j − 1) row of C. ii. We move (aj1 aj2 ⋯ ajn ) back to the (i − 2)th row by interchanging the (j − 1)th and (i − 2)th rows of C, that is, interchanging the two row vectors (aj1 aj2 ⋯ ajn ) and (a(j−1)1 a(j−1)2 ⋯ a(j−1)n ) of C Continuing the process, we can move (aj1 aj2 ⋯ ajn ) back to the ith row by [(j − i) − i] times. The resulting matrix exactly is B. Moreover, we have ∣C ∣ = (−1)[(j−1)−i] ∣B∣ = (−1)(j−j)−1 ∣B∣. ∣ ∣ ∣ ∣ ∣ ∣ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/26 8/30/2020 3.3 Evaluating determinants by row operations This, together with (3.3.4) , implies j−i j−i (j−j)−1 2(j−i)−1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ [(−1) ∣A∣ = (−1) ∣C ∣ = ∣A∣ = (−1) ∣B∣] = (−1) ∣B∣ = −∣B∣ because 2(j − i) − 1 is an odd number and (−1) 2(j−i)−1 = −1. ⎛ fo rD is tri bu tio n 3. We consider the following three cases. i. We first prove if an n × n matrix D has two equal rows, then |D| = 0. We assume that (3.3.5) a11 N ot ⎜ ⋮ ⎜ ⎜ ⎜ a(i − 1)1 ⎜ ⎜ ai1 ⎜ ⎜ ⎜ a(i+1)1 ⎜ ⎜ ⎜ D = ⎜ ⋮ ⎜ ⎜ ⎜ a(j−1)1 ⎜ ⎜ ai1 ⎜ ⎜ ⎜ a (j+1)1 ⎜ ⎜ ⎜ ⎜ ⋮ ⎝ an1 a12 ⋯ ⋮ ⋮ a(i − 1)2 ⋯ ai2 ⋯ a(i+1)2 ⋯ ⋮ ⋮ a(j−1)2 ⋯ ai2 ⋯ a(j+1)2 ⋯ ⋮ ⋮ an2 ⋯ a1n ⎞ ⎟ ⎟ ⎟ a(i − 1)n ⎟ ⎟ ⎟ ain ⎟ ⎟ ⎟ a(i+1)n ⎟ ⎟ ⎟ ⎟. ⋮ ⎟ ⎟ a(j−1)n ⎟ ⎟ ⎟ ain ⎟ ⎟ ⎟ a(j+1)n ⎟ ⎟ ⎟ ⎟ ⋮ ⋮ ann ⎠ Note that after interchanging the ith and jth rows of D, the resulting matrix still is D. Hence, |D| = −|D|. This implies |D| = 0. ii. Expanding by cofactors of D via the jth row of D given in (3.3.5) and using the result (i), we obtain https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/26 8/30/2020 3.3 Evaluating determinants by row operations (3.3.6) n |D| = ∑ aik Djk = 0, k=1 fo rD is tri bu tio n where Djk is the cofactor of D for k ∈ In . iii. Now, we assume that ⎛ a11 ⎜ ⋮ ⎜ ⎜ ⎜ ai1 ⎜ ⎜ ⎜ ⋮ ⎜ ⎜ A = ⎜a (j−1)1 ⎜ ⎜ aj1 ⎜ ⎜ ⎜ ⎜ a(j+1)1 ⎜ ⎜ ⎜ ⋮ and an1 ⋯ a1n ⋮ ⋮ ⋮ ai2 ⋯ ain ⋮ ⋮ ⋮ a(j−1)2 ⋯ a(j−1)n aj2 ⋯ ajn a(j+1)2 ⋯ a(j+1)n ⋮ ⋮ ⋮ an2 ⋯ ann ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ N ot ⎝ a12 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/26 8/30/2020 3.3 Evaluating determinants by row operations a11 ⎛ ⋯ a1n ⋮ ⋮ ⋮ ai2 ⋯ ain ⋮ ⋮ ⋮ a(j−1)2 ⋯ a(j−1)n ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ fo rD is tri bu tio n ⎜ ⋮ ⎜ ⎜ ⎜ ai1 ⎜ ⎜ ⎜ ⋮ ⎜ ⎜ B = ⎜ a(j−1)1 ⎜ ⎜ ⎜ cai1 + aj1 ⎜ ⎜ a(j+1)1 ⎜ ⎜ ⎜ ⎜ ⋮ a12 ⎝ an1 Ri(c) +Rj cai2 + ai2 ⋯ cain + ajn a(j+1)2 ⋯ a(j+1)n ⋮ ⋮ ⋮ an2 ⋯ ann ⎠ Then A −−−−→ B. Let Aik and Bik be the jth row cofactors of A and B, respectively. Then by (3.3.5) , (3.3.7), and (3.3.8), we see that (3.3.9) Ajk = Bjk = Djk N ot for k ∈ In . Expanding by cofactors of B via the jth row of B, we have by (3.3.9) Theorem 3.2.1 , n n n and (3.3.6) and n |B| = ∑(caik + ajk )Bjk = ∑(caik + ajk )Ajk = c ∑ aik Djk + ∑ ajk Ajk k=1 k=1 k=1 k=1 n = 0 + ∑ ajk Ajk = |A|. k=1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/26 8/30/2020 3.3 Evaluating determinants by row operations By Theorem 3.3.1 , we see that there exists a number k such that |A| = k|B|, where B is obtained from A by several row operations, and k can be obtained from the row operations used in Theorem 3.3.1 (i) and (ii). Before we use Theorem 3.3.1 to compute determinants, we try to understand and digest each result given in Theorem 3.3.1 and use it to derive some results. (1), or (3.3.2) , we see that ∣ a11 ∣ ∣ ∣ ⋮ ∣ cai1 a12 ⋯ ⋮ ⋮ cai2 ⋯ fo rD is tri bu tio n By Theorem 3.3.1 a1n ∣ ∣ ⋮ ∣ ⋮ ∣ an1 ⋮ ⋮ an2 ⋯ ∣ ∣ ∣ ∣ ⋮ cain ∣ = c ∣ ai1 ∣ ∣ ∣ a11 ∣ ⋮ ∣ ∣ ∣ ∣ ∣ ∣ ann ∣ a12 ⋯ ⋮ ⋮ ai2 ⋯ a1n ∣ ∣ ∣ ⋮ ∣ ain ∣ . ∣ ⋮ ∣ an1 ⋮ ⋮ an2 ⋯ ∣ ⋮ ∣ ann ∣ Hence, we can take the common factor of one row out of the determinant. Similarly, because ∣∣AT ∣∣ = |A| , we can take a common factor of one column out of the determinant. Example 3.3.1. N ot Compute each of the following determinants. ∣ 3 |A| = ∣ ∣ 12 ∣ 0 −3 4 −2 6 ∣ 16 ∣ ∣ 5 ∣ ∣ −1 |B| = ∣ ∣ ∣ −1 12 ∣ ∣ 2 4 6 0 1 6 ∣ ∣ . Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/26 3.3 Evaluating determinants by row operations ∣ 3(1) ∣ |A| = ∣ (4)(3) ∣ ∣ 0 ∣1 = 3(4) ∣ ∣ ∣ −1 ∣ |B| = ∣ (2)(1) ∣ ∣ 0 ∣ −1 ∣ = ∣ 1 ∣ ∣ 0 3(2) (4)(1) −2 −1 3 ∣0 ∣ ∣ 1 ∣ ∣ (4)(4) ∣ = 3 ∣ (4)(3) ∣ ∣ ∣ ∣ 0 5 3(−1) 1 −2 4 2 1 −2 ∣ ∣ (4)(4) ∣ ∣ ∣ 5 ∣ ∣ = (12)(16) = 192. 5∣ (2)(2) −1 (4)(1) 2 2∣ −1 1 −1 fo rD is tri bu tio n 8/30/2020 ∣ ∣ −1 ∣ ∣ (2)(3) ∣ = 2 1 ∣ ∣ ∣ 0 ∣ 6 12 (3)(4) ∣ ∣ −1 ∣ ∣ (3)(1) ∣ = (2)(3) 1 ∣ ∣ ∣ 0 (3)(2) ∣ −1 12 ∣ ∣ 2 3 1 6 ∣ −1 ∣ 4∣ ∣ 2 1 1 2∣ ∣ = (6)(3) = 18. The following result gives the relation between |cA| and |A|. Theorem 3.3.2. |cA| = c Proof n |A| . N ot Let A be an n × n matrix and c ∈ R. Then The result holds if c = 0. We assume that c ≠ 0. Repeating (3.3.2) implies https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/26 8/30/2020 3.3 Evaluating determinants by row operations ∣ ca11 ca12 ⋯ ca1n ∣ ∣ ∣ ∣ ca21 ca22 ⋯ ca2n ∣ |cA| = ∣ a12 ⋯ a1n ∣ ca22 ⋯ ∣ ca2n ∣ ∣ ∣ ca21 ∣ = c∣ ⋮ ∣ ∣ can1 ∣ a11 ⋮ ⋱ can2 ⋯ n ∣ ∣ ∣ cann ∣ ⋯ a1n ∣ a22 ⋯ ∣ a2n ∣ ∣ ∣ ∣ a12 ∣ ∣ a21 ⋮ ∣ ∣ = c ⋮ ∣ an1 ⋮ ⋱ an2 ⋯ Example 3.3.2. ∣ = ⋯ ⋮ ∣ can1 ⋮ ⋱ can2 ⋯ ⋮ ∣ ∣ cann ∣ fo rD is tri bu tio n ∣ = c ∣ a11 ⋮ n |A| . ∣ ∣ ann ∣ Let A be a 3 × 3 matrix. Assume that |A| = 6. Compute each of the following determinants. T |2A| and ∣ ∣(3A) Solution By Theorem 3.3.2 , 3 By Theorems 3.2.1 and 3.3.2 , |A| = 8 × 6 = 48. N ot |2A| = 2 ∣. ∣ ∣(3A)T ∣ = |3A| = 33 |A| = 27(6) = 162. ∣ ∣ Theorem 3.3.1 (2) means that when interchanging any two different rows of a determinant, the determinant changes its sign. Example 3.3.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/26 8/30/2020 3.3 Evaluating determinants by row operations Evaluate ∣0 |A| = ∣ ∣ 0 Solution ∣0 |A| = ∣ ∣ 0 ∣1 0 1∣ 2 3 −2 Corollary 3.3.1. i. If A has two equal rows, then |A| = 0. ii. If A has two equal columns, then |A| = 0. Proof 1∣ 2 3 ∣ ∣ . 5∣ −2 fo rD is tri bu tio n ∣1 0 ∣ ∣1 R1 , 3 ∣ –––– –––– – – − 0 ∣ 5∣ ∣0 ∣ ∣ 2 3 0 1∣ ∣ = −2. (3) and the result (ii) follows from Theorem N ot The result (i) follows from the result (i) in Theorem 3.3.1 3.2.1 and the result (i). 5∣ −2 Example 3.3.4. Evaluate ∣ −1 ∣1 ∣ A = ∣2 ∣ ∣ 3 ∣1 3 7 9 3 −2 −4 1 −2 4∣ ∣ 8∣ 5 ∣ ∣ 4∣ ∣ ∣ B = ∣ ∣ ∣ ∣ 3 2 0 3 3 1 0 2 1 3 9 1 5 9 2 −3 1 2 −3 1 3 0 4 3 ∣ ∣ ∣ ∣. ∣ ∣ ∣ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/26 8/30/2020 3.3 Evaluating determinants by row operations Solution Because A has two equal rows, rows 1 and 4, |A| = 0. Similarly, because B has two equal columns, columns 2 and 5, |B| = 0. can be generalized to two proportional rows or columns. Theorem 3.3.3. fo rD is tri bu tio n Corollary 3.3.1 1. If A has two proportional rows, then |A| = 0. 2. If A has two proportional columns, then |A| = 0. Proof (1) Assume that A is an n × n matrix and the ith and jth rows are proportional. Without a loss of generalization, we may assume that 1 ≤ i < j ≤ n and Ri (c) = Rj for some c ≠ 0. By Theorem 3.3.1 (1) and Corollary 3.3.1 , we have ∣ a11 a12 ⋯ ∣ ∣ ⋮ ∣ ∣ ai1 ∣ ∣ ∣ ∣ ∣ ⋮ cai1 ⋮ ∣ an1 ∣ ∣ ∣ ∣ ain ∣ ∣ ∣ ∣ ∣ = c ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ⋮ ⋮ ai2 ⋯ ⋮ ⋮ ⋮ cai2 ⋯ cain ∣ ⋮ ⋮ an2 ⋯ ∣ a11 ∣ ⋮ N ot |A| = a1n ∣ ⋮ ann ∣ a12 ⋯ a1n ∣ ∣ ⋮ ⋮ ⋮ ai2 ⋯ ⋮ ⋮ ⋮ ⋮ ai1 ai2 ⋯ ain ⋮ ⋮ ⋮ ⋮ an2 ⋯ ∣ ai1 ⋮ ∣ ∣ ain ∣ ∣ ∣ an1 ∣ = 0. ∣ ∣ ∣ ∣ ann ∣ because the ith and jth rows of the last determinant are the same. (2) The result follows from Theorem 3.2.1 and the result (1). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/26 8/30/2020 3.3 Evaluating determinants by row operations Example 3.3.5. Evaluate ∣ |A| = ∣ ∣ 1 ∣2 Solution 9 3 6 1 −2 ∣ ∣ ∣ ∣ ∣ |B| = −4 ∣ ∣ ∣ 1 3 −2 −2 ∣ 3 9 1 ∣ −6 ∣ −2 6 −1 4 ∣ . ∣ fo rD is tri bu tio n ∣3 ∣ Because rows 2 and 3 of A are proportional, by Theorem 3.3.3 of B are proportional, by Theorem 3.3.3 (2), |B| = 0. 1 1 4 −2 ∣ (1), |A| = 0. Because columns 1 and 4 |A| = |B| = 0. Example 3.3.6. Compute ∣ |A| = ∣ ∣ 2 −3 5 1 7 2 ∣ −4 6 ∣ ∣ ∣ . N ot Theorem 3.3.1 (3) means that when adding the ith row multiplied by a nonzero constant c to the jth row, the determinant remains unchanged. Using Theorem 3.3.1 (3) and Theorem 3.2.2 , one can provide another proof of Theorem 3.3.3 (1). Indeed, we use the row operation Ri (−c) + Rj on A. Then the jth row of the resulting matrix denoted by B is a zero row. by Theorem 3.2.2 , |B| = 0 and by Theorem 3.3.1 (3), −10 ∣ Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/26 8/30/2020 3.3 Evaluating determinants by row operations ∣ |A| = ∣ ∣ 2 1 ∣ −4 −3 7 6 5 ∣ ∣2 ∣ ∣ −3 R1 (2) + R3 1 ∣ ––––––––––– ∣ ––––––––––– ∣0 −10 ∣ 2 5∣ ∣ 7 2 0 0∣ ∣ = 0. fo rD is tri bu tio n Now, we use examples to show how to use row operations to calculate determinants. Before calculating the determinant of a matrix, it is best to check the following: 1. Whether the matrix contains a zero row or a zero column. 2. Whether the matrix contains two equal rows or two equal columns. 3. Whether the matrix contains two proportional rows or two proportional columns. If one of the above occurs, then the determinant is zero. 4. Whether we should use the transpose to compute the determinant. After checking the above four things, we would follow Steps 1-3 given in Section 2.4 for each round and apply Theorem 3.3.1 . Example 3.3.7. Evaluate each of the following determinants. |A| = ∣ ∣ 3 ∣2 Solution 1 −6 6 ∣2 5∣ 9 ∣ ∣ 1∣ ∣ |B| = ∣0 ∣ ∣ 2 6 ∣3 10 4∣ ∣ ∣ ∣ −1 3 4∣ 1 9 6 ∣ ∣ 4 8∣ ∣ N ot ∣0 2 ∣ |C| = 1 0 0 1 0 0 2 1 4 0 1 ∣ −2 ∣ 1 ∣ ∣ −2 ∣ 0 ∣ . ∣ −1 ∣ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/26 3.3 Evaluating determinants by row operations ∣0 |A| = ∣ ∣ 3 −6 6 ∣ 1 ∣ −2 ∣ 10 ∣ ⏐ ⏐ ∣ ⏐ ⏐ 5 ∣ ⏐ ∣ ⏐ ⏐ −5 ∣ ↓ 3 ∣ 0 1 ∣2 6 ∣ ⏐ 3∣ ⏐ ∣ ⏐ ⏐ 5 ∣ ⏐ R1 (−2) + R3 –––––––––––– –––––––––––– ∣ ⏐ – – 1∣ ⏐ ↓ ∣ 1 ∣ ∣ 5 = −3 ∣ 0 ∣ ∣ 1∣ ∣ 2 −2 ∣ 1 ∣ −2 3 ∣ − 3∣ 0 ∣ 1 5 ∣ ∣ 0 0 9∣ −6 R2 (−10) + R3 – ––––––––––––– ––––––––––––– – 1 6 ∣ ∣ −55 ∣ (−3)(1)(1)(−55) = 165. ∣2 ∣ |B| = ∣0 ∣ 6 10 4∣ −1 3 ∣ 4∣ 1 9 6 ∣3 2 4 8∣ ∣ 1 ∣ 3 5 ∣ 0 −1 3 ∣ 0 ∣ −5 −1 ∣ 0 −7 −11 = 2 ∣ 2 ∣ ∣ ∣ 1 ∣ ∣ 0 3 5 −1 3 2 1 9 ∣ 3 2 4 ∣ ∣ ⏐ 2∣ ⏐ ∣ ⏐ ⏐ 4 ∣ ⏐R2 (−5)+R3 ∣ ⏐− −−−− → ⏐R2 (−7)+R4 ∣ 2 ⏐ ∣ ⏐ ⏐ 2∣ ↓ ∣ 1 ∣ 3 ∣ 0 ∣ −1 3 0 −16 0 0 R3 (−2) + R4 – –––––––––––– –––––––––––– –2 ∣ 0 ∣ ∣ 0 ∣1 ∣ |C| = ∣ ∣C T −2 3 5 ∣ 0 ∣ −1 3 ∣ 0 ∣ 0 −16 ∣ 0 0 −32 5 0 4 ∣0 1 2 0 0 0 1 1 −2 0 ∣ = ∣ ∣ ∣ ∣1 ⏐ 2∣ ⏐ ∣ ⏐ 4 ∣ ⏐R1 (−2)+R3 ⏐− −−−− → ∣ ⏐R (−3)+R 4 6 ⏐ 1 ∣ ⏐ ⏐ 8∣ ↓ ∣ 1 ∣ N ot 2∣ − 1∣ 1 ∣ 0 ∣3 R1,2 – ––– – ––– ∣ 9 ∣2 −3 ∣ 0 ∣ = 5∣ 1 fo rD is tri bu tio n 8/30/2020 2 ∣ 4 ∣ ∣ ⏐ ⏐ ∣ ⏐ ∣ ⏐ 4 ⏐ ∣ ⏐ ⏐ −18 ∣ ⏐ ∣ ⏐ ⏐ −26 ∣ ↓ 2 ∣ ∣ = (2)(1)(−1)(−16)(10) = 320. −18 ∣ ∣ 10 ∣ ∣ ∣1 ∣ ∣ ∣ ∣0 R1 (−1) + R4 ∣ ∣ –––––––––––– –––––––––––– – – 0 ∣ ∣ −1 ∣ ∣0 −2 0 4 1 2 0 0 1 1 0 0 ∣ ∣ ∣ ∣ = −5. ∣ −5 ∣ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/26 8/30/2020 3.3 Evaluating determinants by row operations By Theorems 2.6.1 and 3.3.1 , we obtain Corollary 3.3.2. Let A be an n × n matrix and c ∈ R with c ≠ 0. Then the following assertions hold. |Ei (c)A| = |Ei (c)| |A| . ii. ∣∣Ei(c)+j A∣∣ = ∣∣Ei(c)+j ∣∣ |A| . iii. |Ei,j A| = |Ei,j | |A| . Proof fo rD is tri bu tio n i. Ri (c) i. Let B = Ei (c)A. By Theorem 2.6.1 3.3.1 (i), A , |A| = 1 c This implies c = |Ei (c)| and |B| − − − → Ri (c) B and In and |Ei (c)| = 1 c − −− → |In | = Ei (c). 1 By Theorem . c N ot |Ei (c)A| = |B| = c |A| = |Ei (c)| |A| . Ri (c) + Rj ii. Let B = Ei(c)+j A. By Theorem 2.6.1 By Theorem 3.3.1 (iii), − −− − −− → A − Ri (c) + Rj B and In − − −− − −− → Ei(c)+j . , |A| = |B| and ∣ ∣Ei(c)+j ∣ ∣ = |In | = 1. This implies https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/26 8/30/2020 3.3 Evaluating determinants by row operations ∣ ∣Ei(c)+j A∣ ∣ = |B| = |A| = ∣ ∣Ei(c)+j ∣ ∣ |A| . Ri,j iii. Let B = Ei,j A. By Theorem 2.6.1 (ii), A − − → This implies B and In Ei,j . By Theorem 3.3.1 , and |Ei,j | = − |In | = −1. fo rD is tri bu tio n |A| = − |B| Ri,j − − → |Ei,j A| = − |B| = − |A| = |Ei,j | |A| . By using Corollary 3.3.2 repeatedly, it is easy to prove the following result, which gives the relation among the determinants of an n × n matrix, its row echelon matrix, and elementary matrices. Theorem 3.3.4. Let A and B be n × n matrices. Assume that there exist elementary matrices E1 , B = Ek Ek−1 ⋯ E2 E1 A. Then ⋯ ⋯ , Ek such that |E2 | |E1 | |A| . N ot |B| = |Ek | |Ek−1 | E2 , By Theorem 3.3.1 and |I| = 1, it is easy to see that the following result on the determinants of elementary matrices holds. Theorem 3.3.5. The determinant of every elementary matrix is not zero. At the end of the section, we generalize the formula (2.7.1) an n × n matrix for n ≥ 3. of an inverse matrix from a 2 × 2 matrix to https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/26 8/30/2020 3.3 Evaluating determinants by row operations We introduce the concept of the adjoint matrix of an n × n matrix. Definition 3.3.1. Let A be an n × n matrix. The matrix fo rD is tri bu tio n (3.3.10) A11 A12 ⋯ A1n ⎜ A21 ⎜ ⎜ ⎜ ⎜ ⋮ A22 ⋯ ⋮ ⋱ A2n ⎟ ⎟ ⎟ ⎟ ⎟ ⋮ An2 ⋯ Ann ⎛ ⎝ An1 ⎞ ⎠ is called the matrix of cofactors of A. The transpose of the matrix of cofactors of A is called the adjoint of A, denoted by adj(A) By Definition 3.3.1 , the adjoint of A is the following matrix: (3.3.11) A12 ⋯ ⎜ A21 ⎜ adj(A) = ⎜ ⎜ ⎜ ⋮ A22 ⋯ ⋮ ⋱ ⎝ A1n N ot A11 ⎛ An1 An2 ⋯ T ⎞ A2n ⎟ ⎟ ⎟ ⎟ ⎟ ⋮ Ann ⎠ A11 A21 ⋯ An1 ⎜ A12 ⎜ = ⎜ ⎜ ⎜ ⋮ A22 ⋯ ⋮ ⋱ An2 ⎟ ⎟ ⎟. ⎟ ⎟ ⋮ A2n ⋯ Ann ⎛ ⎝ A1n ⎞ ⎠ Example 3.3.8. Find the matrix of cofactors of A and adj(A), where https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/26 8/30/2020 3.3 Evaluating determinants by row operations a b c d A = ( ). Solution ( A11 A12 A21 adj (A) = ( d −c −b a ) = ( A22 and A11 A12 A21 A22 T ) (3.3.12) b N ot a A adj(A)= ( d −b −c a )( c and Corollary 3.3.1 = M22 = a, the d ) d −b −c a = ( Note that (2.7.1) involves the adjoint matrix of A given in Example 3.3.8 Example 3.3.8 , we see that By Theorem 3.2.1 and A22 A12 = −M12 = −c, A21 = −M21 = −b, fo rD is tri bu tio n Because A11 = M11 = d, matrix of cofactors of A is ). . By Theorem 2.7.5 and ) = |A| I. , we generalize (3.3.12) to n × n matrices. Theorem 3.3.6. Let A and adj(A) be the same as in (3.2.1) and (3.3.11) , respectively. Then (3.3.13) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/26 8/30/2020 3.3 Evaluating determinants by row operations A adj(A) = |A| I. Proof j ∈ In and a11 a12 ⋯ a1n ⎜ a21 ⎜ ⎜ ⎜ ⋮ ⎜ ⎜ ⎜ ai1 A = ⎜ ⎜ ⎜ ⋮ ⎜ ⎜ ⎜ aj1 ⎜ ⎜ ⎜ ⋮ a22 ⋯ ⋮ ⋮ ai2 ⋯ ⋮ ⋮ aj2 ⋯ ⋮ ⋮ a2n ⎟ ⎟ ⎟ ⎟ ⋮ ⎟ ⎟ ain ⎟ ⎟. ⎟ ⎟ ⋮ ⎟ ⎟ ajn ⎟ ⎟ ⎟ ⎟ ⋮ an2 ⋯ ⎛ ⎝ By Theorem 3.2.1 ⎞ fo rD is tri bu tio n Let i, an1 ann ⎠ , we expand |A| via the jth row and obtain (3.3.14) N ot n |A| = ∑ ajk Ajk . k=1 Replacing the jth row by the ith row of A, by Corollary 3.3.1 (1) we obtain (3.3.15) n ∑ aik Ajk = 0 for i < j or i > j. k=1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 21/26 8/30/2020 3.3 Evaluating determinants by row operations Let C = (cij ) = A adj(A) a11 ⎜ ⎜ ⋮ ⎜ = ⎜ ai1 ⎜ ⎜ ⎜ ⎜ ⋮ ⎝ an1 a12 ⋯ a1n ⋮ ⋯ ⋮ ai2 ⋯ ain ⋮ ⋯ an2 ⋯ Then n ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⋮ ann ⎠ cij = ∑ aik Ajk k=1 and (3.3.15) C = diag(|A|), By Theorem 3.3.6 Theorem 3.3.7. , cjj = |A| if i = j and cij ⋯ , |A|) = |A|I. , we generalize Theorem 2.7.5 Let A and adj(A) be the same as in (3.2.1) ⋯ Aj1 ⋯ An1 ⎜ A12 ⎜ ⎜ ⎜ ⎜ ⋮ ⋯ Aj2 ⋯ An2 ⎟ ⎟ ⎟. ⎟ ⎟ ⎞ ⎝ A1n ⋮ ⋮⋮ ⋯ Ajn ⋯ Ann ⎠ for i, j ∈ In . = 0 if i ≠ j. Hence, to an n × n matrix. N ot By (3.3.14) A11 ⎛ fo rD is tri bu tio n ⎛ and (3.3.11) −1 A , respectively. If |A| ≠ 0, then 1 = adj(A). |A| Proof Because |A| ≠ 0, by Theorem 3.3.13, we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 22/26 8/30/2020 3.3 Evaluating determinants by row operations 1 A( 1 adj(A)) = |A| 1 [A adj(A)] = |A| The result follows from Definition 2.7.1 (|A|I) = I. |A| . Example 3.3.9. to find the inverse of the following matrix fo rD is tri bu tio n Use Theorem 3.3.7 1 1 1 A = ⎜0 1 1⎟. 0 1 ⎛ ⎝ Solution , we have A11 = ∣1 ∣ ∣0 A21 = ∣1 −∣ ∣0 A31 = ∣1 ∣ ∣1 This, together with (3.3.11) , implies ∣0 1∣ ∣ = 1, A12 = − ∣ 1∣ ∣1 ⎠ ∣0 1∣ ∣ = 1, A13 = ∣ 1∣ ∣1 1∣ ∣ = −1, 0∣ ∣1 1∣ ∣ = −1, A22 = ∣ 1∣ ∣1 ∣1 1∣ ∣ = 0, A23 = − ∣ 1∣ ∣1 1∣ ∣ = 1, 0∣ ∣1 1∣ ∣ = 0, A32 = − ∣ ∣1 1∣ ∣ = −1, A33 = ∣ 1∣ ∣ = 1. 1∣ 1∣ 1∣ N ot By (3.1.9) 1 ⎞ ∣0 ∣0 A11 A21 A31 adj(A) = ⎜ A12 A22 A32 ⎟ = ⎜ A23 A33 ⎛ ⎝ A13 ⎞ ⎠ ⎛ ⎝ 1 −1 1 0 −1 1 0 ⎞ −1 ⎟ . 1 ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 23/26 8/30/2020 3.3 Evaluating determinants by row operations By computation, |A| = 1. It follows from Theorem 3.3.7 −1 A = 1 |A| that 1 −1 1 0 −1 1 ⎛ adj(A) = adj(A) = ⎜ ⎝ Exercises ⎞ −1 ⎟ . 1 ⎠ , where a different method is used. fo rD is tri bu tio n The inverse of A is the same as that of B given in Example 2.7.6 0 1. Evaluate each of the following determinants by inspection. ∣ |A| = ∣ ∣ 1 2 0 1 ∣ −2 −4 −3 ∣ 2 6 ∣ ∣ ∣ ∣2 |B| = ∣ ∣ 2 ∣0 −2 4∣ −2 4 1 ∣ ∣ 2∣ ∣ 1 2 3 1 2 ∣ ∣ |C| = ∣ −2 ∣ ∣ ∣ 1 − 1 2 ∣ ∣ −1 ∣ N ot 2. Use row operations to evaluate each of the following determinants. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 24/26 8/30/2020 3.3 Evaluating determinants by row operations ∣ |A| = ∣ ∣ 1 ∣0 ∣ 0 ∣ |D| = ∣ −2 ∣ ∣ ∣ 0 4 ∣0 ∣ ∣ 1 |F | = ∣ 0 ∣ ∣ 1 ∣2 1 −2 2 4∣ ∣ ∣ ∣ 4 ∣ |B| = 4∣ 0 0 2 0 0 6 −6 0 2 −1 3 −1 2 4 3 1 3 2 ∣ −2 ∣ ∣ ∣ ∣ ∣ ∣ |E| = ∣ ∣ ∣ ∣ |C| = ∣ 3 ∣ ∣ −1 ∣ −1 ∣ −1 ∣ 5 2 −2 ∣ 0 0 ∣ 0 2 −3 2 1 2 0 −2 −3 1 −2 1 −1 2 0 ∣ 0 0 0∣ 3 0 ∣ 1∣ −2 ∣ 1 ∣ ∣ ∣ ∣ ∣ fo rD is tri bu tio n ∣2 ∣ ∣ 0 ∣ −1 0 −1 0 1 2 7 1 −2 3 1 −2 6 10 0 ∣ ∣ ∣ ∣ ∣ ∣ 1 −1 0 0 ∣ ∣ ∣ ∣ ∣ ∣ ∣ 2∣ 1 −1 3 5 0∣ 1 2 0 0 0 0 −1 2 1∣ 3 0 0 −6 2 0 0 0 ∣ |G| = ∣ −2 ∣ 0 6 ∣ ∣ ∣ ∣ 1∣ 3. Evaluate each of the following determinants by using its transpose. ∣1 ∣ ∣2 ∣ 7 0 0 ∣1 0 0 0 0 2 7 0 2 1 |B| = ∣ 0 6 3 0 0 1 3 1 2 −1 ∣0 6 1 2 1 1∣ ∣ ∣ 2∣ ∣ ∣ N ot |A| = 0 ∣ 0 ∣5 4. Use Theorem 3.3.7 6 3 3 1 0 ∣ ∣ ∣ 2∣ ∣ ∣ ∣ ∣. ∣ ∣ ∣ to find the inverse of the following matrix 0 2 −5 A = ⎜1 5 −5 ⎟ ⎛ ⎝ 1 0 8 ⎞ ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 25/26 3.3 Evaluating determinants by row operations N ot fo rD is tri bu tio n 8/30/2020 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 26/26 8/30/2020 3.4 Properties of determinants 3.4 Properties of determinants ut io n A criterion that uses ranks of matrices to verify whether n × n matrices are invertible is Theorem 2.7.6 , which states that an n × n matrix A is invertible if and only if r(A) = n. By Theorem 2.7.5 , a 2 × 2 matrix A is invertible if and only if its determinant is not zero. The following result shows that this is true for any n × n matrix. This provides another criterion that uses determinants to verify whether n × n matrices are invertible. D is tri b Theorem 3.4.1. Let A be an n × n matrix. Then |A | ≠ 0 if and only if A is invertible. Proof , there exist elementary matrices fo r Let C be the reduced row echelon matrix of A. By Theorem 2.7.3 E 1, E 2, ⋯ , E k − 1, E k such that (3.4.1) By Theorem 3.3.4 (3.4.2) , N ot C = E kE k − 1⋯E 2E 1A. |C | = | EkǁEk − 1 | ⋯ | E2ǁE1ǁA | . Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/11 8/30/2020 3.4 Properties of determinants Assume that |A | ≠ 0. By Theorem 3.3.5 and (3.4.2) , |C | ≠ 0. Because C is the reduced row echelon matrix of A, C is an upper triangular matrix and all the entries on the main diagonal that are leading entries are not zero. This implies r(C) = n. By (3.4.1) and Theorem 2.5.4 , r(A) = r(C) = n. By Theorem 2.7.6 , A is invertible. Conversely, assume that A is invertible. By Corollary 2.7.1 (i), C = I n and |C | = 1. It follows from (3.4.2) and Theorem 3.3.5 that |A | ≠ 0. ut io n In Example 2.7.5 , we use ranks of matrices, that is, Theorem 2.7.6 , to verify whether square matrices are invertible. Now we revisit the two matrices and use determinants, that is, Theorem 3.4.1 , to determine whether they are invertible. D is tri b Example 3.4.1. For each of the following matrices, use its determinant to determine whether it is invertible. ( ) ( ) 1 1 1 A= 0 1 1 1 B= 1 0 −1 0 1 0 1 fo r 1 0 1 1 ↓ | | 1 1 1 |A| = 0 1 1 1 0 1 ↓ | | N ot Solution 1 1 1 R 1( − 1) + R 3_ _ 0 1 1 0 −1 0 | | 1 1 1 R 2(1) + R 3_ _ 0 1 1 0 0 1 = 1 ≠ 0. By Theorem 3.4.1 Processing math: 100% , A is invertible. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/11 3.4 Properties of determinants ↓ | | 1 1 ↓ | | 1 |B | = 0 − 1 0 1 0 1 1 1 1 R 1( − 1) + R 3_ _ 0 − 1 0 0 −1 0 R 2( − 1) + R 3_ _ | | 1 1 1 By Theorem 3.4.1 , A is not invertible. Theorem 3.4.2. Let A and B be n × n matrices. Then Proof fo r |AB | = | AǁB | . D is tri b 0 − 1 0 = 0. 0 0 0 ut io n 8/30/2020 , there exist elementary N ot Assume that C is the reduced row echelon matrix of A. By Theorem 2.7.3 matrices E 1, E 2, ⋯ , E k − 1, E k such that C = E kE k − 1⋯E 2E 1A. It follows that (3.4.3) Processing math: 100% A = F 1F 2⋯F k − 1F kC, https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/11 8/30/2020 3.4 Properties of determinants −1 where F i = E i is an elementary matrix for i = 1, 2, ⋯ , k. By Theorem 3.3.4 , (3.4.4) If A is invertible, then by Corollary 2.7.1 (i), C = I n By (3.4.4) By (3.4.3) D is tri b |A | = | F1ǁF2 | ⋯ | Fk − 1ǁFkǁ. ut io n |A | = | F1ǁF2 | ⋯ | Fk − 1ǁFkǁC | . , we obtain AB = F 1F 2⋯F k − 1F kB and by Theorem 3.3.4 , fo r |AB | = | F1F2⋯Fk − 1FkB | = | F1ǁF2 | ⋯ | Fk − 1ǁFkǁB | = | AǁB | . , |A| = 0. N ot If A is not invertible, by Corollary 2.7.1 (ii), the last row of C must be a zero row. By (3.4.4) By (2.2.5) , we see that the last row of CB is a zero row and |CB | = 0. It follows that |AB | = | F1F2⋯Fk − 1FkCB | = | F1ǁF2 | ⋯ | Fk − 1ǁFkǁCB | = 0. Hence, |AB | = 0 = | AǁB | . Example 3.4.2. Processing math: 100% Verify that |AB | = | AǁB| if https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/11 8/30/2020 3.4 Properties of determinants A= ( ) 3 1 2 1 and B = ( ) −1 3 5 8 . Solution |AǁB | = (1)( − 23) = − 23. On the other hand, because 2 17 3 14 2 1 −1 3 5 8 = − 23. Hence, |AB | = | AǁB | . Now, we generalize Theorem 2.7.5 = 2 17 3 14 , fo r | | | | AB = ( )( ) ( ) 3 1 D is tri b AB = ut io n Because |A | = 3 − 2 = 1 and |B | = − 8 − 15 = − 23, we have to an n × n matrix. N ot Theorem 3.4.3. | | If A is invertible, then |A | ≠ 0 and A − 1 1 = |A| . Proof Because AA − 1 = I and |I | = 1, it follows from Theorem 3.4.2 Processing math: 100% that |AA | = | AǁA | = | I | = 1. −1 −1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/11 8/30/2020 3.4 Properties of determinants | | This implies that |A | ≠ 0 and A − 1 1 = |A| . Example 3.4.3. Let A be a 3 × 3 matrix. Assume that |A | = 6. Compute | ut io n | T (2A − 1) . By Theorems 3.2.1 , 3.3.2 , and 3.4.3 | (2A − 1) T | , we obtain D is tri b Solution 1 8 4 = 2A − 1 = 2 3 A − 1 = 8 = = . |A| 6 3 | | | | fo r Theorem 3.4.2 shows that the determinant of the product of two n × n matrices equals the product of the determinants of the two matrices. But in general, it is not true for the addition of two n × n matrices, that is, in general, Example 3.4.4. Let A = ( ) 1 2 2 5 and B = ( ) 3 1 1 3 N ot |A + B| ≠ |A| + |B|. . Show |A + B| ≠ |A| + |B|. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/11 8/30/2020 3.4 Properties of determinants Solution Because A + B = ( ) 4 3 3 8 , |A + B| = 23. On the other hand, |A| = 1, |B| = 8 and |A | + | B| = 9. Hence, |A + B| ≠ | A | + | B | . ut io n The following result provides a method to compute |A | + | B|, where A and B are two n × n matrices and their (n − 1) corresponding rows must be the same. Theorem 3.4.4. ( )( ) a 11 a 12 ⋯ a 1n a 21 a 22 ⋯ a 2n ⋱ ⋮ ⋯ a in ⋱ ⋮ ⋯ a nn ⋮ ⋮ a i1 a i2 ⋮ ⋮ Then N ot a n1 a n2 and B = a 11 a 12 ⋯ a 1n a 21 a 22 ⋯ a 2n ⋱ ⋮ ⋯ b in ⋱ ⋮ ⋯ a nn ⋮ ⋮ b i1 b i2 fo r A= D is tri b Let ⋮ ⋮ a n1 a n2 . (3.4.5) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/11 3.4 Properties of determinants | a 12 ⋯ a 1n a 21 a 22 ⋯ a 2n ⋮ ⋮ ⋱ ⋮ ⋯ a in + b in ⋱ ⋮ ⋯ a nn a i1 + b i1 a i2 + b i2 ⋮ ⋮ a n1 a n2 Proof We denote by |C| the right-side determinant of (3.4.5) || A = ∑ nk = 1a ikA ik, n ∑ (a ik + b ik)A ik = ∑ a ikA ik + ∑ b ikA ik = Compute |A | + | B| if k=1 N ot k=1 |A| + |B|. A= k=1 ( ) ( ) 1 7 5 2 0 3 1 4 7 and B = , n fo r |C | = Example 3.4.5. . By Theorem 3.2.1 | B | = ∑nk = 1bikAik and n | . D is tri b |A | + | B | = a 11 ut io n 8/30/2020 1 7 5 2 0 3 . 0 1 −1 Processing math: 100% Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/11 8/30/2020 3.4 Properties of determinants Because the first two rows of A and B are the same, by Theorem 3.4.4 | || || | | 1 7 5 5 1 5 2 0 3 + 2 0 3 = 2 0 3 1 4 7 0 1 −1 1+0 4+1 7−1 1 7 5 2 0 3 = − 28. 1 5 6 | D is tri b = 7 ut io n |A | + | B| = 1 7 , we have Exercises N ot fo r 1. For each of the following matrices, use its determinant to determine whether it is invertible. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/11 3.4 Properties of determinants E = 0 0 2 1 2 0 −2 −3 1 −2 ( ( 1 −1 0 2 1 4 1 −2 4 −2 −1 −4 0 ) B= ) ( D= 1 −1 3 5 0 0 0 0 0 0 −2 0 −1 2 1 3 0 0 2 0 0 −6 6 0 1 ) ( 0 −1 3 5 2 1 −1 2 4 −1 0 0 1 2 7 1 1 −2 3 1 6 10 4 0 −2 0 1 3 −2 1 2 1 −1 −1 ) ut io n C = ( 0 ) F= ( 1 0 0 0 −2 2 0 0 0 6 −6 4 0 2 fo r A = 1 D is tri b 8/30/2020 0 −1 ) N ot 2. For each of the following matrices, find all real numbers x such that it is invertible. A= Processing math: 100% 3. Let A ( x 0 0 0 2 x 2 0 −2 −3 1 −2 1 0 0 1 ) ( ) 2 B= 1 x 4 −2 4 −2 −1 x be a 4 × 4 matrix. Assume that |A | = 2. Compute each of the following determinants. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/11 8/30/2020 3.4 Properties of determinants 1. | − 3A| | | 3. |A | 4. |A | 2. A − 1 T 3 | 6. (2( − A) T) −1 ut io n T | ( ) ( ) 2 1 3 4. Compute |A | + | B| if A = D is tri b | | 5. (2A − 1) 1 0 3 and B = 1 4 7 2 1 3 1 0 3 0 1 −1 ( ) ( ) 0 0 2 1 0 0 2 −1 0 1 2 3 0 0 1 0 1 4 3 and B = N ot 1 A= fo r 5. Compute |A | + | B| if . 0 1 4 3 −2 2 0 1 . −2 2 0 1 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/11 8/30/2020 Chapter 4 Systems of linear equations ut io n Chapter 4 Systems of linear equations 4.1 Systems of linear equations in n variables D is tri b A linear equation with n variables is of the form (4.1.1) a 1x 1 + a 2x 2 + ⋯ + a nx n = b, fo r where a 1, a 2, ⋯, a n, b are constants and x 1, ⋯, x n are variables. (4.1.2) N ot In particular, a linear equation with two variables can be written as the following form ax + by = c, where a, b, c are constants and x,y are variables. If either a or b is not zero, then (4.1.2) a straight line in the xy-plane. is an equation of A linear equation with three variables can be written in the following form Processing math: 100% (4.1.3) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/16 8/30/2020 Chapter 4 Systems of linear equations ax + by = cz = d, where a, b, c, d are constants and x, y, z are variables. If one of a, b, c is not zero, then (4.1.3) 3 3 equation of a plane in ℝ . We shall study more about planes in ℝ in Section 6.3 is an . ut io n Note that in a linear equation, the power of each variable must be 1 and there are no terms containing the product of two or more variables. Hence, in an equation, if there is a variable whose power is not 1 or if there is a term that contains a product of two or more variables, then the equation is not a linear equation. Determine which of the following equations are linear. 1. 0x + 0y = 1; 2. x 2 + y 2 = 1; 3. XY = 2; Solution fo r 4. √2x 1 + x 2 − x 3 = 1. D is tri b Example 4.1.1. N ot (1) and (4) are linear while (2) and (3) are not linear. Let m, n ∈ ℕ. A set of m linear equations in the same n variables is called a system of linear equations or a linear system, which is of the form: (4.1.4) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/16 8/30/2020 Chapter 4 Systems of linear equations ⋮ a m1x 1 + a m2x 2 + ⋯ + a mnx n = b m. ut io n { a 11x 1 + a 12x 2 + ⋯ + a 1nx n = b 1 a 21x 1 + a 22x 2 + ⋯ + a 2nx n = b 2 The matrix a 1n a 21 a 22 ⋯ a 2n ⋱ ⋮ ⋮ ⋮ a m1 a m2 ⋯ a mn ) and the matrix ( a 12 ⋯ a 21 a 22 (A | b) ⋮ ⋮ ⋯ a m1 a m2 ⋯ → Processing math: 100% ⋯ N ot is called the coefficient matrix of (4.1.4) (4.1.6) ( a 12 fo r A= a 11 D is tri b (4.1.5) a 11 ⋱ |) a 1n b 1 a 2n b 2 ⋮ ⋮ a mn b m https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/16 8/30/2020 Chapter 4 Systems of linear equations is called the augmented matrix of (4.1.4) → , where b = (b 1, b 2, ⋯, b m) T. Example 4.1.2. For each of the following systems of linear equations, find its coefficient and augmented matrices. { 2. (2) 3. { ut io n x1 + x2 = 5 x 1 + x 2 + 3x 3 = 0 2x 1 + 4x 2 − 3x 3 = 0 D is tri b { 3x 1 + 6x 2 − 5x 3 = 0 2x 1 + 3x 3 = 2 4x 2 − 3x 3 = 1 x 1 + 2x 2 − x 3 = 0 fo r 1. x1 − x2 = 7 1. A = ( ) 2. A = ( ) 1 −1 1 1 1 1 → and (A | b) = ( |) 1 −1 7 1 3 2 4 −3 3 6 −5 N ot Solution 1 5 ( |) 1 1 → and (A | b) = . 3 0 2 4 −3 0 . 3 6 −5 0 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/16 8/30/2020 Chapter 4 Systems of linear equations 1 2 −1 ( |) It is useful to write (4.1.4) into other expressions. Let 3. A = 3 0 4 −3 2 0 → and (A | b) = 3 2 0 4 −3 1 . 1 2 −1 0 → → X = (x 1, x 2, ⋯, x n) T ∈ ℝ n and b = (b 1, b 2, ⋯, b m) T. , the system (4.1.4) can be written into the matrix form (4.1.7) → → AX = b. → → with b = 0, that is, (4.1.8) → fo r The system (4.1.7) D is tri b By (2.2.2) ut io n ( ) 2 0 → → N ot AX = 0, → is called a homogeneous system. If b ≠ 0, it is called a nonhomogeneous system. → → → Let a 1, a 2, ⋯, a n be the column vectors of A given in (4.1.5) . By (2.2.3) , (4.1.4) can be written into the linear combination form: (4.1.9) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/16 8/30/2020 Chapter 4 Systems of linear equations → → → → x 1a 1 + x 2a 2 + ⋯ + x na n = b. Example 4.1.3. Consider the system of linear equations x 1 + x 2 + 3x 3 = 2 2x 1 + 4x 2 − 3x 3 = 1 3x 1 + 6x 2 + 5x 3 = 0 Write the above system into the forms (4.1.7) and (4.1.9) Let ( ) Then the system (4.1.10) 3 ( ) () x1 2 x 2 , and → b= 1 . 0 N ot A= . fo r Solution 1 1 D is tri b { ut io n (4.1.10) → 2 4 −3 , X = 3 6 5 → x3 → can be rewritten as AX = b. Let Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/16 8/30/2020 Chapter 4 Systems of linear equations Then the system (4.1.10) () () ( ) () 1 → 2 , a = 2 3 1 → 4 , a = 3 6 3 2 → − 3 , and b = 5 1 . 0 → → → → can be rewritten as x 1a 1 + x 2a 2 + x 3a 3 = b. Now, we introduce the definition of a solution for the system (4.1.4) . Definition 4.1.1. (4.1.11) a 11s 1 + a 12s 2 + ⋯ + a 1ns n = b 1 fo r a 21s 1 + a 22s 2 + ⋯ + a 2ns n = b 2 ⋮ a m1s 1 + a m2s 2 + ⋯ + a mns n = b m, N ot { , that is, D is tri b If (s 1, s 2, ⋯, s n) satisfies each of the linear equations in (4.1.4) ut io n → a1 = then (s 1, s 2, ⋯, s n) is called a solution of (4.1.4) all solutions of the system. . Solving a system of linear equations means finding Note that any solution (s 1, s 2, ⋯, s n) of (4.1.4) must satisfy each equation of (4.1.4) . If (s 1, s 2, ⋯, s n) does not satisfy one of the equations of (4.1.4) , it is not a solution of (4.1.4) Processing math: 100% Also, note that if we consider (4.1.4) . , (s 1, s 2, ⋯, s n) can be treatedasa point in ℝ n or as a vector in ℝ n. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/16 8/30/2020 Chapter 4 Systems of linear equations Example 4.1.4. For each of the following points: P 1(6, − 1), P 2(8, 1), P 3(2, 3), P 4(0, 0), { ut io n verify whether it is a solution of the system of linear equations x − y = 7, x + y = 5. D is tri b Solution (6, − 1) satisfies both equations, so P 1(6, − 1) is a solution. (8, 1) satisfies the first equation but does not satisfy the second one, so P 2(8, 1) is not a solution. (2,3) satisfies the second equation but does not satisfy the the first one, so P 3(2, 3) is not a solution. (0,0) satisfies neither of the two equations, so P 4(0, 0) is not a solution. fo r Example 4.1.5. (4.1.12) N ot Consider the following system of linear equations. { x 1 + x 2 + 3x 3 = 2 x1 − x2 + x3 = − 3 2x 1 + 4x 3 = − 1 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/16 8/30/2020 Chapter 4 Systems of linear equations 5 3 Verify that ( − 2 , 2 , 1) is a solution of (4.1.12) while (1, 3, − 1) is not a solution. Solution 3 3 ( 5 3 3 5 3 x 1 − x 2 + x 3 = − 2 − 2 + 1 = − 3, 5 D is tri b { 5 x 1 + x 2 + 3x 3 = − 2 + 2 + 3(1) = 2, ut io n Let x 1 = 2 , x 2 = 2 , and x 3 = 1. Then 2x 1 + 4x 3 = 2( − 2 ) + 4(1) = − 1. ) and is a solution of (4.1.12) . fo r Hence, − 2 , 2 , 1 satisfies each of the three equations in (4.1.12) Let x 1 = 1, x 2 = 3, and x 3 = − 1. Then N ot x 1 + x 2 + 3x 3 = 1 + 3 + 3( − 1) = 1 ≠ 2 Hence,(1, 3, − 1) does not satisfy the first equation of (4.1.12) (4.1.12) . , and thus it is not a solution of Note that (x 1, x 2, x 3) = (1, 3, − 1) satisfies the second equation but not the third one of (4.1.12) . → By Definition 4.1.1 , we see that the zero vector 0 in ℝ n is always a solution of the homogeneous system (4.1.8) . Sometimes, the zero solution of (4.1.8) is called a trivial solution. A nonzero solution of Processing math: 100% (4.1.8) is called a nontrivial solution. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/16 8/30/2020 Chapter 4 Systems of linear equations We denote by N A the set of all solutions of (4.1.8) , that is, (4.1.13) {X ∈ ℝ : AX = 0 }. → n → → Definition 4.1.2. The set N A is called the solution space of the homogeneous system (4.1.8) ut io n NA = or the nullspace of the D is tri b matrix A. The nullity of A is called the dimension of the solution space N A, denoted by dim (N A), that is, (4.1.14) dim(N A) = null(A). The linear system (4.1.4) it has no solutions. By Definition 4.1.3 is said to be consistent if it has at least one solution, and to be inconsistent if N ot Definition 4.1.3. fo r A solution space that only contains the zero vector is said to be a zero space. , a homogeneous system is consistent because it has at least a trivial solution. By (4.1.9) and Theorem 1.4.3 , we obtain the following relation among the consistency of a linear system, the linear combination, and the spanning space. Theorem 4.1.1. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/16 8/30/2020 Chapter 4 Systems of linear equations Let S = { → → a 1, ⋯, a n } be the same as in (1.4.2) ,A= { → → a 1, ⋯, a n } be the same as in (4.1.5) , and → b = (b 1, b 2, ⋯, b m) T. Then the following assertions are equivalent. → → 1. AX = b is consistent. → ut io n 2. b is a linear combination of S. → 3. b ∈ span S. Proof → (4.1.9) → → , there exists X = (x 1, x 2, ⋯, x n) such that AX = b. By (2.2.3) , → holds and thus, the result (2) holds. Conversely, if (2) holds, then there exist X = (x 1, x 2, ⋯, x n) → → D is tri b If (1) holds, then by Definition 4.1.3 such that (4.1.9) holds. By (2.2.3) , AX = b and the result (1) holds. This shows that (1) and (2) are equivalent. By Theorem 1.4.3 , the results (2) and (3) are equivalent. By Theorem 4.1.1 , we see that the system (4.1.4) → is inconsistent if and only if b is not a linear fo r → combination of S and if and only if b ∉ span S. Example 4.1.6. Solve the following system. N ot For a system of two linear equations, it is not hard to solve it. Here we give examples to show how to solve such systems and provide geometric meanings that exhibit the solution structures. (4.1.15) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/16 8/30/2020 Chapter 4 Systems of linear equations { x 1 − x 2 = 7, x 1 + x 2 = 5. Solution x 2 = x 1 − 7 = 6 − 7 = − 1, so (x 1, x 2) = (6, − 1) is a solution of the system. ut io n Adding the two equations implies 2x 1 = 12 and x 1 = 6. Substituting x 1 = 6 into the first equation implies Example 4.1.7. Solve the following system 2x 1 + 2x 2 = 2 fo r { x 1 + x 2 = 1, D is tri b In geometric terms, each of the equations in (4.1.15) represents a line and the solution (6, − 1) is the intersection point of the two lines. In fact, the solution (6, − 1) is the unique solution of the system. Solution N ot and express the solution with a linear combination. The two equations are the same and the solutions of the first equation are the solutions of the system. Let x 2 = t ∈ ℝ. Then x 1 = 1 − t and ( ) () ( ) x1 Processing math: 100% x2 = 1 0 +t −1 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/16 8/30/2020 Chapter 4 Systems of linear equations is a solution of the system for each t ∈ ℝ. In geometry, the two equations represent the same line and every point on the line is a solution of the system. Hence, the system has infinitely many solutions. Example 4.1.8. x 1 + x 2 = 1, x 1 + x 2 = 2. D is tri b { ut io n Solve the following system Solution It is obvious that any (x 1, x 2) does not satisfy both equations. Hence, the above system has no solutions. fo r In geometric terms, the two equations represent two parallel lines. Hence, the above system has no solutions. N ot The above three examples show that a system of two linear equations with two variables has only one solution, or infinitely many solutions, or no solutions. The same fact holds for a system of linear equations of two or more variables. We shall study the general case in Section 4.5 . The methods used to solve a system of two linear equations are not suitable for a general system (4.1.4) In Sections 4.2–4.4, we shall study approaches to solve (4.1.4) or (4.1.7) , where we do not often mention the linear combinations and spanning spaces. In Sections 4.5 and 4.6, following Theorem 4.1.1 → → → . , → we shall study how to determine whether AX = b is consistent, b is a linear combination of S, and b ∈ span S. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/16 8/30/2020 Chapter 4 Systems of linear equations Exercises 1. Determine which of the following equations are linear. a. x + 2y + z = 6 b. 2x − 6y + z = 1 2 ut io n c. − √2x + 6 − 3 y = 4 − 3z d. 3x 1 + 2x 2 + 4x 3 + 5x 4 = 1 e. 2xy + 3yz + 5z = 8 3. { { 4x 1 + x 2 = 3 − x 1 + 2x 2 + 3x 3 = 4 3x 1 + 2x 2 − 3x 3 = 5 2x 1 + 3x 2 − x 3 = 1 x 1 − 3x 3 − 4 = 0 2x 2 − 5x 3 − 8 = 0 3x 1 + 2x 2 − x 3 = 4 fo r 2. { 2x 1 − x 2 = 6 N ot 1. D is tri b 2. For each of the following systems of linear equations, find its coefficient and augmented matrices. 3. For each of the following augmented matrices, find the corresponding linear system. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/16 8/30/2020 Chapter 4 Systems of linear equations ( | )( | ) 1 2 3 −1 1 0 −2 0 −1 1 1 0 2 1 1 0 −1 −2 0 3 −1 → → → c. { 2x 1 + 6x 2 − 5x 3 = 2 − 3x 1 + 7x 2 − 5x 3 = − 1 D is tri b b. { { − x 1 + x 2 + 2x 3 = 3 x 1 − x 2 + 4x 3 = 0 − 2x 1 + 4x 2 − 3x 3 = 0 3x 1 + 6x 2 − 8x 3 = 0, x 1 − 2x 2 + x 3 = 0 fo r a. ut io n 4. For each of the following linear systems, write it into the form AX = b and express b as a linear combination of the column vectors of A. − x 1 + 3x 2 − 4x 3 = 0 ) N ot ( 1 5. For each of the following points: P 1 6, − 2 , P 2(8, 1), P 3(2, 3), and P 4(0, 0), verify whether it is a solution of the system. { x − 2y = 7, x + 2y = 5. Processing math: 100% 6. Consider the system of linear equations https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/16 8/30/2020 Chapter 4 Systems of linear equations { − x 1 + x 2 + 2x 3 = 3 2x 1 + 6x 2 − 5x 3 = 2 − 3x 1 + 7x 2 − 5x 3 = − 1 2. Verify that ( 5 7 4 , , 6 6 3 ) ut io n 1. Verify that (1, 3, − 1) is a solution of the system. is a solution of the system. 3. Determine if the vector b = (3, 2, − 1) T is a linear combination of the column vectors of the coefficient matrix of the system. D is tri b → → 4. Determine if the vector b = (3, 2, − 1) T belongs to the spanning space of the column vectors of the coefficient matrix of the system. 7. Solve each of the following systems. 2. 3. x1 + x2 = 5 x 1 + 2x 2 = 1 2x 1 + x 2 = 2 x 1 + 2x 2 = 1 x 1 + 2x 2 = 2 fo r { { { N ot 1. x1 − x2 = 6 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/16 8/30/2020 4.2 Gaussian Elimination 4.2 Gaussian Elimination (4.2.1) → → AX = b, D is tri b where the coefficient matrix A is assumed to be a row echelon matrix. : ut io n Before we study Gaussian elimination, we study how to solve (4.1.7) Recall that for each i ∈ I n, the ith column of the coefficients matrix A is from the coefficients of the variable x i in the system. Because A is a row echelon matrix, some columns of A would contain leading entries. A variable x i is said to be a basic variable if the ith column of A contains a leading entry, and to be a free fo r variable if the ith column of A does not contain a leading entry. Example 4.2.1. N ot The method to solve such a system (4.2.1) is to treat the free variables, if any, as parameters, and then solve each linear equation for the basic variable starting from the last equation to the first one, where the parameters and the obtained values of the basic variables are substituted. The method used to solve such a system is called back-substitution. Consider the system of the linear equations (1) Processing math: 78% x 1 + 2x 2 − 3x 3 = 4 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/17 8/30/2020 4.2 Gaussian Elimination (2) 2x 2 − 6x 3 = 5 − x3 = 2 → D is tri b a. Find the coefficient matrix and the augmented matrix of the system. b. Find the basic variables and free variables of the system. c. Solve the system using back-substitution. ut io n (3) d. Determine if the vector b = (4, 5, 2) T is a linear combination of the column vectors of the coefficient matrix of the system. Solution ( ) ( |) 1 2 −3 0 2 − 6 , (A | → b) = 0 0 −1 N ot A= fo r a. The coefficient and the augmented matrices are 1 2 −3 4 0 2 −6 5 . 0 0 −1 2 b. A is a 3 × 3 row echelon matrix. The columns 1,2 3 of A contain leading entries and thus, x 1, x 2, and x 3 are basic variables. There are no free variables. c. To solve the system, we first consider equation (3). Solving equation (3) for the basic variable x 3, we get x 3 = − 2. Substituting x 3 = − 2 into equation (2) and solving equation (2) for the basic Processing math: 78% variable x 2, we obtain https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/17 8/30/2020 4.2 Gaussian Elimination x2 = 1 2 (5 + 6x 3) = 1 2 [5 + 6( − 2)] = − 7 2 . 7 x 1 = 4 − 2x 2 + 3x 3 = 4 − 2( − ( ) 7 7 2 ) + 3( − 2) = 5. D is tri b Hence, (x 1, x 2, x 3) = 5, − 2 , − 2 is a solution of the system. ut io n Substituting x 2 = − 2 and x 3 = − 2 into equation (1) and solving equation (1) for the basic variable x 1, we obtain d. Because the system has a solution, the vector b = (4, 5, 2) T is a linear combination of the column vectors of A, namely, → → 7→ → b = 5a 1 − a 2 − 2a 3, 2 → → 0 , a = 2 0 2 → 2 , and a = 3 0 () fo r () () 1 −3 −6 . −1 N ot → where a 1 = → Example 4.2.1 shows that if r(A) = r(A | b) = n, where n = 3 is the number of variables, then the system has a unique solution. Example 4.2.2. Consider the linear system Processing math: 78% (1) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/17 8/30/2020 4.2 Gaussian Elimination x 1 − x 2 − 3x 3 + x 4 = 1 (2) ut io n − x2 − x3 + x4 = 2 (3) − x4 = 3 D is tri b a. Find the coefficient matrix and the augmented matrix of the system. b. Find the basic variables and free variables of the system. c. Solve the system using back-substitution and express its solution as a linear combination. → d. Determine if the vector b = (1, 2, 3) T is a linear combination of the column vectors of the coefficient matrix of the system. fo r Solution a. The coefficient and the augmented matrices are ( 0 −1 −1 0 ) ( N ot A= 1 −1 −3 0 0 1 1 −1 → , (A | b) = 1 −1 −3 1 0 −1 −1 1 0 0 0 |) 1 2 . −1 3 b. A is a row echelon matrix. The columns 1,2,4 of A contain leading entries and thus, x 1, x 2, and x 4 are basic variables. The third column of A does not contain any leading entries, so x 3 is a free Processing math: 78% variable. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/17 8/30/2020 4.2 Gaussian Elimination c. Because x 3 is a free variable, let x 3 = t, where t ∈ ℝ is an arbitrary real number. Solving equation (3) for the basic variable x 4 implies x 4 = − 3. Substituting x 4 = − 3 into equation (2) and solving equation (2) for the basic variable x 2, we obtain x 2 = − 2 − x 3 + x 4 = − 2 − t + ( − 3) = − 5 − t. ut io n Substituting x 2 = − 5 − t, x 3 = t and x 4 = − 3 into equation (1) and solving equation (1) for the basic variable x 1, we obtain x 1 = 1 + x 2 + 3x 3 − x 4 = 1 + ( − 5 − t) + 3t − ( − 3) = − 1 + 2t. D is tri b Hence, (x 1, x 2, x 3, x 4) = ( − 1 + 2t, − 5 − t, t, − 3) is a solution of the system for each t ∈ ℝ. We can write the solution as a linear combination: ()( )()()()() x1 x3 −5 − t 0+t − 3 + 0t = −5 0 2t + −3 N ot x4 = −1 −t t −1 = fo r x2 − 1 + 2t 0t −5 0 −3 2 +t −1 1 . 0 → d. Because the system has infinitely many solutions, b = (1, 2, 3) T is a linear combination of the column vectors of A. → The above example shows that if r(A) = r(A | b) = 3 < n, where n = 4 is the number of variables, then the system has infinitely many solutions. Example 4.2.3. Processing math: 78% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/17 8/30/2020 4.2 Gaussian Elimination Consider the following system x 1 − x 2 − 3x 3 + x 4 − x 5 = 2 (1) − x 3 + x 4 = 3 (2) ut io n − x 4 + x 5 = 1 (3) a. Find the coefficient matrix and the augmented matrix of the system. b. Find the basic variables and free variables of the system. c. Solve the system using back-substitution and express its solutions as a linear combination. → D is tri b d. Determine if the vector b = (2, 3, 1) T is a linear combination of the column vectors of the coefficient matrix of the system. Solution a. The coefficient and the augmented matrices are −1 0 0 −1 1 0 0 0 0 −1 1 fo r ( ( 1 N ot A = 1 −1 −3 → (A | b) = 1 −1 −3 1 0 0 −1 1 0 0 0 0 −1 1 ) |) , −1 2 3 . 1 b. A is a 3 × 5 row echelon matrix. The columns 1,3,4 of A contain leading entries and thus, x 1, x 3, and x 4 are basic variables. The columns 2,5 of A do not contain any leading entries, so x 2 and x 5 Processing math: 78% are free variables. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/17 8/30/2020 4.2 Gaussian Elimination c. Because x 2 and x 5 are free variables, let x 2 = s and x 5 = t, where s, t ∈ ℝ. Substituting x 5 = t into equation (3) and solving equation (3) for the basic variable x 4, we obtain x 4 = − 1 + x 5 = − 1 + t. Substituting x 4 = − 1 + t into equation (2) and solving equation (2) for the basic variable x 3, we obtain ut io n x 3 = − 3 + x 4 = − 3 + ( − 1 + t) = − 4 + t. Substituting x 2 = s, x 3 = − 4 + t, x 4 = − 1 + t and x 5 = t into equation (1) and solving equation (1) for the basic variable x 1, we obtain Hence, D is tri b x 1 = 2 + x 2 + 3x 3 − x 4 + x 5 = 2 + s + 3( − 4 + t) − ( − 1 + t) + t = − 9 + s + 3t. (x 1, x 2, x 3, x 4, x 5) = ( − 9 + s + 3t, s, − 4 + t, − 1 + t, t) N ot fo r is a solution of the system for each s, t ∈ ℝ. We write the solution as a linear combination: Processing math: 78% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/17 4.2 Gaussian Elimination () ( )( ) x1 − 9 + s + 3t − 9 + s + 3t x2 s 0 + s + 0t x3 −4 + t = − 4 + 0s + t = x4 −1 + t − 1 + 0s + t x5 t 0 + 0s + t ( ) () ( ) = 0 −4 −1 s + 0t 0 + 0 t t 0 t ( ) () () 1 3 0 1 0 fo r −9 −4 +s 0 0 N ot = 3t D is tri b s −9 ut io n 8/30/2020 −1 0 0 +t 1 . 1 1 d. Because the system has infinitely many solutions, b = (2, 3, 1) T is a linear combination of the column vectors of A. → → Example 4.2.3 also shows that if r(A) = r(A | b) = 3 < n, where n = 5 is the number of variables, then the Processing math: 78% system has infinitely many solutions. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/17 8/30/2020 4.2 Gaussian Elimination Example 4.2.4. Consider each of the following linear systems { 0x 3 = 1 x 1 − x 2 − 3x 3 = 1 − x2 + x3 = 3 D is tri b ii. − x2 + x3 = 3 ut io n { i. x 1 − x 2 − 3x 3 = 1 2x 3 = − 1 0x 3 = 5 Solution i. N ot fo r a. Find the coefficient matrix and the augmented matrix of the system. b. Find the basic variables and free variables of the system. c. Solve the system. a. The coefficient and the augmented matrices are ( ) ( |) 1 −1 −3 A= Processing math: 78% 0 −1 1 0 0 0 1 −1 −3 1 → , (A | b) = 0 −1 1 0 0 0 3 . 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/17 8/30/2020 4.2 Gaussian Elimination b. A is a 3 × 3 row echelon matrix. The columns 1,2 of A contain leading entries and thus, x 1 and x 2 are basic variables. The column 3 of A does not contain any leading entries, so x 3 is ii. a. The coefficient and the augmented matrices are ( ) ( |) 0 −1 1 0 0 2 0 0 0 1 −1 −3 1 D is tri b 1 −1 −3 A= ut io n a free variable. c. Because the last equation implies 0 = 1, which shows that any (x 1, x 2, x 3) doesn’t satisfy the equation. Hence, the system has no solutions. → , (A | b) = 0 −1 1 3 0 0 2 −1 0 0 0 . 5 fo r b. A is a 4 × 3 row echelon matrix. The columns 1,2,3 of A contain leading entries and thus, x 1, x 2, and x 3 are basic variables. There are no free variables. Note that in Example 4.2.4 N ot c. Because the last equation implies 0 = 5, which shows that any (x 1, x 2, x 3) doesn’t satisfy the equation. Hence, the system has no solutions. → (i), r(A) = 2 < 3 = n, where n is the number of variables, and r(A | b) = 3. → Example 4.2.4 (i) shows that if r(A) < r(A | b), then the system has no solutions, and even when a system contains free variables, the system may have no solutions. In Example 4.2.4 (ii), r(A) = 3 = n, where n is → → the number of variables and r(A | b) = 4. Example 4.2.4 (ii) also shows that if r(A) < r(A | b), then the system has no solutions, and even when r(A) = n, the system may have no solutions. → → Now, we study how to solve the general system AX = b given in (4.1.7) Processing math: 78% matrix. , where A is not a row echelon https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/17 8/30/2020 4.2 Gaussian Elimination Let B be a row echelon matrix of A. By Theorems 2.6.2 such that B B = EA. By (4.1.7) , we obtain and 2.7.3 , there exists an invertible matrix E (4.2.2) → → → ut io n BX = (EA)X = E b : = → c. Theorem 4.2.1. and (4.2.2) are equivalent. Proof → We have shown above that if X is a solution of (4.1.7) D is tri b (4.1.7) , then it is a solution of (4.2.2) → noting that E is invertible, we obtain that if X is a solution of (4.2.2) Hence, (4.1.7) and (4.2.2) are equivalent. → given in (4.1.6) → , then it is a solution of (4.1.7) . Note that (4.2.2) . can be expressed as → fo r Let (A | b) be the augmented matrix of (4.1.4) . Conversely, E(A | b) = (EA | E b) = (B | → c). → → Example 4.2.5. N ot Hence, to solve (4.1.7) , we first use row operations to change (A | b) to (B | → c) and then to solve BX = → c, where B is a row echelon matrix, by using back-substitution. By Theorem 4.2.1 , the solutions of (4.2.2) are the solutions of (4.1.7) . This method is called Gaussian elimination. Solve each of the following systems by using Gaussian elimination. Processing math: 78% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/17 8/30/2020 4.2 Gaussian Elimination d. { 3x 1 + 5x 2 + x 3 = − 8 − 4x 1 − 7x 2 − 2x 3 = 13 x 1 + 2x 2 − x 3 + 3x 4 = 5 − 2x 1 + 3x 2 − 5x 3 + x 4 = 4 x 1 + 2x 2 + 3x 3 = 4 4x 1 + 7x 2 + 6x 3 = 17 2x 1 + 5x 2 + 12x 3 = 10 Solution a. ut io n { 2x 1 + 2x 2 − 2x 3 = 4 D is tri b c. 3x 1 + x 2 − 2x 3 = 4 fo r b. { { 4x 1 + 5x 2 + 6x 3 = 24 N ot a. 2x 1 + 4x 2 + 6x 3 = 18 Processing math: 78% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/17 4.2 Gaussian Elimination ( |) ( ( ( 2 4 1 4 5 ( | )↓ | ) | )↓ |) 6 24 R 1( )→ 2 3 1 −2 4 R1 ( − 4 ) + R2 → R1 ( − 3 ) + R3 1 R 2( − 3 )→ 2 3 4 5 6 24 3 1 −2 4 3 0 −3 9 − 6 − 12 0 − 5 − 11 − 23 1 2 0 1 9 3 9 2 4 0 − 5 − 11 − 23 1 2 3 0 1 2 9 4 = (B | → c). 0 0 −1 −3 fo r R 2(5) + R 3→ 1 1 2 D is tri b → (A | b) = 6 18 ut io n 8/30/2020 N ot The system of linear equations corresponding to (B | → c) is (1) x 1 + 2x 2 + 3x 3 = 9 (2) x 2 + 2x 3 = 4 Processing math: 78% (3) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/17 8/30/2020 4.2 Gaussian Elimination − x3 = − 3 Now, we use back-substitution to solve the above system. The variables x 1, x 2, and x 2 are basic variables. Solving equation (3) for basic variable x 3 implies x 3 = 3. Substituting x 3 = 3 into equation (2) and solving equation (2) for the basic variable x 2, we obtain x 2 = 4 − 2x 3 = 4 − 2(3) = − 2. x 1, we obtain D is tri b x 1 = 9 − 2x 2 − 3x 3 = 9 − 2( − 2) − 3(3) = 4. ut io n Substituting x 2 = − 2 and x 3 = 3 into equation (1) and solving equation (1) for the basic variable Hence (x 1, x 2, x 3) = (4, − 2, 3) is a solution of the system. ( | ) ( | )↓ ( | ) ( | )↓ ( |) 3 2 −2 4 1 5 1 − 8 R 1( )→ 2 − 4 − 7 − 2 13 Processing math: 78% 1 3 5 −1 2 1 −8 − 4 − 7 − 2 13 1 0 1 1 −1 2 2 −7 1 1 −1 2 N ot 1 1 −1 2 b. R 1 ( − 3 ) + R 2 1 0 2 4 − 14 R 2( )→ → 2 R1 ( 4 ) + R3 0 − 3 − 6 21 R 2(3) + R 3→ 1 fo r → (A | b) = 2 0 1 2 −7 = (B | → c). 0 0 0 0 − 3 − 6 21 0 The system of linear equations corresponding to (B | → c) is (1) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/17 8/30/2020 4.2 Gaussian Elimination x1 + x2 − x3 = 2 (2) x 2 + 2x 3 = − 7. ut io n The variables x 1 and x 2 are the basic variables and x 3 is a free variable. Let x 3 = t, where t ∈ ℝ. Substituting x 3 = t into equation (2) and solving equation (2) for basic variable x 2, we have x 2 = − 7 − 2x 3 = − 7 − 2t. Substituting x2=−7−2t and x3=t into equation (1) and solving equation D is tri b (1) for the basic variable x1, we obtain x1=2−x2+x3=2−(−7−2t)+t=9+3t. We write the solution as a linear combination: (x1x2x3)=(9+3t−7−2tt)=(9−70)+t(3−21). fo r c. (A|b→)=(12−13−23−51|54)↓R1(2)+R2→(12−1307−77|514)R2(17)→(12−1301−11|52)=(B|c→). The system of linear equations corresponding to (B|c→) is (1) (2) N ot x1+2x2−x3+3x4=5 x2−x3+x4=2. The variables x1 and x2 are the basic variables and x3 and x4 are free variables. Let x3=s and x4=t, where s, t∈ℝ. Substituting x3=s and x4=t into equation (2) and solving equation (2) for the basic variable x2, we obtain x2=2+x3−x4=2+s−t. Substituting x2=2+s−t, x3=s and x4=t into equation (1) and solving equation (1) for the basic variable x1, we obtain Processing math: 78% x1=5−2x2+x3−3x4=5−2(2+s−t)+s−3t=1−s−t. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/17 8/30/2020 4.2 Gaussian Elimination We write the above solution as a linear combination: (x1x2x3x4)=(1−s−t2+s−tst)=(1200)+s(−1110)+(−1−101). x1+2x2+3x3=4 D is tri b (2) ut io n d. (A|b→)= (1234762512|41710)↓→R1(−2)+R3R1(−4)+R2(1230−1−6016|412)↓R2(1)+R3→(1230−1−6000|413)= (B|c→). The system of linear equations corresponding to (B|c→) is (1) −x2−6x3=1 (3) 0x1+0x2+0x3=3. N ot Exercises fo r The last equation (3) reads 0=3; this is impossible. Thus the original system has no solutions. 1. Consider each of the following systems. i. { x1+x2−3x3=2−x2−6x3=4−x3=1 ii. { x1−2x2+3x3+x4=−1x2−x3+x4=4−x4=2 iii. { −x1−2x2−2x3+x4−3x5=4−2x3+3x4=1−4x4+2x5=5 iv. { −3x1−2x2−2x3=2x2−x3=20x3=4 Processing math: 78% a. Find the coefficient and augmented matrices of the system. b. Find the basic variables and free variables of the system. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/17 8/30/2020 4.2 Gaussian Elimination c. Solve the system by using back-substitution if it is consistent, and express its solution as a linear combination. d. Determine if the vector b→ in the right side of the system is a linear combination of the column vectors of the coefficient matrix of the system. N ot fo r D is tri b ut io n 2. Solve each of the following systems by using Gaussian elimination and express its solutions as linear combinations if they have infinitely many solutions. a. { x1−2x2+3x3=62x1+x2+4x3=5−3x1+x2−2x3=3 b. { x1+x2−2x3=32x1+3x2−x3=−4−2x1−3x2−x3=4 c. { x1+2x2+3x3=44x1+7x2+6x3=172x1+5x2+12x3=7 d. { x1+2x2−x3+3x4=53x1−x2+4x3+2x4=1−x1+5x2−6x3+4x4=9 e. { x1−2x2=32x1−x2=0−x1+4x2=−1 f. { −x1+2x2−3x3=−42x1−x2+2x3=2x1+x2−x3=2 Processing math: 78% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/17 8/30/2020 4.3 Gauss-Jordan Elimination 4.3 Gauss-Jordan Elimination → → → → → D is tri bu tio n When we use Gaussian elimination to solve the system AX = b, the augmented matrix (A | b) is changed only to a row echelon matrix. In fact, we can continue changing the row echelon matrix to a reduced row echelon matrix → (C | d) and then solve the system CX = d, where C is a reduced row echelon matrix. This method is called → → Gauss-Jordan elimination. The advantage of using Gauss-Jordan elimination to solve AX = b is that when we → → solve the system CX = d, each equation in the system only contains one basic variable and can be easily solved. By Theorem 4.2.1 → → → → → → → , we see that AX = b and CX = d are equivalent, that is, the solutions of CX = d are the → solutions of AX = b. Example 4.3.1. b. { { 4x 1 + 5x 2 + 6x 3 = 24 3x 1 + x 2 − 2x 3 = 4 x1 + x2 − x3 = 1 − 3x 1 + 5x 2 − x 3 = − 1 N ot a. 2x 1 + 4x 2 + 6x 3 = 18 fo r Solve each of the following systems by using Gauss-Jordan elimination. 3x 1 + 7x 2 − 5x 3 = 4 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/20 8/30/2020 4.3 Gauss-Jordan Elimination d. { { x 1 + x 3 − 2x 4 = 2 − 3x 1 + 5x 2 − 5x 4 = − 5 x 1 + x 2 − 2x 3 = 1 − 2x 1 − x 2 + x 3 = − 2 Solution ( | ) ( | )↓ ( | ) ( ) ( | )↓ ( | ) ( | )↑ ( | )↑ ( | ) 2 4 (A | b) = 1 2 1 4 5 6 24 R 1( )→ 2 3 1 −2 4 3 9 1 − 6 − 12 R 2 − → 3 0 − 5 − 11 − 23 a. 3 9 4 5 6 24 1 2 3 9 0 1 2 1 2 3 9 0 1 2 4 0 1 2 4 0 0 1 3 1 2 0 0 0 1 0 −2 0 0 1 3 4 R 2(5) + R 3→ 0 − 5 − 11 − 23 3 9 0 0 −1 −3 → R1 ( − 3 ) + R3 3 1 −2 4 1 2 R 3( − 1)→ R1 ( − 4 ) + R2 N ot 0 −3 1 2 fo r → 6 18 D is tri bu tio n c. 2x 2 + 3x 3 + x 4 = 4 R3 ( − 2 ) + R2 → R3 ( − 3 ) + R1 1 0 0 4 R 2( − 2) + R 1→ 0 1 0 −2 . 0 0 1 3 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/20 8/30/2020 4.3 Gauss-Jordan Elimination From the last augmented matrix, we obtain x 1 = 4, x 2 = − 2, and x 3 = 3. Hence, (x 1, x 2, x 3) = (4, − 2, 3) is a solution of the system. (A | b) = 1 −1 1 −3 5 −1 −1 3 7 −5 4 R1 ( 3 ) + R2 → R1 ( − 3 ) + R3 1 1 −1 1 R 2( 2 )→ b. 1 R 2( 4 )→ ( 0 4 −2 1 R 2( − 1) + R 3→ 0 0 ↑| ) ( | ) 1 0 1 −2 4 0 0 4 −2 1 0 0 1 1 1 −1 1 0 0 0 4 −2 1 1 1 −1 1 0 4 −2 1 1 0 8 −4 2 0 3 1 0 −2 4 R 2( − 1) + R 1→ 1 1 0 1 −2 4 . fo r 1 ( |) ( |) 1 1 −1 1 D is tri bu tio n ( | )↓ ( | )↓ 1 → 0 0 0 0 N ot The system of linear equations corresponding to the last augmented matrix is { 1 3 1 1 x1 − 2 x3 = 4 x2 − 2 x3 = 4 , where x 1, x 2 are basic variables and x 3 is a free variable. Let x 3 = t ∈ ℝ. Then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/20 8/30/2020 4.3 Gauss-Jordan Elimination 3 1 1 1 (x 1, x 2, x 3) = ( + t, + t, t) 4 2 4 2 is a solution of the system. ( 0 2 3 1 −2 2 0 2 3 1 4 0 1 − 3 − 13 − 7 1 0 R 2( − 2) + R 3→ ( 1 1 0 2 3 1 −2 2 R 2 , 3→ 1 1 −2 2 2 0 1 − 3 − 13 − 7 0 2 3 1 −2 5 1 −2 5 3 1 1 0 4 −3 5 0 −5 −5 0 1 − 3 13 − 3 R ( )→ 3 9 0 0 9 27 18 0 1 − 3 13 − 3 0 0 0 1 −2 2 4 R ( − 2) + R → 2 3 0 5 3 − 11 1 1 0 1 0 R 1 , 2→ 1 D is tri bu tio n 0 1 −2 2 1 0 1 c. 4 −3 5 0 −5 −5 R 3(3) + R 3→ ↓ |) |) ( ( |) | ) ( | )↓ ( |) | )↑ ( | ) 1 4 fo r ( 1 2 3 N ot → (A | b) = 0 R3 ( 3 ) + R2 → R3 ( − 1 ) + R1 1 0 0 −5 3 0 1 0 22 3 . 0 0 1 3 2 The system of linear equations corresponding to the last matrix is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/20 8/30/2020 4.3 Gauss-Jordan Elimination { x 1 − 5x 4 = 3 x 2 − 22x 4 = 3 x 3 + 3x 4 = 2, D is tri bu tio n where x 1, x 2, and x 3 are basic variables and x 3 is a free variable. Let x 4 = t ∈ ℝ. Then (x 1, x 2, x 3, x 4) = (3 + 5t, 3 + 22t, 2 − 3t, t) is a solution of the system. → (A | b) = d. ( 1 1 −2 −1 R 2( − 1) + R 1→ ( | )↓ −2 1 1 −2 1 0 |) 1 1 0 1 −3 0 R 1(2) + R 2→ ( | )↑ 1 1 −2 1 0 1 −3 0 = (B | → c). fo r The system of linear equations corresponding to (B | → c) is x 2 − 3x 3 = 0 N ot { x1 + x3 = 1 where x 1 and x 2 are basic variables and x 3 is a free variable. Let x 3 = t. Then x 1 = 1 − x 3 = 1 − t and x 2 = 3x 3 = 3t. Hence, (4.3.1) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/20 8/30/2020 4.3 Gauss-Jordan Elimination ( ) ( ) () ( ) () ( ) x1 x2 1−t 3t = t x3 1 = 0 −t + 1 3t 0 = t −1 0 +t 0 3 . 1 D is tri bu tio n Remark 4.3.1. The system a) in Example 4.3.1 is the same as the system (a) in Example (4.2.5) and we have used two methods to solve the system. The solution (4.3.1) of the system d) in Example 4.3.1 is the parametric equation of the line of intersections of the two planes x 1 + x 2 − 2x 3 = 1 and − 2x 1 − x 2 + x 3 = − 2. We shall study the lines and planes in more detail in Sections 6.2 For a homogeneous system (4.1.8) and 6.3 . , we can use Gauss-Jordan elimination to solve it. Assume that we use row → → fo r operations to change the augmented matrix (A | 0) to the reduced row echelon matrix denoted by (B | 0) from which we see that there are null(A) free variables. Theorem 4.3.1. N ot The following result gives the expression of the solution space of (4.1.8) Let k = null(A). Let x n1, x n2, ⋯ , x nk be free variables of (4.1.8) there exist k vectors . ,where 1 ≤ n 1 < n 2 < ⋯ < n k ≤ n. Then (4.3.2) → → → v 1 , v 2 , ⋯ , v k in ℝ m https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/20 8/30/2020 4.3 Gauss-Jordan Elimination that satisfy the following properties. → → (P 1) for each i ∈ {1, 2, ⋯ , k}, the ith component of v i is 1 and the component on n j of v i is zero for j ≠ i and j ∈ {1, 2, ⋯, k}. { } D is tri bu tio n → → → (P 1)N A = span v 1 , v 2 , ⋯ , v k . Proof → Let x n = t i for i ∈ {1, 2, ⋯, k}. We solve the system corresponding to the augmented matrix (B | 0) and we i can rewrite the solution into the following form (4.3.3) → → → X = t1v1 + t2v2 + ⋯ + tk vk , { Definition 4.3.1. } satisfies (P 1). By (4.3.3) , we see that (P 2) holds. N ot where → → → v1, v2, ⋯ , vk fo r → Let α 1, ⋯ , α k be nonzero numbers. The vectors → → → α1v1, α2v2, ⋯ , αk vk https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/20 8/30/2020 4.3 Gauss-Jordan Elimination → → → are called the basis vectors of the solution space N A, where the vectors v 1 , v 2 , ⋯ , v k are given in (4.3.2) . Example 4.3.2. c. { 4x 1 + 5x 2 + 6x 3 = 0 3x 1 + x 2 − 2x 3 = 0. x 1 + 2x 2 − x 3 = 0 3x 1 − 3x 2 + 2x 3 = 0 − x 1 − 11x 2 + 6x 3 = 0. x 1 − 2x 2 − 3x 3 + x 4 = 0 x 2 − 2x 3 + 2x 4 = 0. Solution a. fo r b. { { 2x 1 + 4x 2 + 6x 3 = 0 N ot a. D is tri bu tio n For each of the following homogeneous systems, solve it by using Gauss-Jordan elimination, write its solution as a linear combination, and find its solution space N A and dim(N A). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/20 8/30/2020 4.3 Gauss-Jordan Elimination (A b) ( | ) ( | )↓ = 6 0 R1 4 5 6 0 3 1 −6 0 ( | ) ( | )↓ ( | ) ( | )↑ ( | )↑ ( | ) | 2 0 −3 ( ) 1 3 0 R2 − 3 −6 0 → 0 − 5 − 11 0 1 2 3 0 1 2 0 1 0 0 3 0 0 1 2 0 0 R ( −1) 1 2 3 0 3 0 0 1 2 0 → 0 0 1 0 R2 ( − 2 ) + R1 → 2 1 2 3 0 → 4 5 6 0 3 1 −2 0 R1 ( − 4 ) + R2 → R1 ( − 3 ) + R3 R2 ( 5 ) + R3 → R3 ( − 2 ) + R → R3 ( − 3 ) + R1 1 0 0 0 0 1 0 0 → = (D 0). 0 0 1 0 N ot 0 0 1 0 2 0 − 5 − 11 0 0 0 −1 0 1 2 0 0 1 1 fo r 1 () D is tri bu tio n | → 2 4 | → The system corresponding to the augmented matrix (D 0) only has the zero solution (0, 0, 0). Hence, the original system has only the zero solution. The solution space is N A = {0 }. Because → r(A) = 3, null(A) = 3 − 3 = 0. Hence, dim(N A) = 0. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/20 8/30/2020 4.3 Gauss-Jordan Elimination (A | b) = ( 2 −1 0 3 −3 − 1 − 11 R2 ( − 1 ) + R3 → b. ( | )↓ ( |) 1 1 2 −1 5 2 6 0 0 R1 ( − 3 ) + R2 → R1 ( 1 ) + R3 ( | )↓ 1 2 −1 0 0 −9 5 0 0 −9 5 0 −1 0 R ( − 1 ) 2 9 0 −9 5 0 → 1 2 0 0 0 0 | )↑ ( | ) 0 0 1 −9 0 0 0 0 0 R2 ( − 2 ) + R1 → 1 0 1 9 0 5 0 0 1 −9 0 0 0 0 → = (D | 0). D is tri bu tio n → | → 1 N ot { fo r The system corresponding to the augmented matrix (D 0) is 1 5 x1 + 9 x3 5 x2 − 9 x3 = 0 = 0. 1 5 This implies x 1 = − 9 x 3 and x 2 = 9 x 3. Let x 3 = t ∈ ℝ. Then x 1 = − 9 t and x 2 = 9 t. We write the solution as a linear combination as follows: https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/20 4.3 Gauss-Jordan Elimination ()( x1 x2 = x3 1 − 9t 5 t 9 t )() 1 −9 { . 9 1 → 1 5 Let v 1 = ( − 9 , 9 , 1). Then the solution space is NA = 5 =t → tv1 : t ∈ ℝ } D is tri bu tio n 8/30/2020 → = span{ v 1 }. Because r(A) = 2, null(A) = 3 − 2 = 1. Hence, dim(N A) = 1. c. (A | b) = ( | )↑ 1 −2 −3 1 0 0 1 −2 2 0 R2 ( 2 ) + R1 → ( |) 1 0 −7 5 0 0 1 −2 2 0 . fo r → The system corresponding to the last augmented matrix is = 0 x 2 − 2x 3 + 2x 4 = 0. N ot { x 1 − 7x 3 + 5x 4 This implies x 1 = 7x 3 − 5x 4 and x 2 = 2x 3 − 2x 4. Let x 3 = s and x 4 = t for s, t ∈ ℝ. Then x 1 = 7s − 5t and x 2 = 2s − 2t. We write the solution as a linear combination of basis vectors as follows: https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/20 8/30/2020 4.3 Gauss-Jordan Elimination ( ) ( ) () ( ) x2 x3 x4 7s − 5t = 2s − 2t s 7 =s t → → Let v 1 = (7, 2, 1, 0) and v 2 = ( − 5, − 2, 0, 1). Then NA = { → → s v 1 + t v 2 : s, t ∈ ℝ Because r(A) = 2, null(A) = 4 − 2 = 2 and dim(N A) = 2. Example 4.3.3. 2 1 −5 +t 0 −2 0 . 1 D is tri bu tio n x1 } { } → → = span v 1 , v 2 . N ot fo r There are three types of food provided to a lake to support three species of fish. Each fish of Species 1 weekly consumes an average of one unit of Food 1, two units of Food 2, and one unit of Food 3. Each fish of Species 2 consumes, each week, an average of two units of Food 1, five units of Food 2, and 3 units of Food 3. The average weekly consumption for a fish of Species 3 is two units of Food 1, two unit of Food 2, and six units of Food 3. Each week, 10000 units of Food 1, 15000 units of Food 2, and 25000 units of Food 3 are supplied to the lake. Assuming that all food is eaten, how many fish of each species can coexist in the lake? Solution We denote by x 1, x 2, and x 3 the number of fish of the three species in the lake. Using the information in the problem, we see that x 1 fish of Species 1 consume x 1 units of Food 1, x 2 fish of Species 2 consume 2x 2 units of Food 1, and x 3 fish of Species 3 consume 2x 3 units of Food 1. Hence, x 1 + 2x 2 + 2x 3 = 10000 = total weekly https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/20 8/30/2020 4.3 Gauss-Jordan Elimination supply of Food 1. Similarly, we can establish the equations for the other two foods. This leads to the following system 2x 1 + 5x 2 + 2x 3 = 15000 2x 1 + 3x 2 + 6x 3 = 25000. D is tri bu tio n { x 1 + 2x 2 + 2x 3 = 10000 We solve the above system by using Gauss-Jordan elimination. ( | )↓ ( | )↓ ↑ ( | ) ( | ) 1 2 2 10000 2 5 2 15000 2 3 6 25000 → 1 2 6 10000 0 0 1 0 −1 2 10000 − 2 − 5000 2 5000 R 2 ( − 21 ) + R 1 0 1 − 2 − 5000 0 0 1 0 2 R1 ( − 2 ) + R3 2 → 0 N ot R2 ( 1 ) + R3 → 1 fo r → (A | b) = R1 ( − 2 ) + R2 2000 0 1 − 2 − 5000 . 0 0 0 0 | → The system of linear equations corresponding to (D d) is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/20 8/30/2020 4.3 Gauss-Jordan Elimination { x 1 + 6x 3 = 20000 x 2 − 2x 3 = − 5000, { D is tri bu tio n where x 1, x 2 are basic variables and x 3 is a free variable. Let x 3 = t ∈ ℝ. Then x 1 = 20000 − 6t x 2 = − 5000 + 2t x 3 = t, which is a solution of the system. Noting that the number of fish must be nonnegative, we must have x 1 ≥ 0, x 2 ≥ 0, and x 3 ≥ 0. Hence, fo r x 2 = − 5000 + 2t ≥ 0 x 3 = t ≥ 0. N ot { x 1 = 2000 − 6t ≥ 0 Solving the above system of inequalities, we get 2500 ≤ t ≤ 20000 / 6 ≈ 3333.3. Because t is an integer, so we take 2500 ≤ t ≤ 3333. This means that the population that can be supported by the lake with all food consumed is { x 1 = 20000 − 6x 3 x 2 = − 5000 + 2x 3 2500 ≤ x 3 ≤ 3333. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/20 8/30/2020 4.3 Gauss-Jordan Elimination For example, if we take x 3 = 2500, then x 1 = 20000 − 6(2500) = 5000 and x 2 = − 5000 + 2(2500) = 0. This means that if there are 2500 fish of Species 3, then there are no fish of Species 2. Otherwise, the lake cannot support all three species with the current supply of food. If we take x 3 = 3000, then x 1 = 20000 − 6(3000) = 2000 and x 2 = − 5000 + 2(3000) = 1000. This provides the suggestion that with the D is tri bu tio n current supply of food, if 2000 fish of Species 1, 1000 fish of Species 2, and 3000 fishes of Species 3 can coexist in the lake. Example 4.3.4 (Leontief input-output model with two industries) In an economic system with steel and automobile industries, let x 1 and x 2 denote the outputs of the steel and automobile industries, respectively, and b 1 and b 2 represent the external demand from outside the steel and automobile industries. Let a 11 represent the internal demand placed on the steel industry by itself, that is, the steel industry needs a 11 units of the output of the steel industry to produce one unit of its output. Thus, a 11x 1 is the total amount the steel industry needs from itself. Let a 12 represent the internal demand placed on the steel industry by the automobile industry, that is, the automobile industry needs a 12 units of the output of the steel industry to produce one unit of output of the automobile industry. a 12x 2 is the total amount the (4.3.4) fo r automobile industry needs from the steel industry. Thus, the total demand on the steel industry is N ot a 11x 1 + a 12x 2 + b 1. Let a 21 represent the internal demand placed on the automobile industry by the steel industry, that is, the steel industry needs a 21 units of the output of the automobile industry to produce one unit of output of the steel industry. Hence, a 21x 1 is the total amount the steel industry needs from the automobile industry. Let a 22 represent the internal demand placed on the automobile industry by itself, that is, the automobile industry needs a 22 units of the output of the automobile industry to produce one unit of its output. Hence, a 22x 2 is the total amount the automobile industry needs from itself. Thus, the total demand on the automobile industry is (4.3.5) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/20 8/30/2020 4.3 Gauss-Jordan Elimination a 21x 1 + a 22x 2 + b 2. Now we assume that the output of each industry equals its demand, that is, there is no overproduction. Then, by hypotheses, (4.3.4) and (4.3.5) , we obtain the following system D is tri bu tio n { a 11x 1 + a 12x 2 + b 1 = x 1 a 21x 1 + a 22x 2 + b 2 = x 2, or equivalently, (4.3.6) − a 21x 1 + (1 − a 22)x 2 = b 2. Example 4.3.5. In the economic system given in Example 4.3.4 9 1 fo r { (1 − a 11)x 1 − a 12x 2 = b 1, , suppose that the external demands are 100 and 200, N ot respectively. Suppose that a 11 = 10 , a 12 = 2 , a 21 = 110, and a 22 = 15. Find the output in each industry such that supply exactly equals demand. Solution By (4.3.6) , we obtain (4.3.7) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/20 8/30/2020 4.3 Gauss-Jordan Elimination | → 1 1 − 10 x 1 + 1 − 5 x 2 = 20. ( |) 1 (A b) = 1 10 1 − 10 1 D is tri bu tio n { ( ) ( ) 9 1 − 10 x 1 − 2 x 2 = 10, − 2 100 R 1 ( 10 ) 1 − 5 100 R 1 ( 1 ) + R 2 → → 4 200 − 1 8 200 R ( 10 ) 5 1 ( 2 | ) ( | ) ( | ) ( | ) R2 ( 3 ) 1 − 5 100 0 → 3 300 1 0 600 . fo r 0 1 100 1 − 5 100 R 2 ( 5 ) + R 1 → 0 1 100 Hence, x 1 = 600 and x 2 = 100. The outputs needed for supply to equal demand are 600 and 100 for the steel N ot and automobile industries, respectively. Exercises 1. Solve each of the following systems using Gauss-Jordan elimination. a. { x1 − x2 + x3 = 2 2x 1 − 3x 2 + x 3 = − 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/20 8/30/2020 4.3 Gauss-Jordan Elimination e. f. g. x 1 + 3x 2 − x 3 = 2 4x 1 − 6x 2 + 6x 3 = 14 − 3x 1 − x 2 − 2x 3 = 3 D is tri bu tio n d. { { { { { − 4x 1 + 2x 2 − 4x 3 = 2 x 1 − 2x 2 + 2x 3 = 2 − 2x 1 + 3x 2 − 4x 3 = − 2 − 3x 1 + 4x 2 + 6x 3 = 0 x2 − x3 = 1 − x 1 + 3x 2 − x 3 = − 2 3x 1 + 3x 2 − 2x 3 = 4 x 1 + x 2 + 2x 3 = 1 − 2x 1 − 4x 2 − 5x 3 = − 1 3x 1 + 6x 2 + 5x 3 = 0 2x 2 + 3x 3 − 2x 4 = 3 fo r c. { N ot b. 3x 1 − x 2 + 3x 3 = − 1 2x 1 + 3x 3 + 2x 4 = 1 − 3x 1 + x 2 − 2x 3 = 3 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/20 8/30/2020 4.3 Gauss-Jordan Elimination { − x 1 + 3x 2 − 4x 3 = − 3 x2 − x3 = − 2 x 1 − 2x 2 + 3x 3 = 1 D is tri bu tio n h. x 1 − x 2 + 2x 3 = − 1 2. For each of the following homogeneous systems, solve it by using Gauss-Jordan elimination, write its solution as a linear combination, and find its solution space N A and dim(N A). b. { c. d. 2x − y + 3y = 0 2x − y + 3z = 0 4x − 2y + 6z = 0 − 6x + 3y − 9z = 0 { { x 1 + 4x 2 + 6x 3 = 0 2x 1 + 6x 2 + 3x 3 = 0 − 3x 1 + x 2 + 8x 3 = 0. − x 1 + 2x 2 + x 3 = 0 3x 1 − 3x 2 − 9x 3 = 0 fo r { x + 2y − z = 0 N ot a. − 2x 1 − x 2 + 12x 3 = 0. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/20 8/30/2020 4.3 Gauss-Jordan Elimination x 1 + x 2 − 2x 3 − x 5 = 0 x3 + x4 + x5 = 0 x 1 + x 2 − 2x 3 = 0 − 3x 1 + 2x 2 − x 3 = 0 − 2x 1 + 3x 2 − 3x 3 = 0 4x 1 − x 2 − x 3 = 0. N ot fo r f. { { − x 1 − x 2 + 2x 3 − 3x 4 + x 5 = 0 D is tri bu tio n e. 2x 1 + 2x 2 − x 3 + x 5 = 0 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/20 8/30/2020 4.4 Inverse matrix method and Cramer’s rule 4.4 Inverse matrix method and Cramer’s rule ut io n Both Gaussian elimination and Gauss-Jordan elimination can be used to solve a system of linear equations with an m × n coefficient matrix, where m and n do not necessarily equal. Of course, they can be used to solve the following system (4.4.1) → → D is tri b AX = b, where A is an n × n matrix. However, if A is an invertible matrix, then there are two other methods that can be used to solve such systems. One method is to use the inverse matrix of A to solve the system, called the inverse matrix method, and another is to use the determinant of A to solve the system, called Cramer’s rule. fo r We start with the inverse matrix method. Theorem 4.4.1. → (4.4.2) N ot If A is an invertible n × n matrix, then for each b ∈ ℝ n, the system (4.4.1) → has the unique solution → X = A − 1 b. Proof → Because A is invertible, A − 1 exists and X given in (4.4.2) because → is defined. With X given in (4.4.2) , https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule → → → → → AX = A(A − 1 b) = (AA − 1) b = I n b = b, → where I n is the identity matrix, and X given in (4.4.2) (4.4.1) → → is a solution of (4.4.1) → → . Now we assume that → has a solution Y, that is, AY = b. Multiplying both sides of AY = b by A − 1, we obtain → → → → → → → This implies Y = A − 1 b and Y = X. Hence, (4.4.1) has only one solution X given in (4.4.2) Solve the following system using the inverse matrix method . D is tri b Example 4.4.1. x 1 + 2x 2 + x 3 = 2 2x 1 + 5x 2 + 2x 3 = 6 x1 − x3 = 2 fo r { ut io n A − 1(AY) = A − 1 b. Solution N ot From the system, we obtain the coefficient matrix A= ( ) () 1 2 1 2 5 2 2 → and b = 1 0 −1 We find A − 1 by using the method given in Section 2.7 6 . 2 . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule 2 2 1 1 0 0 1 0 −2 1 0 ( | ) ( | ) 1 0 0 R (−1) 1 2 1 3 2 −2 0 −2 1 0 0 1 0 → 1 2 1 0 1 1 0 0 −2 −5 2 1 0 0 1 ( | ) ( | ) 3 R3 ( − 1 ) + R1 → 1 2 0 −2 1 0 1 0 −2 1 0 0 1 5 2 5 1 0 0 −1 2 0 1 0 −2 0 0 1 5 2 1 1 0 0 → 1 −1 − 2 = (I 3 | A − 1). 1 −1 − 2 R2 ( − 2 ) + R1 2 0 2 0 1 1 −1 − 2 2 By (4.4.2) 1 5 0 ut io n 1 0 − 1 0 0 1 R1 ( − 1 ) + R3 0 − 2 − 2 − 1 0 1 D is tri b → 2 5 1 0 0 R ( −2) +R 1 i 2 0 1 0 0 → fo r R2 ( 2 ) + R3 ( | ) ( | ) 1 N ot (A | I 3) = 1 2 , https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule () → X= x2 ( )( ) ( ) 5 x1 2 → = A − 1b = −2 5 x3 1 1 2 2 0 6 1 −1 − 2 2 0 = 2 −2 ut io n 2 −1 is the solution of the system. and 3.4.1 , we have D is tri b By Theorems 4.4.1 Corollary 4.4.1. If either |A | ≠ 0 or r(A) = n, then (4.4.1) has a unique solution for each b ∈ ℝ n. → Example 4.4.2. Solution Because |A | = x + y = b1 2x + ay = b 2 N ot { fo r Find all the numbers a ∈ ℝ such that the following system has a unique solution for each (b 1, b 2) ∈ ℝ 2. | | 1 1 2 a = a − 2, we have if a ≠ 2, | A | ≠ 0. By Corollary 4.4.1 , if a ≠ 2, then the system has a unique solution for each (b 1, b 2) ∈ ℝ 2. Example 4.4.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule Show that the following system has a unique solution for each (b 1, b 2, b 3) ∈ ℝ 3. 2x 1 + 5x 2 + 2x 3 = b 2 x1 − x3 = b3 ut io n { x 1 + 2x 2 + x 3 = b 1 Solution D is tri b We find the rank of A. ( ) ( ) ( ) 1 2 5 2 R ( − 2 ) + R2 → 1 0 −1 1 2 1 0 1 0 Hence, r(A) = 3. By Corollary 4.4.1 0 2 1 1 0 R2 ( 2 ) + R3 → 0 −2 −2 N ot 0 0 −2 . R1 ( − 1 ) + R3 1 fo r A= 1 2 , the system has a unique solution for each (b 1, b 2, b 3) ∈ ℝ 3. By Corollary 4.4.1 , we see that if |A | ≠ 0, then the system (4.4.1) has a unique solution. The Cramer’s rule we study below gives the expression of the unique solution by using determinants. Theorem 4.4.2. (Cramer’s rule) Let A be an n × n matrix given in (3.2.1) . If |A | ≠ 0, then the unique solution of (4.4.1) is (4.4.3) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule |A1 | x1 = |A| |A 2 | |A n | , x2 = , ⋯, x n = , |A| |A| → (4.4.4) ( ⋯ a1 ( i − 1 ) b1 a1 ( i + 1 ) a 21 a 22 ⋯ a2 ( i − 1 ) b2 a2 ( i + 2 ) ⋮ ⋮ a n1 a n2 ⋱ ⋮ ⋮ , we have ⋯ a 2n ⋱ ⋮ ⋯ a nn ) . N ot (4.4.5) and (4.4.4) a 1n fo r Proof By Theorem 3.2.1 ⋮ an ( i − 1 ) bn an ( i + 1 ) ⋯ ⋯ D is tri b Ai = a 11 a 12 ut io n where A i is obtained by replacing the ith column of A by b, namely, n |A i | = By (3.3.11) ∑ b kA kj. k=1 , (4.4.6) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule ⋮ ⋮ ⋮ A ni bi ⋮ ⋮ ⋮ bn ∑ nk = 1b kA km A 11 A 21⋯ A n1 ⋮ A 1i A 2i⋯ ⋮ ⋮ A 1n A 2n⋯ A nn By Theorems 3.3.7 and 4.4.1 , the solution of system (4.4.1) (4.4.7) → → X = A − 1b = Example 4.4.4. , and (4.4.7) → for each b ∈ ℝ n. , we see that (4.4.3) holds. fo r , (4.4.6) 1 → adj(A) b |A| equals . N ot By (4.4.5) ∑ nk = 1b kA ki = D is tri b → adj(A) b = ⋮ ut io n ( )( ) ( ) b1 ∑ nk = 1b kA k1 Solve each of the following systems by using Cramer’s rule. a. { 2x − 3y = 7 x + 5y = 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule { − 3x 1 + 4x 2 + 6x 3 = 30 − x 1 − 2x 2 + 3x 3 = 8 Solution a. a)Let 1 5 , we have A1 = ( )( ) b 1 a 12 b 2 a 22 = 7 −3 1 5 and b = and A 2 = N ot | | = 13, | A1 | = 38, and |A2 | = By computation, A → (x, y) = ( 38 13 ( ) () b1 b2 = 7 1 . ( )( ) a 11 b 1 a 21 b 2 fo r By (4.4.4) ( )( ) = 2 −3 D is tri b A= a 11 a 12 a 21 a 22 ut io n b. x 1 + 2x 3 = 6 = 2 7 1 1 − 5. By Theorem 4.4.2 , − 5 13 . , ) is the solution of the system. b. Let https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule ( a 31 a 32 a 33 By (4.4.4) )( ) ()() = 1 0 2 −3 4 6 b1 → and b = −1 −2 3 b2 6 = b3 30 . 8 , we obtain A1 = ( )( ) ( )( ) ( )( ) b 1 a 12 a 13 b 2 a 22 a 23 = a 11 a 12 b 1 a 21 a 22 b 2 N ot A3 = a 31 a 32 b 3 By computation, 30 4 6 1 = | | | | −2 3 −1 = 6 2 − 3 30 6 fo r a 31 b 3 a 33 2 D is tri b A2 = 0 8 b 3 a 32 a 33 a 11 b 1 a 13 a 21 b 2 a 23 6 ut io n A= a 11 a 12 a 13 a 21 a 22 a 23 8 3 1 0 −3 4 6 30 . −1 −2 8 | | |A| = 44, A 1 = − 40, A 2 = 72, and A 3 = 152, and https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule |A| Hence, by (4.4.3) = − 40 44 10 = − 18 10 |A2 | 72 18 152 38 , x2 = = = , x3 = = = . 11 |A| 44 11 |A| 44 11 38 , ( − 11 , 11 , 11 ) is the solution of the system. Exercises 1. Solve each of the following systems using the inverse matrix method. c. d. D is tri b { { { x 1 − x 2 = 14 x1 + x2 + x3 = 2 2x 1 + 3x 2 + x 3 = 2 x1 − x3 = 1 x 1 − 2x 2 + 2x 3 = 1 2x 1 + 3x 2 + 3x 3 = − 1 x 1 + 6x 3 = − 4 x 1 + 2x 2 + 3x 3 = 0 fo r b. { 3x 1 + 4x 2 = 7 N ot a. |A 3 | ut io n x1 = |A1 | 2x 1 + 5x 2 + 3x 3 = 1 x 1 + 8x 3 = − 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule { x 2 + x 3 + 2x 4 = 1 − x1 − x2 + x3 = − 1 2x 1 + 9x 4 = 1 ut io n e. x 1 − 2x 3 + x 4 = 0 2. Show that if a ≠ 3, then the following system has a unique solution for (b 1, b 2) ∈ ℝ 2. D is tri b { x − y = b1 ax − 3y = b 2 3. For each (b 1, b 2, b 3) ∈ ℝ 3, show that the following system has a unique solution. fo r − 3x 1 + 4x 2 + 6x 3 = b 2 − x 1 − 2x 2 + 3x 3 = b 3 N ot { x 1 + 2x 3 = b 1 4. Solve each of the following systems by using Cramer’s rule. a. { x − 3y = 6 2x + 5y = − 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/12 8/30/2020 4.4 Inverse matrix method and Cramer’s rule e. { ut io n x1 + x2 − x3 = 1 2x 1 + x 2 − x 3 = 0 x 2 − 2x 3 = − 2 x1 + x2 + x3 = 0 D is tri b d. − x 1 − 2x 2 + 4x 3 = 5 2x 1 − x 2 − x 3 = 1 x 2 + 2x 3 = 2 x1 − x4 = 0 x2 + x3 = 0 x1 − x2 = 0 x3 − x4 = 1 fo r c. { { { − 2x 1 + 3x 2 + x 3 = 16 N ot b. x 1 + 2x 3 = 2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/12 8/30/2020 4.5 Consistency of systems of linear equations 4.5 Consistency of systems of linear equations | → In this section, we use ranks of r(A) and (A b) to determine whether a linear system (4.1.7) (4.5.1) → , that is, fo rD is tri b → ut io n In Sections 4.2–4.4, we study the methods to find all the solutions of systems of linear equations. However, in some cases, for example, see Section 4.6, we do not want the exact solutions of the systems but we just want to know whether the systems have solutions, that is, whether the systems are consistent or inconsistent. AX = b, where the coefficient matrix A is an m × n matrix, has a unique solution, infinitely many solutions,, or no solutions without solving the system. Theorem 4.5.1. has a unique solution. | 2. r(A) = r(A | b) < n if and only if the system (4.5.1) has infinitely many solutions. 3. r(A) < r(A | b) if and only if the system (4.5.1) has no solutions. → 1. r(A) = r(A b) = n if and only if the system (4.5.1) N ot → → Proof | | → → Note that A is an m × n matrix and (A b) is an m × (n + 1) matrix. Use row operations to change (A b) to a row echelon matrix, denoted by (B | → c). Then B is a row echelon matrix of A and | → | r(A) = r(B) and r(A b) = r(B → c). Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/12 8/30/2020 4.5 Consistency of systems of linear equations , the linear system corresponding to (B | → c) is equivalent to (4.5.1) By Theorem 4.2.1 . We first prove that the following assertions hold: | | → i. If r(A) = r(A b) = n, then the system (4.5.1) → has a unique solution. Indeed, if r(A) = r(A b) = n, then by Theorem , we have n ≤ m and r(B) = r(B | → c) = n. We write B = (b ij) . This implies that b ii ≠ 0 for i = 1, 2, ⋯, n. If m×n m > n, then all rows of the matrix (B | → c) from n + 1 to m are zero rows. Using back-substitution to solve the system 2.5.2 → BX = → c, we obtain only one solution. → | has infinitely many solutions. ut io n | ii. If r(A) = r(A b) < n, then the system (4.5.1) | Indeed, because r(A) = r(A b) < n, r(B) = r(B → c) < n. This implies that if r(B) < m, then all rows of the matrix (B | → c) → → → | → iii. If r(A) < r(A b), then the system (4.5.1) fo rD is tri b from r(B) + 1 to m are zero rows and the system BX = → c has n − r(B) free variables. Hence, BX = → c has infinitely many solutions. | has no solutions. In fact, if r(A) < r(A b), then r(B) < r(B | → c) and c j ≠ 0, → → where j = r(B) + 1. This implies that the jth equation of the system BX = → c is 0x 1 + ⋯ + 0x n = c j, → from which we see that the system BX = → c has no solutions. | → | → | → Note that r(A) ≤ r(A b) and r(A) ≤ min{n, m} ≤ n. Hence, there are only three cases: r(A) < r(A b), r(A) < r(A b) = n, | → By Theorem 4.5.1 N ot and r(A) < r(A b) < n, from which we see that the converses of (i)-(iii) hold. , we obtain the following result, which gives the structure of the solutions of (4.1.4) Corollary 4.5.1. . One of the following assertions must occur. 1. (4.5.1) 2. (4.5.1) 3. (4.5.1) has a unique solution. has infinitely many solutions. has no solutions. Processing math: 30% Proof https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/12 8/30/2020 4.5 Consistency of systems of linear equations | | → | → → | → Because r(A) ≤ r(A b) and r(A) ≤ n, there are only three cases: r(A) = r(A b) = n, r(A) = r(A b) < n, and r(A) = r(A b). The result follows from Theorem 4.5.1 . Example 4.5.1. Use Theorem 4.5.1 solutions. c. d. ut io n 4x 1 + 5x 2 + 6x 3 = 24 3x 1 + x 2 − 2x 3 = 4 2x 1 + 2x 2 − 2x 3 = 4 3x 1 + 5x 2 + x 3 = − 8 − 4x 1 − 7x 2 − 2x 3 = 13 x 1 + 2x 2 + 3x 3 = 4 4x 1 + 7x 2 + 6x 3 = 17 2x 1 + 5x 2 + 12x 3 = 10 x1 − x2 = 1 − 2x 1 − 5x 2 = 5 − 3x 1 + 10x 2 = − 10 fo rD is tri b b. { { { { 2x 1 + 4x 2 + 6x 3 = 18 N ot a. to determine whether the following systems have a unique solution, infinitely many solutions, or no Solution a. By the solution of Example 4.2.5 | → a), we see that the augmented matrix (A b) is changed to the row echelon matrix Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/12 8/30/2020 4.5 Consistency of systems of linear equations | ( (B → c) = 1 2 3 9 0 1 2 4 0 0 −1 −3 from which we see that r(A) = r(B | → c) = 3 = n. By Theorem 4.5.1 (1), the system has a unique solution. | | ( 1 1 −1 2 0 1 2 −7 0 0 0 0 from which we see that r(A) = r(B | → c) = 2 < n. By Theorem 4.5.1 | → ) fo rD is tri b (B → c) = c. By the solution of Example 4.2.5 → b), we see that the augmented matrix (A b) is changed to the row echelon matrix ut io n b. By the solution of Example 4.2.5 ) (2), the system has infinitely many solutions. d), (A b) is changed to the row echelon matrix | ( (B → c) = 1 2 3 4 0 −1 −6 1 0 0 0 3 d. | → N ot from which we see that r(A) = 2 < 3 = r(B | → c). By Theorem 4.5.1 (A b) = ( 1 −2 −5 5 − 3 10 − 10 → (3), the system has no solutions. ) ( ) ( ) 1 −1 R2 ( 1 ) + R3 ) R1 ( 2 ) + R2 → R1 ( − 3 ) + R3 1 −1 1 0 −7 7 0 7 −7 1 −1 1 0 −7 7 . 0 0 0 Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/12 8/30/2020 4.5 Consistency of systems of linear equations Hence, we have r(A) = r(B | → c) = 2 = n ≤ 3 = m. By Theorem 4.5.1 The following result shows that if n > m, then it is impossible for (4.5.1) unique solution, then n ≤ m, see Example 4.5.1 (a) and (d). (1), the system has a unique solution. to have a unique solution. Hence, if (4.5.1) has a Corollary 4.5.2. either has infinitely many solutions or no solutions. Proof ut io n If n > m, then (4.5.1) Because n > m, by Theorem 4.5.1 (1), (4.5.1) doesn’t have a unique solution. Hence, by Corollary 4.5.1 that (4.5.1) either has infinitely many solutions or no solutions. | → (4.5.1) doesn’t have a unique solution by comparing m and n. Example 4.5.2. . Hence, one can easily determine that fo rD is tri b The ranks of A and (A b) are not directly involved in the conditions of Corollary 4.5.2 , we see a. b. { { x 1 − x 2 − 3x 3 − 4x 4 = 1 − x1 + x2 − x3 = − 1 x1 + x2 + x3 = 1 x1 + x2 + x3 = 2 N ot For each of the following systems, determine whether it has a unique solution. Solution a. Because m = 2 < 4 = n, by Corollary 4.5.2 b. Because m = 2 < 3 = n, by Corollary 4.5.2 By Theorem 4.5.1 , the system doesn’t have a unique solution. , the system doesn’t have a unique solution. , we obtain the following result, which can be used to determine whether a linear system is consistent. Processing math: 30% Corollary 4.5.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/12 8/30/2020 4.5 Consistency of systems of linear equations | → 1. (4.5.1) is consistent if and only if r(A) = r(A b). 2. (4.5.1) is inconsistent if and only if r(A) < r(A b). | will be given in Section 4.6 Note that a homogeneous system (4.1.8) system, the third case of Theorem 4.5.1 systems. . is always consistent because it has a zero solution. Hence, for a homogeneous never occurs. By Theorem 4.5.1 , we obtain the following result on homogeneous ut io n Applications of Corollary 4.5.3 → Theorem 4.5.2. → → 1. AX = 0 has a unique solution if and only if r(A) = n. → → fo rD is tri b 2. (2)AX = 0 has infinitely many solutions if and only if r(A) = n. Remark 4.5.1. → → By Theorem 4.5.2 (2), if n > m, then the homogeneous system AX = 0 has infinitely many solutions because by Theorem 2.5.2 , r(A) ≤ min{m, n} = m < n. By Theorems 2.7.6 matrix. and 3.4.1 , we obtain the following result for a homogeneous system AX→=0→, where A is an n×n N ot Theorem 4.5.3. Let A be an n×n matrix. Then the following are equivalent. 1. The system AX→=0→ only has the zero solution. 2. r(A)=n. 3. |A|≠0. 4. A is invertible. Example 4.5.3. Determine whether each of the following systems has a unique solution or infinitely many solutions. Processing math: 30% a. { x1−2x2=0−2x1+3x2=0−3x1+4x2=0. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/12 8/30/2020 4.5 Consistency of systems of linear equations b. { x1+3x2−x3=0−x1+x2+6x3=0−3x1−x2+2x3=0. c. { x1−2x2−x3+x4=0x2−x3+2x4=0. d. { x1−3x2+x3=0−x1+2x2−2x3=02x1−5x2+3x3=03x1−7x2+5x3=0. Solution a. Because A=(1−2−23−34)→R1(3)+R3R1(2)+R2(1−20−10−2)→R2(−2)+R3(1−20−100), ut io n r(A)=2. Note that m=3 and n=2. Hence, r(A)=2=n and by Theorem 4.5.2 (1), the system has a unique solution. b. Because |A|=| 13−1−116−3−12 |13−11−3−1=(2−54−1)−(3−6−6)=−44≠0, fo rD is tri b it follows from Theorem 4.5.3 that the system has a unique solution. c. Because m=2<4=n, it follows from Theorem 4.5.2 (2) that the system has infinitely many solutions. d. Because A= (1−31−12−22−533−75)→R1(−3)+R4R1(1)R2R1(−2)+R3(1−310−1−1011022)→R2(2)+R4R2(1)+R3(1−310−1−1000000), r(A)=2. Hence, n=3<4=m and r(A)=2<3=n. It follows from Theorem 4.5.2 solutions. (2) that the system has infinitely many N ot In Section 4.4, we show that if A is an n×n matrix and r(A)=n, then the system AX→=b→ has a unique solution for every b→∈ℝn, see Corollary 4.4.1 . Now we generalize the result to the general system (4.5.1) . Let A be an m×n matrix. We study the following two problems. i. How to determine A(ℝn)=ℝm, where A(ℝn) is the same as in (2.2.1) , that is, how to determine whether AX→=b→ is consistent for every b→∈ℝm? ii. If A(ℝn)≠ℝm, that is, AX→=b→ is not consistent for every b→∈ℝm? then how to find all the vector b→ in ℝm such that AX→=b→ is consistent? or how to find all the vector b→ in ℝm such that AX→=b→ is inconsistent? By Corollary 4.5.3 , we see that for a given vector b→∈ℝm, AX→=b→ is consistent if r(A)=r(A|b→). Here we want AX→=b→ to be consistent for every b→∈ℝm, that is, for every b→∈ℝm, r(A)=r(A|b→). Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/12 8/30/2020 4.5 Consistency of systems of linear equations The following result shows that r(A)=r(A|b→) for every b→∈ℝm is equivalent to r(A)=m. Theorem 4.5.4. 1. The following assertions are equivalent. i. The system (4.5.1) is consistent for each b→∈ℝm. ii. A(ℝn)=ℝm. iii. r(A)=m. has no solutions in Rn if and only if r(A)<m. Proof ut io n 2. There exists a vector b→∈ℝm such that the system (4.5.1) (4.5.2) fo rD is tri b It is obvious that the results (i) and (ii) are equivalent, and the results (i) and (2) are equivalent. So we only prove that results (i) and (iii) are equivalent. Assume that (i) holds. We change (A|b→) to (B|c→) by using row operations, where B is a row echelon matrix of A. By Theorem 4.2.1 , for each c→∈ℝm, the linear system BX→=c→ N ot has at least one solution in ℝn. If we assume r(A)<m, then r(B)=r(A)<m, the rows from r(A)+1 to m of B are zero rows. Let c0→∈ℝm be a vector whose (r(A)+1)th component is 1 and other components are zero. Then r(B|c0→)=r(B)+1 and r(B) <r(B|c0→). By Theorem 4.5.1 (3), we obtain that the following system BX→=c0→ has no solutions in ℝn. Because the system (4.5.1) is equivalent to the system (4.5.2) , it follows that the system (4.5.1) has no solutions in ℝn, which contradicts the fact that system (4.5.1) is consistent for each b→∈ℝm. Hence, r(A)=m and (iii) holds. Conversely, we assume r(A)=m. We use row operations to change A to a row echelon matrix B. Then r(B)=r(A)=m. For each c→∈ℝm, m=r(B)≤r(B|c→)≤min{m, n+1}=m. Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/12 8/30/2020 4.5 Consistency of systems of linear equations Hence, r(B)=r(B|c→). By Corollary 4.5.3 for each b→∈ℝm and (i) holds. Theorem 4.5.4 one solution. , (4.5.2) is consistent. By Theorem 4.2.1 , the system (4.5.1) is consistent provides a criterion to justify whether a linear system is consistent, that is, whether a linear system has at least The following result gives criteria to determine whether a linear system has infinitely many solutions or a unique solution. Proof has infinitely many solutions for each b→∈ℝm if and only if r(A)=m<n. has a unique solution in ℝn for each b→∈ℝm if and only if m=n, r(A)=n if and only if m=n and |A| fo rD is tri b 1. The system (4.5.1) 2. The system (4.5.1) ≠0. ut io n Theorem 4.5.5. 1. Assume that the system (4.5.1) has infinitely many solutions for each b→∈ℝm. By Theorem 4.5.4 (1), r(A)=m. If m=n, then A is an n×n matrix and is invertible. By Corollary 4.4.1 , the system (4.5.1) with m=n has a unique solution for each b→∈ℝm, which contradicts the fact that the system (4.5.1) has infinitely many solutions for each b→∈ℝm. Hence, m<n. Conversely, assume that r(A)=m and m<n. Let b→∈ℝm. We change (A|b→) to (B|c→) by using row operations, where B is a row echelon matrix of A. Because r(A)=m, by Theorem 4.2.1 , (4.5.3) N ot BX→=c→ has at least one solution for every c→∈ℝm. Because r(A)=m, we have r(B)=m. Because m<n, v(B)=n−m≥1, so the system (4.5.3) has at least one free variable. Hence, the system (4.5.3) has infinitely many solutions. It follows from Theorem 4.2.1 that the system (4.5.1) has infinitely many solutions. 2. If m=n, r(A)=n, then by Theorems 4.5.4 , the system (4.5.1) has a unique solution in ℝn for each b→∈ℝm=ℝn. Conversely, assume that the system (4.5.1) has a unique solution in ℝn for each b→∈ℝm. By Theorem 4.5.4 (1), r(A)=m and m≤n. Because the system (4.5.1) has a unique solution in ℝn for each b→∈ℝm, by the above result (1), m=n; otherwise, if m<n, then the system (4.5.1) would have infinitely many solutions in ℝn for each b→∈ℝm, a contradiction. Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/12 8/30/2020 4.5 Consistency of systems of linear equations Corollary 4.5.2 shows that if (4.5.1) has a unique solution, then n≤m. But Theorem 4.5.5 (2) shows that if the system (4.5.1) has a unique solution in ℝn for each b→∈ℝm, then A must be a square matrix. In other words, if A is an m×n matrix with m≠n, then there exists a vector b→∈ℝm such as AX→=b→ has infinitely many solutions or has no solutions. Example 4.5.4. Consider the following system 1. Determine if the system is consistent for all b1, b2∈ℝ. 2. Determine if the system has infinitely many solutions for all b1, b2∈ℝ. Because fo rD is tri b Solution ut io n { x1−x2+x3=b13x1−2x2−x3=b2 A=(1−113−2−1)→R1(−3)+R2(1−1101−4), r(A)=2. Because A is a 2×3 matrix, m=2 and n=3. Example 4.5.5. N ot 1. Because r(A)=2=m, by Theorem 4.5.4 (1), the system is consistent for all b1, b2∈ℝ. 2. Because r(A)=2=m and m=2<3=n, by Theorem 4.5.5 (1), the system has infinitely many solutions for all b1, b2∈ℝ. For each of the following systems, determine whether the system is consistent for all b1, b2, b3∈ℝ. If not, find conditions on b1, b2, b3 such that the system is consistent. a. { x1−x2+2x3+x4=b12x1+4x2−4x3−x4=b23x1+3x2−2x3=b3 b. { x1−x2+2x3=b1−3x1+2x2+x3=b24x1−3x2+x3=b3. c. { x1+x2+2x3=b1x1+x3=b23x1+4x2+7x3=b3 Solution a. (4.5.4) Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/12 8/30/2020 4.5 Consistency of systems of linear equations (A|b→)= (1−121b124−4−1b233−20b3)→R1(−3)+R3R1(−2)+R2(1−121b106−8−3b2−2b106−8−3b3−3b1)→R2(−1)+R3(1−121b106−8−3b2−2 ut io n From the left side of the last matrix in (4.5.4) , we have r(A)=2<3=m. By Theorem 4.5.4 (2), the system is not consistent for all b1, b2, b3∈ℝ. To choose b1, b2, b3∈ℝ such that the system with these b1, b2, b3 are consistent, by Corollary 4.5.3 , we need r(A)= (A|b→). By (4.5.4) , we see that if b3−b2−b1=0, then r(A)=(A|b→). Hence, the condition on b1, b2, b3 is b3−b2−b1=0, that is, when b3=b1+b2, the system is consistent. b. (4.5.5) (A|b→)= (1−12b1−321b24−31b3)→R1(−4)+R3R1(3)+R2(1−12b10−17b2+3b101−7b3−4b1)→R2(1)+R3(1−12b10−17b2+3b1000b3+b2−b1). Exercises , if −4b1+b2+b3=0, the system is consistent. N ot By Theorem 4.5.1 fo rD is tri b From the left side of the last matrix in (4.5.5) , we have r(A)=2<3=m. It follows from Theorem 4.5.4 (2) that the system is not consistent for all b1, b2, b3∈ℝ. By (4.5.5) , we see that if b3+b2−b1=0, then r(A)=2=(A|b→). Hence, when b3+b2−b1≠0, the system is consistent. c. (A|b→)= (112b1101b2347b3)→R1(−3)+R3R1(−1)+R2(112b10−1−1b2−b1011b3−3b1)→R2(1)+R3(112b1011b1−b2000−4b1+b2+b3)= (B|c→). 1. For each of the following systems, determine whether it has a unique solution, infinitely many solutions, or no solutions. a. { x1−2x2+3x3=62x1+x2+4x3=5−3x1+x2−2x3=3 b. { x1+x2−2x3=32x1+3x2−x3=−4−2x1−3x2x3=4 c. { x1+2x2+3x3=44x1+7x2+6x3=172x1+5x2+12x3=7 d. { x1−2x2−x3+2x4=0−x2−2x3+x4=0 e. { x1−x2+3x3=2−2x1+2x2−6x3=52x1−2x2+6x3=4 f. { x1−x2−3x3−4x4=1−x1+x2−x3=−1 2. For each of the following systems, determine if it has a unique solution. a. { 2x1−3x2−x3−5x4=−1−3x1+6x2−2x3+x4=3x1+x4=−2 Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/12 8/30/2020 4.5 Consistency of systems of linear equations b. { x1−3x2+2x3=−1x1+4x2−x3=2 4. Consider the following system { 6x1−4x2=b13x1−2x2=b2 fo rD is tri b 1. Find b1 and b2 such that the system is consistent. 2. Find b1 and b2 such that the system is inconsistent. 3. Determine whether the system with b→=(4, 1)T is inconsistent. ut io n 3. For each of the following homogeneous systems, determine whether it has a unique solution or infinitely many solutions. a. { x1−2x2−x3=0−x1+3x2+2x3=0−x1+x2=0x2+x3=0. b. { x1+x2−2x3=02x1−x2+4x3=0−x1−2x2−3x3=0. c. { x1−x2+x3=02x2−4x3+x3=0. d. { x1−x2+2x3=0−2x1−2x2+x3=0−x1−3x2+3x3=03x1+x2+x3=0. 5. Determine if the following system is consistent for all b1, b2, b3∈ℝ. { x1−x2+2x3+x4=b12x1+4x2−4x3=b2−3x1+4x3−x4=b3 6. Find conditions on b1, b2, b3 such that the following system is consistent. { x1−x2+2x3=b12x1+4x2−4x3=b2−3x1−2x3=b3 N ot 7. Determine whether the following system has infinitely many solutions for all b1, b2∈ℝ. { x1−2x2+2x3=b1−3x1+3x2+x3=b2 Processing math: 30% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/12 8/30/2020 4.6 Linear combination, spanning space and consistency 4.6 Linear combination, spanning space and consistency → ut io n From Theorem 4.1.1 , we see three equivalent assertions about linear systems, linear combinations and spanning spaces. By Examples 1.4.2 , 1.4.4 , 4.2.1 , 4.2.2 , and 4.2.3 , we see that we need to → solve the system AX = b to determine whether one of the three assertions holds. However, Corollary 4.5.3 provides another method to justify whether these assertions hold by comparing the two ranks, r(A) | → → → Combining Theorem 4.1.1 and Corollary 4.5.3 D is tri b and r(A b), without solving the system AX = b , we obtain the following criteria. Theorem 4.6.1. { } → → be the same as in A = (a 1, ⋯, a n) be the same as in (4.1.5) , and fo r Let S : = → → a 1, ⋯, a n → b = (b 1, b 2, ⋯, b m) T. Then the following assertions are equivalent. → → → 2. b is a linear combination of S. → 3. b ∈ span S. | → 4. r(A) = r(A b). By Theorem 4.6.1 N ot 1. AX = b is consistent. → → | | → → → (4), we see that if b = 0, then r(A) = r(A 0) = r(A b). It follows that 0 is a linear combination of S. This provides another proof of Theorem 1.4.1 . Processing math: 100% Example 4.6.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/11 8/30/2020 4.6 Linear combination, spanning space and consistency { } → → → and A = (a 1, a 2, a 3), where → a1 = → () ( ) () ( ) 1 → a2 = 2 , 3 −2 1 , −1 → → a3 = 3 4 , 7 6 → b= 5 . 11 → 1. Determine whether AX = b is consistent, where X = (x 1, x 2, x 3) T. → 2. Determine whether the vector b is a linear combination of S. D is tri b → 3. Determine whether the vector b ∈ span S. Solution | → We find r(A) and r(A b). → ) ( ( ) 1 −2 3 6 R ( −2) +R 1 −2 3 6 1 2 2 1 4 5 0 5 −2 −7 → 3 − 1 7 11 R 1 ( − 3 ) + R 3 0 5 − 2 − 7 N ot ( fo r | (A b) = R2 ( − 1 ) + R3 → | → Hence, r(A) = r(A b) = 2. By Theorem 4.6.1 ut io n Let S = → → → a 1, a 2, a 3 1 −2 3 ) 6 0 5 −2 −7 . 0 0 0 0 , we have the following conclusions: Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/11 8/30/2020 4.6 Linear combination, spanning space and consistency → → 1. The system AX = b is consistent. → 2. The vector b is a linear combination of S. → 3. The vector b ∈ span S. Example 4.6.2. Let S and A be the same as in Example 4.6.1 → → ut io n → and let b 1 = (6, 5, 10) T. → D is tri b 1. Determine whether AX = b is consistent, where X = (x 1, x 2, x 3) T. → 2. Determine whether the vector b is a linear combination of S. → 3. Determine whether the vector b 1 ∈ span S. Solution | → We find r(A) and r(A b). ( ) ( ( ) 2 1 4 5 3 − 1 7 10 R1 ( − 2 ) + R2 R2 ( − 1 ) + R3 → 1 −2 3 6 → 0 5 −2 −7 R1 ( − 3 ) + R3 0 5 −2 −8 N ot (A b) = 1 −2 3 6 fo r | → 1 −2 3 ) 6 0 5 −2 −7 . 0 0 0 −1 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/11 8/30/2020 4.6 Linear combination, spanning space and consistency | | → → Hence, r(A) = 2 and r(A b 1) = 3. Because r(A) < r(A b 1), by Theorem 4.6.1 , we have the conclusions. → 1. The system AX = b 1 is inconsistent. → ut io n → 2. The vector b is not a linear combination of S. → 3. The vector b ∉ span S. Let S = D is tri b Example 4.6.3. {( ) ( )} 6 3 , −4 −2 . → 1. Determine whether b = (2, 1) is in span S. → Solution → N ot Let b = (b 1, b 2) T. fo r 2. Determine whether b = (1, 2) is in span S. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/11 8/30/2020 4.6 Linear combination, spanning space and consistency | → (A b) = ( ) () 1 R1 6 − 4 b1 3 − 2 b2 6 → ( ) 2 b1 1 −3 6 R1 ( − 3 ) + R2 → 3 − 2 b2 ( ) b1 6 b1 ut io n 2 1 −3 . b1 | → D is tri b 0 0 b2 − 2 1. If b 2 − 2 = 0, then r(A) = (A b) and by Theorem 4.6.1 b1 → 2 → → , b ∈ span S. Hence, if b = (b 1, b 2) = (2, 1), then b 2 − 2 = 1 − 2 = 0 and b = (2, 1) ∈ span S. b1 | → 2. If b 2 − 2 ≠ 0, then r(A) < (A b) and by Theorem 4.6.1 → b1 1 3 → , b ∉ span S. Hence, if → fo r b = (b 1, b 2) = (1, 2), then b 2 − 2 = 2 − 2 = 2 ≠ 0 and b = (1, 2) ∉ span S. N ot Example 4.6.4. → → → → T T Let a 1 = (1, 1, 3) , a 2 = (1, 0, 4) , a 3 = (2, 1, 7) T, and b = (b 1, b 2, b 3) T. { { } } → → → 1. Find conditions on b 1, b 2, b 3 such that b ∈ span a 1, a 2, a 3 . → → → → → 2. Find conditions on b , b , b such that b ∉ span a 1, a 2, a 3 . 1 2 3 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/11 8/30/2020 4.6 Linear combination, spanning space and consistency { } → → → 3. Find a vector b = (b 1, b 2, b 3) such that b ∉ span a 1, a 2, a 3 . → → T Solution →→→ Let b = (b 1, b 2, b 3) and A = (a 1 a 2 a 3). ( ) ( ) ( ) | | → 1 0 1 b2 3 4 7 b3 R2 ( 1 ) + R3 → 1 1 R1 ( − 1 ) + R2 → R1 ( − 3 ) + R3 2 b1 0 − 1 − 1 b2 − b1 D is tri b 1 1 2 b1 (A b) = ut io n → 0 1 1 1 2 b1 0 1 1 b1 − b2 1 b 3 − 3b 1 = (B → c). fo r 0 0 0 − 4b 1 + b 2 + b 3 | → { } → → → b ∈ span a 1, a 2, a 3 . → N ot 1. If − 4b 1 + b 2 + b 3 = 0, that is, b 3 = 4b 1 − b 2, then r(A) = (A b). By Theorem 4.6.1 | → 2. If b 3 ≠ 4b 1 − b 2 ≠ 0, then r(A) < (A b). By Theorem 4.6.1 { , } → → → , b ∉ span a 1, a 2, a 3 . → { } → → → Processing math: 100% 3. Let b 1 = b 2 = 0 and b 3 = 1. Then b 3 ≠ 4b 1 − b 2. Hence b = (0, 0, 1) ∉ span a 1, a 2, a 3 . → https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/11 8/30/2020 4.6 Linear combination, spanning space and consistency Theorem 4.6.1 → provides a criterion to determine whether a given vector b belongs to span S. In the → following, we study how to determine whether every vector b in ℝ m belongs to span S, that is, to determine whether span S = ℝ m. By (2.2.1) Theorem 4.5.4 , we have span S = A(ℝ n). This, together with → → (1), implies that span S = A(ℝ n) = ℝ m if and only if the system AX = b is consistent for every → ut io n b ∈ ℝ m. This leads us to obtain the following criterion to determine whether span S = ℝ m by using ranks. Theorem 4.6.2. The following assertions are equivalent. D is tri b 1. span S = ℝ m 2. The system AX = b is consistent for every vector b in ℝ m. 3. r(A) = m. → → → Proof By Theorems 4.5.4 and 4.6.1 → → , span S = ℝ m if and only if every vector b in ℝ n belongs to span S if → → fo r and only if the system AX = b is consistent for every vector b in ℝ n if and only if r(A) = m. Example 4.6.5. N ot Determine whether span S = ℝ 3, where {( ) ( ) ( )} 1 S= −4 , 4 −2 5 −5 1 , 2 . 3 Solution Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/11 8/30/2020 4.6 Linear combination, spanning space and consistency ( ) ( ) ( ) 1 −2 1 R (4) +R 1 2 −4 5 2 → A = 1 −2 −1 R (1) +R 2 3 0 −3 6 → 4 − 5 3 R1 ( − 4 ) + R3 0 3 − 1 1 −2 −1 0 0 . 5 Example 4.6.6. Determine whether span span S = ℝ 3, where {( ) ( ) ( ) ( )} 1 0 0 , 1 1 , 0 −2 0 , 3 . −3 fo r −1 −2 N ot S= D is tri b , span S = ℝ 3. Hence, r(A) = 3. By Theorem 4.6.2 Solution ut io n 0 −3 6 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/11 8/30/2020 4.6 Linear combination, spanning space and consistency A = ( 1 0 −2 0 0 1 0 3 1 −1 −2 −3 ) ( R1 ( − 1 ) + R3 → 1 0 −2 0 0 1 0 3 0 −1 0 −3 ) R2 ( 1 ) + R3 → ( ) 1 0 −2 0 , span S ≠ ℝ 3. D is tri b Hence, r(A) = 2. Because r(A) = 2 < 3, by Theorem 4.6.2 Exercises } → a1 = () ( ) () ( ) () 1 2 , 3 2 → b= → → a2 = fo r { → → → → and A = (a 1, a 2, a 3, a 4), where −2 1 −1 , → a3 = N ot 1. Let S = → → → → a 1, a 2, a 3, a 4 ut io n 0 1 0 3 . 0 0 0 0 3 4 , 7 → a1 = −1 −4 , −5 2 . 4 → → 1. Determine whether AX = b is consistent, where X = (x 1, x 2, x 3, x 4) T. → Processing math: 100% 2. Determine whether the vector b is a linear combination of S. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/11 8/30/2020 4.6 Linear combination, spanning space and consistency → 3. Determine whether the vector b ∈ span S. → → → → T T 2. Let a 1 = (1, − 2, 2) , a 2 = ( − 1, 0, − 10) , a 3 = (1, − 1, 6) T, and b = (b 1, b 2, b 3) T. { { { } } } → → → 2. Find conditions on b 1, b 2, b 3 such that b ∉ span a 1, a 2, a 3 . → 3. Let S = {( ) ( )} 2 1 −4 , −2 D is tri b → → → → → 3. Find a vector b = (b 1, b 2, b 3) T such that b ∉ span a 1, a 2, a 3 . . 1. Determine whether span S = ℝ 2. → ut io n → → → → 1. Find conditions on b 1, b 2, b 3 such that b ∈ span a 1, a 2, a 3 . fo r 2. Determine whether b = (4, 2) T is in span S. → 3. Determine whether b = (6, 1) T is in span S. i. S = ii. S = iii. S = Processing math: 100% N ot 4. Determine whether span S = ℝ 2, where {( ) ( ) ( ) ( )} {( ) ( ) ( )} {( ) ( ) ( ) ( )} −1 0 1 −1 0 1 , 1 , 0 2 , −2 2 1 1 , , −2 3 , 5 −5 , . 5 3 −2 , . 4 1 . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/11 8/30/2020 4.6 Linear combination, spanning space and consistency 5. Determine whether span S = ℝ 3, where {( ) ( ) ( )} 2 3 2 , 1 5 , 0 4 . 5 ut io n S= 1 6. Determine whether span S = ℝ 4, where {( ) ( ) ( )} S= 0 4 , 0 0 −1 1 , 2 3 . 4 N ot fo r 1 1 D is tri b 1 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/11 8/30/2020 Chapter 5 Linear transformations ut io n Chapter 5 Linear transformations 5.1 Linear transformations D is tri b In this section, we use the product of a matrix and a vector studied in Section 2.4 to define linear transformations. Definition 5.1.1. Let A be an m × n matrix given in (2.1.1) defined by . A linear transformation from Rn to Rm , denoted by T, is fo r (5.1.1) a11 a12 ⋯ a1n ⎜ a21 → → ⎜ T (X ) = AX = ⎜ ⎜ ⎜ ⋮ a22 ⋯ ⎝ where → X = (x1 , x2 , ⋯ , xn ) T n ) ∈ R am1 ⋮ ⋱ an2 ⋯ amn → . ⎞ ⎛ x1 ⎞ a2n ⎟ ⎟ ⎟ ⎟ ⎟ ⋮ N ot ⎛ ⎜ x2 ⎜ ⎜ ⎜ ⋮ ⎠⎝ xn ⎟ ⎟, ⎟ ⎟ ⎠ The vector T ( X ) is said to be the image of → X under T and → X is → called the inverse image of T ( X ). The space Rn is the domain of T and Rm is the codomain of T. The matrix A is said to be the standard matrix for T and T is called the multiplication by A. A linear transformation from Rn to Rn is said to be a linear operator. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/11 8/30/2020 Chapter 5 Linear transformations If T is a linear transformation from Rn to Rm , then we say that T maps Rn into Rm , which is denoted by the symbol T : Rn → Rm . When m = n = 1, the linear transformation is a function T : R → R defined by T (x) = a11 x. Hence, linear transformations are generalizations of such functions to higher dimensional spaces. ut io n By Definition 5.1.1 , we see that every matrix determines a linear transformation. Hence, for every linear transformation, there is a unique matrix corresponding to it. Sometimes, we write T as TA in order to emphasize that the linear transformation is determined by the standard matrix A. Example 5.1.1. A1 = ( determine the domain and codomain of TA 2 3 1 0 1 1 ⎛ ) , A2 = ⎜ −1 ⎝ , i = 1, 2. 3 0 1 1 2⎟, −2 0 Moreover, compute TA 1 ⎞ ⎠ (1, 1, 1) and TA 2 (−1, 0, 1) . fo r i 1 D is tri b For each of the following matrices Solution N ot Because the size of A1 is 2 × 3, the domain and codomain of TA are R3 and R2 , respectively. ⎛ 1 ⎞ TA1 ⎜ 1 ⎟ = ( ⎝ 1 ⎠ 1 1 2 3 1 0 1 ⎛ 1 ⎞ )⎜1⎟ = ( ⎝ 1 ⎠ 6 ). 2 Because the size of A2 is 3 × 3, both the domain and codomain of TA are R3 . 2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/11 8/30/2020 Chapter 5 Linear transformations ⎛ TA2 ⎜ ⎝ −1 0 1 ⎞ 1 ⎛ ⎟ = ⎜ −1 ⎠ ⎝ 3 0 1 1 2⎟⎜ −2 0 ⎞⎛ ⎠⎝ −1 0 1 ⎞ ⎛ ⎟ = ⎜ ⎠ ⎝ 0 ⎞ ⎟. 3 −3 ⎠ Example 5.1.2. 4 7 Let A = ⎜ 2 5 8⎟. 6 9 ⎝ 3 ⎞ → → → Find T ( e1 ), T ( e2 ), T ( e3 ), where → → → e1 , e2 , e3 are the standard vectors in R3 . ut io n 1 ⎛ ⎠ 1 4 7 T ⎜0⎟ = ⎜2 5 8 ⎟⎜0⎟ = ⎜2⎟. 3 6 9 1 4 7 T ⎜1⎟ = ⎜2 5 8 ⎟⎜1⎟ = ⎜5⎟. 3 6 9 1 4 7 T ⎜0⎟ = ⎜2 5 8 ⎟⎜0⎟ = ⎜8⎟. 6 9 ⎛ = ⎝ → T ( e1 ) ⎛ = ⎝ → T ( e1 ) ⎛ = 0 0 0 0 ⎞ ⎛ ⎠ ⎝ ⎞ ⎛ ⎠ ⎞ ⎝ ⎠ ⎛ ⎝ N ot ⎝ 1 1 ⎞⎛ ⎠⎝ ⎞⎛ ⎠⎝ fo r → T ( e1 ) D is tri b Solution 3 ⎞⎛ ⎠⎝ 1 0 0 0 0 1 ⎞ ⎛ ⎠ ⎝ ⎞ ⎛ ⎠ ⎝ ⎞ ⎛ ⎠ ⎝ 1 3 4 6 7 9 ⎞ ⎠ ⎞ ⎠ ⎞ ⎠ By Example 5.1.2 , we see that the column vectors of the standard matrix A are the images of the standard vectors in R3 under T. In fact, the result holds for any linear transformations. Theorem 5.1.1. → → Let T : Rn → Rm be a linear transformation and e1 , ⋯ , en be the standard vectors in Rn given in (1.1.1) . Then the standard matrix A of T is given by https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/11 8/30/2020 Chapter 5 Linear transformations → → → A = (T ( e 1 ), T ( e 2 ), ⋯ , T (e n )). By (5.1.1) , we obtain the following expressions for linear transformations. (5.1.2) → T (X ) = ⎜ a21 x1 + ⎜ ⎜ ⎜ ⎞ a22 x2 + ⋯ + a2n xn ⎟ ⎟ , ⎟ ⎟ ⋮ am1 x1 + am2 x2 + ⋯ + amn xn ⎠ D is tri b ⎝ a12 x2 + ⋯ + a1n xn ut io n a11 x1 + ⎛ or (5.1.3) T (x1 , ⋯ , xn ) = (a11 x1 + ⋯ + a1n xn , ⋯ , am1 x1 + ⋯ + amn xn ). fo r Example 5.1.3. a. T (x, y, z) = ( 12 x, y, 4z); b. T (x, y, z) = ( 12 x + 1 2 z, 1 3 y+ . N ot For each of the following linear transformations, express it in form (5.1.1) 1 3 z). Solution ⎛ a. T x 1 ⎞ ⎛ 2 ⎜y ⎟ = ⎜ 0 ⎝ z ⎠ ⎝ 0 0 0 1 0⎟⎜y ⎟; 0 4 ⎞⎛ ⎠⎝ x z ⎞ ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/11 8/30/2020 Chapter 5 Linear transformations ⎛ b. T x 1 ⎞ ⎜y ⎟ = ( ⎝ z 2 0 ⎠ 1 0 2 1 1 3 3 ⎛ x ⎞ )⎜y ⎟. ⎝ z ⎠ Definition 5.1.2. n : R n → R satisfying (5.1.4) → → → n for each X ∈ R D is tri b T (X ) = kX ut io n A linear operator T is called a contraction with factor k if 0 ≤ k < 1, and a dilation with factor k if k ≥ 1 . By Definition 5.1.2 matrix kIn . , we see that the standard matrix for a contraction or dilation with factor k is a diagonal , we obtain the following properties. Proposition 5.1.1. n : R m → R be a linear transformation and let → → X , Y N ot Let T fo r By Theorem 2.2.1 (5.1.5) → → n ∈ R → and α, β ∈ R. Then → T (α X + β Y ) = αT ( X ) + βT ( Y ). Proof Let A be the m × n standard matrix of T. Then by Definition 5.1.1 → → T(Y ) = AY . By Theorem 2.2.1 → → , we have T ( X ) = A X and , we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/11 8/30/2020 Chapter 5 Linear transformations → → → → → → → → T (α X + β Y ) = A(α X + β Y ) = α(A X ) + β(A Y ) = αT ( X ) + βT ( Y ) and (5.1.5) holds. Example 5.1.4. → → X = (1, 1, −1), Y and = (−1, 2, 1), 0 1 A = ⎜1 2 ⎛ → Compute TA (3 X → − 2Y ) 1 → . , we have → → TA (3 X − 2 Y ) = 3TA ( X ) − 2TA ( Y ) 1 = 3⎜1 2 ⎝ ⎛ = 3⎜ ⎝ 1 1 −1 0 4 −3 ⎞⎛ 1 0 1 ⎟ − 2⎜1 2 ⎞ ⎛ N ot 0 ⎛ ⎠ 3 fo r → −1 ⎟ . −1 Solution By Proposition 5.1.1 ⎞ D is tri b ⎝ 1 ut io n Let ⎞ −1 ⎟ ⎜ 3 ⎠⎝ ⎛ 3 ⎞ 1 −1 ⎠ ⎛ 0 ⎝ ⎞ 1 −1 ⎛ 6 ⎞ 1 ⎞⎛ −1 ⎟ ⎜ 3 ⎠⎝ ⎛ ⎟ − 2 ⎜ 2 ⎟ = ⎜ 12 ⎟ − ⎜ 4 ⎟ = ⎜ ⎠ ⎝ 0 ⎠ ⎝ −9 ⎠ ⎝ 0 ⎠ ⎝ −6 8 9 −1 2 1 ⎞ ⎟ ⎠ ⎞ ⎟. ⎠ Now, we introduce operations on linear transformations. Definition 5.1.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/11 8/30/2020 Chapter 5 Linear transformations 1. If S, T n : R m → R i. (Addition) (T are linear transformations, then we define the following operations: → → → + S)( X ) = T ( X ) + S( X ) ii. (Subtraction) (T → → . → − S)( X ) = T ( X ) − S( X ) → . → iii. (Scalar multiplication) (αT )( X ) = α(T ( X )) for α ∈ R . m : R are linear transformations, then we define the following l → R → → D is tri b (ST )( X ) = S(T ( X )). ut io n 2. If T : Rn → Rm and S composition of T and S: (5.1.6) By Definition 5.1.1 and the above operations, we see that if A and B are the standard matrices for TA , TB : R → R , respectively, then A + B, A − B, αA are the standard matrices for TA + TB , TA − TB , αTA , respectively, that is, (ii)′ (iii)′ → → (TA + TB )( X ) = (A + B)( X ) → → (TA − TB )( X ) = (A − B)( X ) → → (αTA )( X ) = (αA)( X ) . fo r (i)′ m . for α ∈ R . If A and B are the standard matrices for TA standard matrix for TB TA , that is, N ot n n : R m → R and TB → m : R l → R , respectively, then BA is the → (TB TA )( X ) = (BA)( X ). Note that the standard matrix for TB TA is BA instead of AB and the composition linear transformation TB TA is defined under the condition that the domain of TB must be the same as the codomain of TA . Example 5.1.5. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/11 8/30/2020 Chapter 5 Linear transformations Compute (2T − 3S)(x1 , x2 ), where T (x1 , x2 ) = (3x1 − 2x2 , −x1 − x2 , 2x1 + 3x2 ), S(x1 , x2 ) = (x1 + x2 , x1 − x2 , −2x1 + 4x2 ). (Method 1). We write T and S as follows. ⎛ 3 ) = ⎜ −1 x2 ⎝ 2 −2 ⎞ −1 ⎟ ( 3 x1 ) and S ( x2 ⎠ x1 ) = x1 2T ( x2 ) − 3S ( x2 ⎛ = 3 ⎝ ⎞ −1 ⎟ ( 2 3 3 ⎠ −2 N ot ⎛ = [2 ⎜ −1 ⎝ ⎛ = ⎜ −5 ⎝ (Method 2) By Definition 5.1.3 3 10 1 ⎞ −1 ⎟ ( 1 −2 x1 ). x2 ⎠ 4 x1 ) x2 −2 2 ⎜ −1 ⎝ x1 ) − 3⎜ x2 ⎞ ⎛ −1 ⎟ − 3 ⎜ 2 3 −7 1 −6 ⎠ ⎞ ⎟( ⎠ ⎝ x1 x2 1 ⎛ fo r (2T − 3S) ( ) = ⎜ x2 Then 1 ⎛ x1 D is tri b T ( x1 ut io n Solution ⎝ 1 1 −2 ⎛ 1 −2 1 1 −1 ⎟ ( −1 ⎟] ( 4 ⎠ 4 ⎞ ⎠ ⎞ x1 x1 ) x2 ) x2 3x1 − 7x2 ⎞ ) = ⎜ −5x1 + x2 ⎟ . ⎝ 10x1 − 6x2 ⎠ , we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/11 8/30/2020 Chapter 5 Linear transformations (2T − 3S)(x1 , x2 ) = 2T (x1 , x2 ) − 3S(x1 , x2 ) = 2(3x1 − 2x2 , −x1 − x2 , 2x1 + 3x2 ) − 3(x1 + x2 , x1 − x2 , −2x1 + 4x2 ) = (6x1 − 4x2 , −2x1 − 2x2 , 4x1 + 6x2 ) − (3x1 + 3x2 , 3x1 − 3x2 , −6x1 + 12x2 ) = (3x1 − 7x2 , −5x1 + x2 , 10x1 + 6x2 ). ut io n Example 5.1.6. Let TA , TB be linear transformations given by TA (x1 , x2 ) = (x1 + x2 , 2x1 − 3x2 , 3x1 − 2x2 ) D is tri b and TB (x1 , x2 , x3 ) = (x1 − x2 + x3 , x1 − 2x3 ). Find (TB TA )(x1 , x2 ) . Because ⎛ ⎛ ) = ⎜2 x2 x1 ⎝ ⎞ TB ⎜ x2 ⎟ = ( ⎝ 1 1 x3 ⎠ ⎞ −3 ⎟ ( N ot TA ( x1 fo r Solution ⎠ 3 −2 1 −1 1 1 0 −2 x1 ) and x2 ⎛ x1 ⎞ ) ⎜ x2 ⎟ , ⎝ x3 ⎠ we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/11 8/30/2020 Chapter 5 Linear transformations (TB TA ) ( x1 ) = ( x2 −1 1 1 0 −2 2 = ⎛ 1 2 ( )( −5 5 −2 2 1 1 1 −3 ⎟ 1 0 −2 1 )⎜2 x1 ⎝ 3 1 ⎞ −3 ⎟ ( −2 ⎠ ). ut io n 1. For each of the following matrices, 4 −1 2 3 0 0 1 2 −1 1 −2 0 3 1 −2 3 2 −1 A1 = ⎜ −1 ⎝ ⎛ ⎜ −2 A3 = ⎜ ⎜ 3 ⎝ 1 ⎞ ⎛ ⎟ A2 = ⎜ ⎠ ⎝ −1 1 ⎜ 2 A4 = ⎜ ⎜ −1 0 ⎞ ⎟ ⎟ ⎟ ⎛ ⎠ ⎝ −1 0 determine the domain and codomain of TA i 1 ⎞ D is tri b −2 ⎠ ⎞ −2 ⎟ ⎟ 1 ⎟ 0 1 ⎠ fo r 2 ) x2 x2 Exercises ⎛ x1 , i = 1, 2, 3, 4. Moreover, compute TA3 (0, 1, 1, 2), TA2 (1, −2, 1), TA3 (−1, 0, 1), TA4 (−1, 1, 1). 3. Let A = ( 1 2 −1 3 −1 2 1 3 0 6 ). Find TA ( ). 1 0 0 N ot 2. Let A = ( ) → and TA ( → ) . 1 → Find TA ( e1 ), TA ( e2 ), TA ( e3 ), where → → → e1 , e2 , e3 are the standard vectors in . 4. For each of the following linear transformations, express it in (5.1.1) . a. T (x1 , x2 ) = (x1 − x2 , 2x1 + x2 , 5x1 − 3x2 ). b. T (x1 , x2 , x3 , x4 ) = (6x1 − x2 − x3 + x4 , x2 + x3 − 2x4 , 3x3 + x4 , 5x4 ). c. T (x1 , x2 , x3 ) = (4x1 − x2 + 2x3 , 3x1 − x3 , 2x1 + x2 + 5x3 ). d. T (x1 , x2 , x3 ) = (8x1 − x1 + x3 , x2 − x3 , 2x1 + x2 , x2 + 3x3 ). 3 R https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/11 8/30/2020 Chapter 5 Linear transformations 5. Let → → X = (2, −2, −2, 1), Y = (1, −2, 2, 3) and ⎛ A = ⎜ ⎝ → 1 0 1 0 1 −2 1 1⎟. −1 1 2 0 ⎞ ⎠ → be linear transformations such that T (1, 2, 3) = (1, 1, 0, −1) and S(1, 2, 3) = (2, −1, 2, 2). Compute (3T − 2S)(1, 2, 3) . 8. Compute (T + 2S)(x1 , x2 , x3 , x4 ), where 3 : R 4 → R D is tri b 7. Let S, T ut io n Compute TA ( X − 3 Y ) . 6. Let S, T : R3 → R3 be linear operators such that T (1, 2, 3) = (1, −4) and S(1, 2, 3) = (4, 9). Compute (5T + 2S)(1, 2, 3) . T (x1 , x2 , x3 , x4 ) = (x1 − x2 + x3 , x1 − x2 + 2x3 − x4 ), S(x1 , x2 , x3 , x4 ) = (x2 − x3 + 2x4 , x1 + x3 − x4 ). + 3x2 + x3 ). N ot fo r 9. For each pair of linear transformations TA and TB , find TB TA . a. TA (x1 , x2 ) = (x1 + x2 , x1 − x2 ), TB (x1 , x2 ) = (x1 − x2 , 2x1 − 3x2 , 3x1 + x2 ). b. TA (x1 , x2 , x3 ) = (−x1 − x2 + x3 , x1 + x2 ), TB (x1 , x2 ) = (x1 − x2 , 2x1 + 3x2 ). c. TA (x1 , x2 , x3 ) = (−x1 + x2 − x3 , x1 − 2x2 , x3 ), TB (x1 , x2 , x3 ) = (x1 − x2 + x3 , 2x1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/11 8/30/2020 5.2 Onto, one to one, and inverse transformations 5.2 Onto, one to one, and inverse transformations Let T : ℝ n → ℝ m be a linear transformation with standard matrix A given in (5.1.1) → . We collect all images → together as a set denoted by T(ℝ ), that is, all the images T(X) as X takes all the vectors in ℝ n. ut io n n Definition 5.2.1. (5.2.1) T(ℝ n) = {T(X) ∈ ℝ → m By Definitions 5.1.1 and 5.2.1 → : X ∈ ℝn and Theorem 2.2.2 , we obtain the following result, which Let S = { → → → a 1, a 2, ⋯, a n } and the spanning space given in N ot establishes the relations among the range of T, A(ℝ n) given in (2.2.1) (1.4.8) . Theorem 5.2.1. } fo r is called the range of T. D is tri b Let T : ℝ n → ℝ m be a linear transformation. The set be the set of the column vectors of A given in (2.1.2) Then (5.2.2) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/12 8/30/2020 5.2 Onto, one to one, and inverse transformations T A(ℝ n) = A(ℝ n) = span S. By Theorems 4.6.1 and 5.2.1 → , we obtain the following result, which uses the ranks of r(A) and → r(A | b) to determine whether b ∈ T(ℝ n). Let S be the same as in Theorem 5.2.1 ut io n Theorem 5.2.2. . Then the following assertions are equivalent. → i. b ∈ T(ℝ n); → → ii. AX = b is consistent; → D is tri b iii. b is a linear combination of S; → iv. b ∈ span S; → v. r(A) = r(A | b), Example 5.2.1. fo r Let → → Solution N ot T(x 1, x 2, x 3) = (x 1 − x 2, − 2x 1 + 2x 2 + x 3, − x 1 + x 2 + x 3) and b = (, , 0, − 1). Determine whether b ∈ T(ℝ 3). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/12 5.2 Onto, one to one, and inverse transformations A= ( ( 1 −1 0 1 −2 2 1 0 ) ( ) R1 ( 2 ) + R2 → − 1 1 1 − 1 R1 ( 1 ) + R3 1 −1 0 1 0 0 1 2 0 0 0 −2 ) 1 −1 0 1 0 0 1 2 R ( − 1) + R → 2 3 0 0 1 0 . → , b ∉ T(ℝ 3). → D is tri b This implies r(A) = 2 < 3 = r(A | b). By Theorem 5.2.2 ut io n 8/30/2020 Definition 5.2.2. A linear transformation T : ℝ n → ℝ m is said to be onto if T(ℝ n) = ℝ m. By Theorem 5.2.1 , T is onto if and only if span S = ℝ m. By Theorem 4.6.2 result, which uses the rank of A to determine whether T is onto. fo r Theorem 5.2.3. , we obtain the following i. T is onto. ii. span S = ℝ m. → → N ot Let T : ℝ n → ℝ m be a linear transformation with standard matrix A. Then the following assertions are equivalent. → iii. The system T(X) = b is consistent for each b ∈ ℝ m. iv. r(A) = m. By Theorem 2.5.2 , r(A) ≤ min{m, n}. Hence, if T : ℝ n → ℝ m is onto, then by Theorem 5.2.3 , m = r(A) ≤ min{m, n} ≤ n. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/12 8/30/2020 5.2 Onto, one to one, and inverse transformations Hence, the necessary condition for T to be onto is m ≤ n and if m > n, then T is not onto. Example 5.2.2. a. T(x 1, x 2, x 3, x 4) = (x 1 + 2x 2 − 2x 3 + x 4, 3x 2 + 4x 3, x 1 + 2x 3 + x 4). b. T(x 1, x 2, x 3, ) = (x 1 + x 2, − x 1 + 2x 2 + 4x 3, 2x 1 − x 2 + 3x 3, 5x 1 + 3x 2 + 6x 3). Solution ( 1 1 0 2 2 ) ( ) ) ( ) 1 3 0 −2 −2 1 R (1) +R 3 2 4 0 → 4 −2 1 R (2) +R 1 2 −2 1 3 0 1 8 0 0 1 8 0 . → 0 −2 4 0 0 0 20 0 D is tri b a. ( 2 0 fo r A = 1 2 −2 1 R ( −1) +R 1 1 3 0 3 4 0 0 → ut io n Determine which of the following linear transformations are onto. In (4.1.13) N ot Hence, r(A) = 3 = m. By Theorem 5.2.3 , T is onto. b. Because the standard matrix A of T is a 4 × 3 matrix, by Theorem 5.2.3 r(A) ≤ 3 = min{4, 3} < 4 = m. → , T is not onto because → , we introduce the solution space N A of a homogeneous system AX = 0. For a linear → → transformation T, the solution space of TX = 0 is called a kernel for T and is denoted by ker(T), that is, (5.2.3) ker(T) = {X ∈ ℝ : T(X) = 0 }. → n → → https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/12 8/30/2020 5.2 Onto, one to one, and inverse transformations → The kernel ker(T) is the set of all inverse images of zero vector 0 in ℝ m under T. If A is the standard matrix of T, then by (4.1.13) and Definition 5.1.1 , we have (5.2.4) Definition 5.2.3. ut io n ker(T A) = N A. A linear transformation T : ℝ n → ℝ m is said to be one to one if T(X) = 0 implies X = 0 . → → → → → → → → → → → D is tri b Because T(X) = AX, if T is one to one, then AX = 0 implies X = 0. That is, T one to one if and only if the → system AX = 0 only has the zero solution. The latter can be justified by Theorem 4.5.2 (1). Hence, we obtain the following criterion, which uses the rank of A to determine whether T is one to one. Theorem 5.2.4. i. T is one to one. → → N ot ii. AX = 0 only has the zero solution. iii. r(A) = n. fo r Let T : ℝ n → ℝ m be a linear transformation with standard matrix A. Then the following assertions hold. If T : R n → ℝ m is one to one, then by Theorems 2.5.2 and 5.2.4 , n = r(A) ≤ min{m, n} ≤ m. Hence, what’s necessary for T to be one to one is n ≤ m and if n > m, then T is not one to one. Example 5.2.3. Determine which of the following linear transformations are one to one. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/12 8/30/2020 5.2 Onto, one to one, and inverse transformations a. T(x, y) = (x − y, 2x + y, 4x − y). b. T(x, y) = (x − 2y, 2x − 4y, 4x − 8y). c. T(x, y, z) = (x − y − z, 2x + y + 3y). Solution ( ) ( ) ( ) 1 −1 R ( −2) +R 1 −1 R ( −1) +R 1 −1 1 2 2 3 2 1 0 3 0 3 , → → 4 − 1 R1 ( − 4 ) + R3 0 3 0 0 r(A 3 × 2) = 2 and by Theorem 5.2.4 D is tri b A3 × 2 = ut io n a. Because , T is one to one. b. Because , T is not one to one. N ot r(A 3 × 2) = 1 < 2 and by Theorem 5.2.4 ( ) ( ) fo r A3 × 2 = 1 −2 R ( −2) +R 1 −2 1 2 2 −4 0 0 , → R1 ( − 4 ) + R3 4 −8 0 0 { } = 2 < n = 3, by Theorem 5.2.4 c. Because r(A 2 × 3) ≤ min 2, 3 , T is not one to one. Let T : ℝ n → ℝ m be a linear transformation. By Theorem 5.2.3 , T is onto if and only if r(A) = m and by Theorem 5.2.4 , T is one to one if and only if r(A) = n. Hence, when m = n, we see that onto and one to one linear transformations are the same. This, together with Theorems 2.7.6 and 2.7.5 , implies the following result that uses the rank or determinant of A to determine whether a linear operator T : ℝ n → ℝ n is onto or one to one. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/12 8/30/2020 5.2 Onto, one to one, and inverse transformations Theorem 5.2.5. Let T be a linear operator from ℝ n to ℝ n with standard matrix A. Then the following assertions are equivalent. ut io n i. T is one to one. ii. T is onto. iii. |A | ≠ 0. iv. r(A) = n. v. A is invertible. D is tri b Let T : ℝ n → ℝ n be a linear operator with standard matrix A. If A is invertible, then A − 1 exists. The linear operator with standard matrix A − 1 is called the inverse of T, denoted by T − 1, that is, (5.2.5) → → Example 5.2.4. fo r T − 1(X) = A − 1X. N ot Determine whether the following linear operators are invertible. If so, find their inverses. a. T(x, y) = (xcosθ − ysinθ, xsinθ + ycosθ), where θ ∈ ℝ is an angle in radians. b. T(x 1, x 2) = (2x 1 + x 2, 3x 1 + 4x 2). c. T(x 1, x 2, x 3) = (x 1 + x 2 − x 3, − x 1 + 2x 2, x 2 + x 3). Solution a. Because https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/12 5.2 Onto, one to one, and inverse transformations |A | = by Theorem 5.2.5 sinθ cosθ | = cos 2θ + sin 2θ = 1 ≠ 0, , T is invertible. By Theorem 2.7.5 A −1 = 1 |A| sinθ cosθ | ( () ( cosθ sinθ = | cosθ − sinθ = cosθ sinθ − sinθ cosθ , b. Because |A | = | | 2 1 3 4 x = y − sinθ cosθ = 5 ≠ 0, by Theorem 5.2.5 1 ( 4 |A| − 3 −1 2 N ot A −1 = )( ) x ) . D is tri b T −1 By (5.2.5) (2), . y , T is invertible. By Theorem 2.7.5 fo r By (5.2.5) | cosθ − sinθ ut io n 8/30/2020 ) ( ) 4 = 5 3 −5 (2), 1 −5 2 . 5 , we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/12 5.2 Onto, one to one, and inverse transformations T −1 () x1 x2 ( ) 4 = 1 −5 5 3 2 −5 5 () x1 x2 c. By computation, | | 1 −1 |A | = − 1 2 0 1 1 = 4 ≠ 0. N ot fo r Hence, T is invertible. 0 D is tri b 1 . ut io n 8/30/2020 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/12 8/30/2020 5.2 Onto, one to one, and inverse transformations (A | I 3) = ) 1 1 −1 1 0 0 R ( −3) +R 2 3 0 1 1 0 0 1 → 0 3 −1 1 1 0 1 1 −1 1 0 0 0 1 1 0 0 1 0 0 −4 1 1 −3 ) 1 R( − 4 ) → 1 1 −1 1 0 0 0 1 1 0 1 0 1 ) R3 ( − 1 ) + R2 1 3 → R3 ( 1 ) + R1 0 0 1 −4 −4 4 R ( − 1 ) + R1 → 1 0 1 0 4 1 0 1 0 4 1 1 1 3 −4 4 1 1 4 4 1 3 0 0 1 −4 −4 4 ( ) 1 1 0 0 2 ) ( ) 3 1 1 0 4 N ot ( 0 1 1 0 0 1 ut io n ( ( ) ( D is tri b → 0 1 1 0 0 1 fo r R2 , 3 ( 1 1 −1 1 0 0 R (1) +R 1 1 −1 1 0 0 1 2 −1 2 0 0 1 0 0 3 −1 1 1 0 → 1 1 −2 2 1 1 4 4 = (I 3 | A − 1). 1 3 0 0 1 −4 −4 4 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/12 8/30/2020 5.2 Onto, one to one, and inverse transformations , we have () x1 T − 1 x2 = 1 1 2 −2 1 1 1 4 4 4 x2 . 3 x3 1 2 1 −4 −4 4 x1 D is tri b x3 ( )( ) 1 ut io n By (5.2.5) Exercises 1. For each of the following linear transformations, determine whether it is onto. a. T(x 1, x 2, x 3) = (2x 1 − 2x 2 + x 3, 3x 1 − 2x 2 + x 3, x 1 + 2x 2 − 4x 3, 2x 1 − 3x 2 + 2x 3). fo r b. T(x 1, x 2, x 3) = (x 1 + 2x 2 − x 3, − x 1 − 2x 2, − x 1 + x 2 + 2x 3). c. T(x 1, x 2, x 3, x 4) = (x 1 − 2x 2 + x 4, 3x 1 + 2x 4, x 1 − 2x 3 + 3x 4). d. T(x 1, x 2, x 3, x 4) = (x 1 − x 2 + x 3, x 1 − x 4, 2x 1 − x 2 + x 3 − x 4). N ot 2. Let T : ℝ 2 → ℝ 2 be defined by T(x, y) = (x − y, 2x − 2y). → → 3. Show that T is not onto and find a vector b ∈ ℝ 2 such that b ∉ T(ℝ 2). 4. For each of the following linear transformations, determine whether it is one to one. a. T(x 1, x 2, x 3) = (x 1 − x 2 + 2x 3, − x 1 − x 2, x 1 + x 2 − x 3, − 2x 1 − x 2 + x 3). b. T(x 1, x 2, x 3) = ( − x 1 + x 2 + x 3, x 1 − x 2 − 2x 3, + x 1 + x 2). c. T(x 1, x 2, x 3, x 4) = (x 1 − x 2 + x 3, 2x 1 + x 2 + x 3, x 1 + x 2 + 2x 4). d. T(x 1, x 2, x 3) = (x 1 − x 2 + 3x 3, x 1 − 2x 3, 2x 1 − x 2 + x 3). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/12 8/30/2020 5.2 Onto, one to one, and inverse transformations 5. For each of the following linear operators, determine whether it is invertible. If so, find its inverse. a. T(x 1, x 2) = (x 1 + x 2, 6x 1 + 7x 2). b. T(x 1, x 2) = (3x 1 − 9x 2, 4x 1 − 12x 2). c. T(x 1, x 2, x 3) = (x 1 + 3x 2 + 2x 3, x 2 − x 3, x 3). N ot fo r D is tri b ut io n d. T(x 1, x 2, x 3) = (x 1 + 3x 2, 3x 1 + 7x 2 − 4x 3, x 1 + 5x 2 + 5x 3). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/12 8/30/2020 5.3 Linear operators in Math Equation and Math Equation 5.3 Linear operators in ℝ 2 and ℝ 3 ut io n In this section, we seek some special linear operators in ℝ 2 and in ℝ 3, called projection, reflection, and rotation operators. Projection operators in ℝ 2 D is tri b Theorem 5.3.1. Let T : ℝ 2 → ℝ 2 be the orthogonal projection on a line l passing through the origin (0, 0) and let the angle between l and the x-axis be θ (see Figure 5.3 ). Then (5.3.1) Proof From Figure 5.3 , we see that (5.3.2) y ( sinθcosθ )( ) x fo r () = cos 2θ cosθsinθ sin 2θ y . N ot T x (w 1, w 2) = (rcosθ, rsinθ). We prove the following assertion. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation (5.3.3) r = xcosθ + ysinθ. Indeed, by Figure 5.3 , r = scosβ = scos(θ − α) and (x, y) = (scosα, ssinα). N ot fo r D is tri b Figure 5.1: Projection operator ut io n (5.3.4) Because cos(θ − α) = cosαcosθ + sinαsinθ, by (5.3.4) we have r = scos(θ − α) = (scosα)cosθ + (ssinα)sinθ = xcosθ + ysinθ and (5.3.3) we obtain holds. By (5.3.2) and (5.3.3) , https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation (w 1, w 2) = (rcosθ, rsinθ) = ((xcosθ + ysinθ)cosθ, (xcosθ + ysinθ)sinθ) = (xcos 2θ + ysinθcosθ, xsinθcosθ + ysin 2θ) holds. The result (5.3.3) see that can be proved by using the dot product. Indeed, by Figure 5.3 → BA = (x − rcosθ, y − rsinθ). (x − rcosθ, y − rsinθ) , we D is tri b → → → → Because BA ⊥ OB, the dot product BA ⋅ OB = 0, that is, and (5.3.2) ut io n and (5.3.1) ( ) rcosθ rsinθ fo r This, together with sin 2θ + cos 2θ = 1, implies = 0. 0 = (x − rcosθ)(rcosθ) + (y − rsinθ)(rsinθ) [ N ot = r[(x − rcosθ)cosθ + (y − rsinθ)sinθ] ] = r xcosθ + ysinθ − r(cos 2θ + sin 2θ) = r[xcosθ + ysinθ − r]. This implies (5.3.3) . Figure 5.2: (I), (II),(III),(IV) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation π π ut io n By Theorem 5.3.1 with θ = 0, 2 , 4 , we obtain Example 5.3.1. T D is tri b i. The projection operator on the x-axis is ( ) ( )( ) ( ) x = y 1 0 x 0 0 y ii. The projection operator on the y-axis is x 0 . fo r ( ) ( )( ) ( ) x = y 0 0 x 0 1 y = N ot T = 0 y . iii. The projection operator on the line y = x is T () x y = () () 1 1 2 2 1 1 2 2 () x y x+y = 2 x+y . 2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation Solution i. The projection operator on the x-axis is () y = ( cos 20 sin0cos0 cos0sin0 sin 20 ) ( )( ) ( ) = 1 0 x 0 0 y = ii. The projection operator on the y-axis is () x y = ( π π π sin 2 cos 2 π cos 2 sin 2 2 π sin 2 Reflection operators in ℝ 2 () x y = ( π π cos 2 4 π π sin 4 cos 4 π 2 π N ot T = ( )( ) ( ) 0 0 x 0 1 y cos 4 sin 4 sin 4 = 0 y . )( ) ( ) fo r iii. The projection operator on the line is ) . 0 D is tri b T π cos 2 2 x ut io n T x = 1 1 2 2 1 1 2 2 () x y x+y = 2 x+y . 2 Theorem 5.3.2. Let T : ℝ 2 → ℝ 2 be the reflection operator about a line l passing through the origin (0, 0) and let the angle between l and the x-axis be θ (see Figure 5.3 ). Then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation (5.3.5) T () ( x = y cos2θ sin2θ sin2θ − cos2θ )( ) x . y and Theorem 5.3.1 () w1 + x 2 w2 + y 2 = ( , from Figure 5.3 cos 2θ sinθcosθ cosθsinθ sin 2θ () y = = ( ) () ( ( )( ) w2 = − x y +2 . sinθcosθ cosθsinθ sin 2θ 2cos 2θ − 1 2sinθcosθ x 2sin 2θ − 1 y 2cosθsinθ y cos 2θ N ot T w1 x fo r This implies x )( ) , we have D is tri b By the midpoint formula (1.2.14) ut io n Proof )( ) x y . This, together with cos2θ = 2cos 2θ − 1 = 1 − 2sin 2θ and sin2θ = 2sinθcosθ, implies (5.3.5) . Figure 5.3: Reflection operator https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/19 5.3 Linear operators in Math Equation and Math Equation By Theorem 5.3.2 D is tri b ut io n 8/30/2020 π π with θ = 0, 2 , 4 , we obtain fo r Example 5.3.2. i. The reflection about the x-axis is ( ) ( )( ) ( ) = 1 0 N ot T x y 0 −1 ii. The reflection about the y-axis (see Figure 6.2 T x y = x −y . (II)) is ( ) ( )( ) ( ) x y = −1 0 0 1 iii. The reflection about the line y = x (see Figure 6.2 x y = −x y . (II)) is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation T ( ) ( )( ) ( ) x = y 0 1 x 1 0 y y = x . Solution () ( x = y cos2(0) sin2(0) sin2(0) − cos2(0) ) ( )( ) ( ) ii. The reflection about the y-axis is ( 0 0 −1 ) () () ( () () () ( )( ) ( ) x y = π sin2 2 π π sin2 2 −1 0 0 1 y = x −y . − cos2 2 = cosπ sinπ sinπ − cosπ ) x y = −x y . N ot = π cos2 2 x fo r T 1 = D is tri b T ut io n i. The reflection about the x-axis is iii. The reflection about the line y = x is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/19 5.3 Linear operators in Math Equation and Math Equation T ( () () () () () ( )( ) ( ) x y = = π π cos2 4 sin2 4 π π sin2 4 − cos2 4 0 1 x 1 0 y = y x )( ) π = cos 2 π sin 2 π sin 2 π − cos 2 . Example 5.3.3. ut io n 8/30/2020 D is tri b 1. Find the linear operator on ℝ 2 that reflects about the line y = x and then projects on the y-axis. 2. Find the linear operator on ℝ 2 that projects on the y-axis and then reflects about the line y = x. Solution 1. The operator on ℝ 2 is T 2T 1, that is, fo r The reflection about the line y = x is T 1(x, y) = (y, x) and the projection on the y-axis is T 2(x, y) = (0, y). N ot T 2T 1(x, y) = T 2(T 1(x, y)) = T 2(y, x) = (0, x). 2. The operator on ℝ 2 is T 1T 2 and T 1T 2(x, y) = T 1(0, y) = (y, 0). The rotation operators in ℝ 2 through angle θ A rotation operator on ℝ 2 is an operator that rotates each vector in ℝ 2 through a fixed angle θ (see Figure 6.2 (IV)). Theorem 5.3.3. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation The rotation operator in ℝ 2 through an angle θ is (5.3.6) () ( x y = cosθ − sinθ sinθ cosθ )( ) x y . ut io n T Proof We rotate (x, y) to (b 1, b 2) through a counterclockwise angle θ Then D is tri b (5.3.7) T(x, y) = (b 1, b 2). b2 = = and (5.3.6) rcos(θ + ϕ) rsin(θ + ϕ) ) ( ) ( = rcosθcosϕ − rsinθsinϕ rsinθcosϕ + rsinϕcosθ N ot () ( b1 fo r Let r be the length of the segment from the origin to the point (x, y) and let ϕ denote the angle between the half-line beginning at the origin (0, 0) and passing through (x, y) and the x-axis. Then (x, y) = (rcosϕ, rsinϕ) and ( xcosθ − ysinθ xsinθ + ycosθ = cosθ − sinθ sinθ cosθ ) )( ) x y holds. Example 5.3.4. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation () 1 π Find the rotation operator with the rotation angle 6 and the image of x = under the operator. 1 Solution π () 1 ( ) ( ) π sin 6 ( ) () √3 = y = cos 6 − sin 6 1 2 −2 1 √3 2 2 Example 5.3.5. () 1 1 π cos 6 () x y √3 − 1 = 2 1 + √3 2 . √3 = 1 2 −2 1 √3 2 2 () x y . fo r and T 1 () x π D is tri b T π ut io n with θ = 6 , N ot By (5.3.6) Find the standard matrix for the linear operator on ℝ 2 that rotates by θ 1 and then rotates by θ 2. Solution By (5.3.6) with θ = θ 1 and θ = θ 2, the standard matrix, denoted by A, for the linear operator is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation = sinθ 2 cosθ 2 )( sinθ 1 cosθ 1 ) cosθ 2cosθ 1 − sinθ 2sinθ 1 − cosθ 2sinθ 1 − sinθ 2cosθ 1 sinθ 2cosθ 1 + cosθ 2sinθ 1 − sinθ 2sinθ 1 + cosθ 2cosθ 1 cos(θ 1 + θ 2) − sin(θ 2 + θ 1) sin(θ 2 + θ 1) cos(θ 1 + θ 2) ) ) ut io n = ( ( ( cosθ 1 − sinθ 1 . D is tri b A = cosθ 2 − sinθ 2 N ot fo r Figure 5.4: (I), (II) Projection operators in ℝ 3 Theorem 5.3.4. i. The projection on the xy-plane (see Figure 6.3 (II)) is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation ( ) ( )( ) x T y z = 1 0 0 x 0 1 0 y , z 0 0 0 ( ) ( )( ) x = x 0 0 0 y . z 0 0 1 D is tri b T y z 1 0 0 ut io n ii. The projection on the xz-plane is iii. The projection on the yz-plane is ( ) ( )( ) x 0 1 0 y . z 0 0 1 N ot Reflection operators in ℝ 3 Theorem 5.3.5. = x fo r T y z 0 0 0 1. The reflection about the xy-plane (see Figure 6.3 (I)) is ( ) ( )( ) x T y z = 1 0 0 x 0 1 0 y . z 0 0 −1 2. The reflection about the xz-plane is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation ( ) ( )( ) x T y z 1 0 0 x 0 −1 0 = 0 0 y . z 1 ( ) ( )( ) T y z −1 0 0 = 0 1 0 0 0 1 x y . z D is tri b x ut io n 3. The reflection about the yz-plane is The rotation operators in ℝ 3 through angle θ of rotation The axis of rotation is a ray emanating from the origin. The rotation is counterclockwise looking toward the origin along the axis of rotation. fo r An angle of rotation is said to be positive if the rotation is counterclockwise looking toward the origin along the axis of rotation, and said to be negative if the rotation is clockwise. N ot Theorem 5.3.6. 1. The rotation operator in ℝ 3 about the positive z-axis with a positive angle θ of rotation is (5.3.8) () ( x T y z = cosθ − sinθ 0 sinθ cosθ 0 0 0 1 )( ) x y . z https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation 2. The rotation operator in ℝ 3 about the positive x-axis with a positive angle θ of rotation is (5.3.9) T y z = 1 0 0 0 cosθ − sinθ 0 sinθ cosθ )( ) x y . z ut io n () ( x () ( x T y z cosθ 0 = D is tri b 3. The rotation operator in ℝ 3 about the positive y-axis with a positive angle θ of rotation is (5.3.10) 0 sinθ 1 0 − sinθ 0 cosθ )( ) x y . z b2 = b3 = [a (1 − cosθ) + cosθ ]x + [ab(1 − cosθ) − csinθ]y + [ac(1 − cosθ) + bsinθ]z, [ab(1 − cosθ) + csinθ]x + [b (1 − cosθ) + cosθ ]y + [bc(1 − cosθ) − asinθ]z, [ac(1 − cosθ) − bsinθ]x + [bc(1 − cosθ) + asinθ]y + [c (1 − cosθ) + cosθ ]z. 2 N ot b1 = fo r 4. If the axis of rotation is parallel to the vector → u = (a, b, c), then the rotation operator with a positive angle θ is T(x, y, x) = (b 1, b 2, b 3), where 2 2 Proof 1. The rotation operator in the xy-plane in ℝ 2 is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation T(x, y) = (xcosθ − ysinθ, xsinθ + ycosθ). D is tri b ut io n Figure 5.5: (I) 2. So, the rotation operator in the xy-plane in ℝ 3 (see Figure 6.4 (I)) is T(x, y, 0) = (xcosθ − ysinθ, xsinθ + ycosθ, 0). fo r Hence, the rotation operator in ℝ 3 about the positive z-axis with θ is and (5.3.8) holds. N ot T(x, y, z) = (xcosθ − ysinθ, xsinθ + ycosθ, z) 3. The rotation operator in the yz-plane in ℝ 2 is obtained by replacing x and y in (5.3) by y and z, respectively, that is, T(y, z) = (ycosθ − zsinθ, ysinθ + zcosθ). So, the rotation operator in the yz-plane in ℝ 3 is T(0, y, z) = (0, ycosθ − zsinθ, ysinθ + zcosθ). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation Hence, the rotation operator in ℝ 3 about the positive x-axis with θ is T(x, y, z) = (x, ycosθ − zsinθ, ysinθ + zcosθ). and (5.3.9) holds. T(z, x) = (zcosθ − xsinθ, zsinθ + xcosθ) or equivalently The rotation operator in the zx-plane in ℝ 3 is D is tri b T(x, z) = (zsinθ + xcosθ, zcosθ − xsinθ). ut io n 4. The rotation operator in the zx-plane in ℝ 2 is obtained by replacing x and y in (5.3) by z and x, respectively, that is, T(x, 0, z) = (zsinθ + xcosθ, 0, zcosθ − xsinθ). fo r The rotation operator in ℝ 3 about the positive y-axis with θ is T(x, y, z) = (zsinθ + xcosθ, y, zcosθ − xsinθ) N ot and (5.3.10) holds. 5. We omit the derivation of the linear operator. It is easy to see that Theorem 5.3.6 (1), (2), and (3) can be obtained by Theorem 5.3.6 with (a, b, c) = (1, 0, 0), (0, 1, 0), (0, 0, 1), respectively. (4) Example 5.3.6. Find the standard matrix for the linear operator on ℝ 3 that rotates by θ about the positive z-axis and then reflects about the yz-plane and then projects on the xy-plane. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation Solution Let A, B, and C be the standard matrices, respectively, for the linear operator that rotates by θ about the positive z-axis, reflects about the yz-plane, and projects on the xy-plane. Then ( sinθ cosθ 0 0 ) ( ) ( ) −1 0 0 0 ,B = 1 0 0 1 0 0 1 0 ,C = 0 1 0 1 0 . 0 0 0 ut io n A= cosθ − sinθ 0 ( )( )( ( ) 1 0 0 CBA = −1 0 0 0 1 0 0 1 0 0 0 0 0 0 1 − cosθ sinθ 0 sinθ 1. In Theorem 5.3.1 cosθ 0 0 0 1 ) N ot π π sinθ cosθ 0 . 0 0 0 Exercises cosθ − sinθ 0 fo r = D is tri b Hence, the standard matrix for the required operator is ( ) 4 , if θ = 6 , 3 , find T . −4 2. Find the linear operator T = T 2T 1 on ℝ 2, where T 1 is a dilation with factor k = 2 on ℝ 2 and T 2 is a projection on the line y = x. 3. 1 1. Find the linear operator on ℝ 2 that contracts with factor 2 and then reflects about the line y = x. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/19 8/30/2020 5.3 Linear operators in Math Equation and Math Equation 2. Find the linear operator on ℝ 2 that reflects about the y-axis and then reflects about the x-axis, and the linear operator on ℝ 2 that reflects about the x-axis and then reflects about the y-axis. ( ) 1 π 4. Find the rotation operator with the rotation angle of 4 and the image of x = under the operator. −1 π ut io n π 5. Find the linear operator on ℝ 2 that rotates by 3 radians about the origin. N ot fo r D is tri b 6. Find the linear operator on ℝ 3 that rotates by 4 radians about the z-axis and then reflects about the yz-plane. 7. Prove Theorem 5.3.1 by using Theorem 5.1.1 . 8. Prove Theorem 5.3.2 by using a method similar to the proof of Theorem 5.3.1 . 9. Prove Theorem 5.3.2 by using Theorem 5.1.1 . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/19 8/30/2020 Chapter 6 Lines and planes in Math Equation ut io n Chapter 6 Lines and planes in ℝ 3 6.1 Cross product and volume of a parallelepiped D is tri b In Section 1.6, we see that the dot product of two vectors in ℝ n is a real number but not a vector. Now, we define a product of two vectors in ℝ 3 called the cross product, which is a vector in ℝ 3. → (6.1.1) x 1a + x 2b + x 3c = 0 N ot { fo r Let → a = (x 1, x 2, x 3) and b = (y 1, y 2, y 3) be two nonzero vectors in ℝ 3. We want to find a vector → n = (a, b, c) in → → ℝ 3 such that → n⊥→ a and → n ⊥ b . By Theorem 1.2.6 , → n⋅→ a = 0 and → n ⋅ b = 0. This implies that → n satisfies the following system of linear equations y 1a + y 2b + y 3c = 0. Lemma 6.1.1. Let → a = (x 1, x 2, x 3) and b = (y 1, y 2, y 3) be two nonzero vectors in ℝ 3. Then → (6.1.2) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/10 8/30/2020 Chapter 6 Lines and planes in Math Equation (a, b, c) = is a solution of (6.1.1) (| | | | | |) x2 x3 x1 x3 , − , y2 y3 y1 y3 x1 x2 y1 y2 . ut io n Substituting a, b, c given in (6.1.2) to (6.1.1) shows that (a, b, c) defined in (6.1.2) is a solution of (6.1.1) . Here, we provide a proof of Lemma 6.1.1 that shows where the solution (a, b, c) given in (6.1.2) comes from. → If → a is parallel to b , then the three determinants in (6.1.2) solution of (6.1.1) D is tri b Proof are zero, and (a, b, c) = (0, 0, 0) is a → . If → a is not parallel to b , then one of the three determinants in (6.1.2) | | is not zero. x1 x2 Without loss of generalization, we assume that c = ≠ 0. Solving the following system by using y1 y2 x 1a + x 2b = − x 3c y 1a + y 2b = − y 3c, N ot { fo r Cramer’s rule we obtain | || | − x 3c x 2 x3 y3 a= c = − y 3c y 2 x2 y2 1 Hence, (6.1.2) is a solution of (6.1.1) | || | x 1 − x 3c x1 y1 and b = c = . y 1 − y 3c x3 y3 1 . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/10 8/30/2020 Chapter 6 Lines and planes in Math Equation Definition 6.1.1. → → → Let → a = (x 1, x 2, x 3) and b = (y 1, y 2, y 3) be two vectors. The cross product → a × b of → a and b is defined by → → a×b= → → (| | | | | |) x2 x3 x1 x3 , − , y2 y3 y1 y3 x1 x2 y1 y2 . → ut io n (6.1.3) → expansion, we rewrite → a × b into the following two forms: (6.1.4) → a×b = | | | | | | x2 x3 y2 y3 → i − x1 x3 y1 y3 → j + x1 x2 y1 y2 → k fo r → D is tri b Let i = (1, 0, 0), j = (0, 1, 0), , and k = (0, 0, 1) be the standard vectors in ℝ 3. Following the minor | | → i → j → k N ot = x1 x2 x3 . y1 y2 y3 Let → c = (z 1, z 2, z 3). By Theorem 3.1.1 and (6.1.3) , we have (6.1.5) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/10 8/30/2020 Chapter 6 Lines and planes in Math Equation | | x1 x2 x3 → (→ a × b) ⋅ → c = y1 y2 y3 . z1 z2 z3 → By (6.1.5) and Corollary 3.3.1 → → → ut io n → The value (→ a × b) ⋅ → c is called the scalar triple product of → a, b, → c. , we see that (→ a × b) ⋅ → a = 0 and (→ a × b) ⋅ b = 0. By Theorem 1.2.6 → → the cross product → a × b is perpendicular to both vectors → a and b (see Figure 6.1 (I)), that is, N ot fo r D is tri b Figure 6.1: (I) , (6.1.6) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/10 8/30/2020 Chapter 6 Lines and planes in Math Equation → → → (→ a × b) ⊥ → a and (→ a × b) ⊥ b. Example 6.1.1. → → → → → Let → a = (1, 2, − 2) and b = (3, 0, 1). Find → a × b and verify that (→ a × b) ⊥ → a and (→ a × b) ⊥ b . i j → → → k a × b = 1 2 −2 = 3 0 1 → → | | | | | | 2 −2 0 1 → i − 1 −2 3 1 → j + 1 2 3 0 → k D is tri b | | → → ut io n Solution → = 2 i − 7 j − 6 k = (2, − 7, 6). Because → and → → fo r (→ a × b) ⋅ → a = (2)(1) + ( − 7)(2) + ( − 6)( − 2) = 0 → → → Hence, (→ a × b) ⊥ → a and (→ a × b) ⊥ b . N ot (→ a × b) ⋅ b = (2)(3) + ( − 7)(0) + ( − 6)( − 1) = 0 → By Corollary 1.3.1 (2) and Definition 6.1.3, we obtain the following result, which shows that the norm of → a×b → is equal to the area of a parallelogram in ℝ 3 determined by → a and b . Theorem 6.1.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/10 8/30/2020 Chapter 6 Lines and planes in Math Equation → Let → a = (x 1, x 2, x 3) and b = (y 1, y 2, y 3) be two vectors. Then the area of the parallelogram determined by → a and b is → | | a × b | |. → → → → ut io n Let → a, b, and → c be three vectors in ℝ 3 that are not in the same plane but their initial points are the same. Then these vectors determine a parallelepiped (see Figure 6.2 (I)). The purpose of this section is to derive formulas for the volume of such a parallelepiped. It is known that the volume of a parallelepiped is equal to the product of the area of its base and its height. The base of the parallelepiped is determined by any two of the vectors → a, b, and → c and is a parallelogram in ℝ 3. Here we use the base determined by → a and → b. By Theorem 6.1.1 → , the area of the parallelogram determined by → a and b is | | a × b | | . To find the → → → D is tri b height of the parallelepiped, we use the projection of the vector → c on the vector → a × b because the norm of the projection is equal to the height of the parallelepiped (see Figure 6.2 (I)). Based on the above ideas, → we show the following two formulas for the volume of the parallelepiped determined by → a, b, and → c. N ot fo r Figure 6.2: (I) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/10 Chapter 6 Lines and planes in Math Equation N ot fo r D is tri b ut io n 8/30/2020 Theorem 6.1.2. → Let → a = (x 1, x 2, x 3) and b = (y 1, y 2, y 3), → c = (z 1, z 2, z 3) be three vectors. Then the volume of the parallelepiped determined by the three vectors is given by https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/10 8/30/2020 Chapter 6 Lines and planes in Math Equation (6.1.7) x1 x2 x3 → V = |→ c ⋅ (→ a × b) | = ǁ y 1 y 2 y 3 ǁ. z1 z2 z3 , we only prove V = | c ⋅ (a × b) | . Let n = a × b. By Theorem 6.1.1 → → → parallelogram determined by → a and b is → → | |n | |. → → the projection of → c on → a × b (see Figure 6.2 → → , the area of the The height of the parallelepiped is equal to the norm of (I)). By (1.3.1) , D is tri b By (6.1.5) ut io n Proof → → → c ⋅ n| | c ⋅ n → → h = ǁproj n cǁ = ǁ → 2 nǁ = . ǁ→ nǁ ǁ nǁ → → Hence, → Example 6.1.2. → → ǁ nǁ → = |→ c⋅→ n | = |→ c ⋅ (→ a × b) | . N ot V = ǁ nǁh = ǁ nǁ |→ c⋅→ n| fo r → Let → a = (3, − 2, − 5), b = (1, 4, − 4), and → c = (0, − 3, − 2). Use the two formulas in Theorem 6.1.2 → to find the volume of the parallelepiped determined by → a, b, and → c. Solution By (6.1.3) , we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/10 8/30/2020 Chapter 6 Lines and planes in Math Equation | | → i → → j → k b×→ c = 1 4 −4 = 0 −3 −2 | 4 −4 −3 −2 → | | | | | → i − 1 −4 0 −2 → → j + 1 4 0 −3 → k → ut io n = ( − 8 − 12) i − ( − 2 − 0) j + ( − 3 − 0) k = ( − 20, 2, − 3). Hence, we obtain → , we have V= | 3 −2 −5 1 4 −4 0 −3 −2 Exercises → → → | | 3 −2 1 4 0 −3 = | − 49 | = 49. fo r By (3.1.4) D is tri b V = |→ a ⋅ (b × → c) | = | (3)( − 20) + ( − 2)(2) + ( − 5)( − 3) | = | − 49 | . → → 1. → a = (1, 2, 3); b = (1, 0, 1), → N ot 1. Find → a × b and verify (→ a × b) ⊥ → a and (→ a × b) ⊥ b . 2. → a = (1, 0, − 3), b = ( − 1, 0, − 2), → 3. → a = (0, 2, 1); b = ( − 5, 0, 1), → 4. → a = (1, 1, − 1); b = ( − 1, 1, 0), → → 2. Let → a = (1, − 2, 3), b = ( − 1, − 4, 0), and → c = (2, 3, 1). Find the scalar triple product of → a, b, → c. → 3. Find the area of the parallelogram determined by → a and b . → 1. → a = (1, 0, 1), b = ( − 1, 0, 1); https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/10 8/30/2020 Chapter 6 Lines and planes in Math Equation → 2. → a = (1, 0, 0), b = ( − 1, 1, − 2). → 4. Find the volume of the parallelpiped determined by a, b, and → c. → → 1. → a = (1, − 1, 0), b = ( − 1, 0, 2), → c = (0, − 1, − 1). → N ot fo r D is tri b ut io n 2. → a = ( − 1, − 1, 0), b = ( − 1, 1, − 2), → c = ( − 1, 1, 1). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/10 8/30/2020 6.2 Equations of lines in Math Equation 6.2 Equations of lines in ℝ 3 In the xy-plane, we can obtain the equation of a (straight) line if we know a point on the line and the slope of ut io n the line. The slope of the line provides the direction of the line. In ℝ 3, the basic idea is the same. To establish an equation of a line in ℝ 3, we need a point on the line and a vector parallel to the line (see Figure 6.3 (I)). N ot fo r D is tri b Figure 6.3: (I) Definition 6.2.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/8 8/30/2020 6.2 Equations of lines in Math Equation → → A vector υ in ℝ 3 is said to be parallel to a line in ℝ 3 if it is parallel to PQ for any two different points P and Q on the line. ut io n → → → Note that if P 1 and Q 1 are two other different points on the line, then PQ is parallel to P 1Q 1 because PQ and → → P 1Q have the same or opposite direction. By Definition 1.1.6 , if a vector is parallel to PQ, it is parallel to 1 → P 1Q 1. Hence, in Definition 6.2.1 → , υ is parallel to PQ for any two different points P and Q on the line if and → → only if there exist two different points P and Q on the line such that υ is parallel to PQ. D is tri b → Definition 6.2.2. → Let P 0(x 0, y 0, z 0)be a point in ℝ 3 and let υ = (a, b, c) be a nonzero vector in ℝ 3. Then the set of all points → → P(x, y, z) in ℝ 3 satisfying P 0Pǁ υ constitutes a line in ℝ 3. and 6.2.2 the same line in Definition 6.2.2 → , we see that a line passing through the point P 0 that is parallel to υ is fo r By Definitions 6.2.1 . → vector υ . Theorem 6.2.1. N ot The following result gives a parametric equation of a line passing through a point P 0 that is parallel to a given → Let P 0(x 0, y 0, z 0) be a point in ℝ 3 and let υ = (a, b, c) be a nonzero vector in ℝ 3. Then the parametric → equation of the line passing through P 0 that is parallel to υ is (6.2.1) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/8 8/30/2020 6.2 Equations of lines in Math Equation { x = x 0 + ta y = y 0 + tb −∞ < t < ∞ z = z 0 + tc ut io n Proof By (6.2.1) , we see that the solution (4.3.1) equation of a line. D is tri b → → Let P(x, y, z) be any point on the line. Then P 0P = (x − x 0, y − y 0, z − z 0) and P 0P is parallel to → → υ P 0P = (x − x 0, y − y 0, z − z 0). By Definition 1.1.6 , there exists a constant t ∈ ℝ that depends on P → → such that P 0P = t υ. This, together with Definition 1.1.4 , implies (6.2.1) holds. of the system d) in Example 4.3.1 is a parametric N ot fo r By the proof of Theorem 6.2.1 , we see that for each point (x, y, z) on the line, there exists a unique number t ∈ ℝ such that (6.2.1) holds. Conversely, by (6.2.1) , we see that for each t ∈ ℝ, there exists a unique point (x, y, z)on the line that satisfies (6.2.1) . Hence, there is a one to one relation between the points on the line and the numbers t ∈ ℝ. The steps to find a parametric equation of a line are: 1. Step 1. Find a point P 0(x 0, y 0, z 0) on the line. → 2. Step 2. Find a vector υ that is parallel to the line. 3. Step 3. Write the equation by using (6.2.1) . Example 6.2.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/8 8/30/2020 6.2 Equations of lines in Math Equation a. Find the parametric equation of the line passing through the point P 0(1, 2, 3) that is parallel to the → vector υ = (2, 3, − 1). b. Where does the line intersect the yz-plane? Solution , the parametric equation of the line is x = 1 + 2t, y = 2 + 3t, z = 3 − t. ut io n a. By (6.2.1) (6.2.2) 1 1 D is tri b b. Because the line intersects the yz-plane, x = 0. By the first equation of (6.2.2) with x = 0, we obtain t = − 2 . Substituting t = − 2 into the second and third equations of (6.2.2) 1 ( 7 1 7 implies that ) y = 2 and z = 2 . Hence, the intersection point is 0, 2 , 2 . fo r Example 6.2.2. Solution N ot Find an equation of the line that passes through P 1(1, − 1, 1) and P 2(2, 1, 0). → Let P 0(x 0, y 0, z 0) = P 1(1, − 1, 1) and υ = (a, b, c) = P 1P 2 = (1, 2, − 1). By (6.2.1) x = 1 + t, y = − 1 + 2t, z = 1 − t. → , we have Now, we discuss the distance from a point to a given line in ℝ 3, that is, the distance between the point and the intersection point of the given line and the line passing through the point that is orthogonal to the given line. Theorem 6.2.2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/8 8/30/2020 6.2 Equations of lines in Math Equation The distance D from Q(x 1, y 1, z 1) to the line (6.2.1) → → D= is ǁ υ × PQ ǁ → ǁυǁ for any point P on the line. Proof ut io n → → Let P be a point on the line. Then D = ǁPQ − proj υPQǁ. It follows from (1.3.4) and Theorem 6.1.1 → that D is tri b → → → → ǁ υ × PQǁ D = ǁPQ − proj υPQǁ = . → ǁ υǁ → Example 6.2.3. Find the distance from the point Q(1, 0, − 1) to the line fo r (6.2.3) N ot x = t, y = − 1 + t, z = 1 − t. Solution → Let t = 0. Then P(0, − 1, 1) is on the line and PQ = (1, 1, − 2). By (6.2.3) → parallel to the line. By computation, ǁ υǁ = → υ × PQ = → → , υ = (1, 1, − 1), which is √3 , (| | | | | |) 1 −1 1 −2 , − 1 −1 1 −2 , 1 1 1 1 = ( − 1, 1, 0) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 5/8 8/30/2020 6.2 Equations of lines in Math Equation √2. By Theorem 6.2.2 , D= By Theorem 6.2.2 → → ǁ υ × PQǁ → ǁ υǁ √6 = 3 . , we derive the distance from a point in ℝ 2 to a line (6.2.4) D is tri b Ax + By = C, where A ≠ 0 or B ≠ 0 in the xy-plane. Theorem 6.2.3. The distance D from Q(x 1, y 1) to the line (6.2.4) is fo r (6.2.5) |Ax1 + By1 − C | N ot D= Proof ut io n → → and ǁ υ × PQǁ = √A 2 +B 2 . Without loss of generalization, we assume that A ≠ 0. Let y = t. Solving (6.2.4) x= for x, we obtain C B − t. A A https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 6/8 8/30/2020 6.2 Equations of lines in Math Equation in ℝ 3 is Hence, the parametric equation of the line (6.2.4) (6.2.6) C B x = A − A t, y = t, z = 0 + 0t. → ( B ) ( ) C υ = − A , 1, 0 and the point P * A , 0, 0 is on the line (6.2.6) ǁ υǁ = |A| → ǁυ × P Q * ǁ * → ǁ υǁ = |Ax1 + By1 − C | √A 2 +B 2 . N ot D= Example 6.2.4. → | Ax1 + By1 − C | * * and ǁ υ × P Q ǁ = . |A| → , → . By computation, fo r By Theorem 6.2.2 √A 2 + B 2 , D is tri b → . From (6.2.6) ut io n Note that D is equal to the distance from Q * (x 1, y 1, 0) to the line (6.2.6) Find the distance from Q(1, 1) to the line y = x + 1. Solution Rewrite y = x + 1 as − x + y = 1. By Theorem 6.2.3 , we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 7/8 8/30/2020 6.2 Equations of lines in Math Equation D= |Ax1 + By1 − C | √ A2 + B2 = |( − 1)(1) + 1 − 1| √ ( − 1) 2 + 1 2 = 1 √2 . 1. ut io n Exercises a. Find the parametric equation of the line passing through the point P 0( − 1, 2, 0) that is parallel → to the vector υ = (1, − 1, 1). b. Where does the line intersect the xy-plane? 3. Find the distance from the point Q( − 1, 2, 1) to the line D is tri b 2. Find the parametric equation of the line that passes through P 1(2, 1, − 1) and P 2( − 2, 3, − 5). x = 1 + t, y = 2 − 4t, z = − 3 + 2t. N ot fo r 4. Find the distance from the point Q( − 1, 2) to the line y = − 2x + 4. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 8/8 8/30/2020 6.3 Equations of planes in Math Equation 6.3 Equations of planes in ℝ 3 ut io n In Section 6.2 , we see that the parametric equation of a line in ℝ 3 is established according to a given point on the line and a given vector that is parallel to the line. In this section, we establish an equation of a plane in ℝ 3. Definition 6.3.1. → n ⋅ P 0P = 0 → D is tri b Let P 0(x 0, y 0, z 0) be a point in ℝ 3 and → n = (a, b, c) a vector in ℝ 3. Then the set of all points P(x, y, z)in ℝ 3 satisfying fo r constitutes a plane in ℝ 3. This vector → n is called a normal vector of the plane. Theorem 6.3.1. N ot Let P 0(x 0, y 0, z 0) be a point in ℝ 3 and let → n = (a, b, c) be a normal vector of a plane in ℝ 3. We derive the equation of the plane passing through P 0(x 0, y 0, z 0) and having the normal vector → n = (a, b, c), which is called the point-normal form equation of the plane. The equation of the plane passing through P 0(x 0, y 0, z 0) with the normal vector → n = (a, b, c) is (6.3.1) Processing math: 100% ax + by + cz = d, https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/22 8/30/2020 6.3 Equations of planes in Math Equation where (6.3.2) () d = (a, b, c) ⋅ y0 = ax 0 + by 0 + cz 0. z0 D is tri b Proof ut io n x0 Let P(x, y, z) be a point in the plane. Then → P 0P = (x − x 0, y − y 0, z − z 0). fo r By Definition 6.3.1 → , n ⋅ P 0P = 0, which implies → a(x − x 0) + b(y − y 0) + c(z − z 0) = 0. N ot This implies ax + by + cz = ax 0 + by 0 + cz 0 Note that equation (6.3.1) is the same as (6.3.1) , so it is a linear equation with three variables x, y, z. The coefficients a, b, c of x, y, z of (6.3.1) are the components of a normal vector of the plane. Example 6.3.1. Find a normal vector of the plane 2x + y − z = 3.. Processing math: 100% Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/22 8/30/2020 6.3 Equations of planes in Math Equation By (6.3.1) ,→ n = (2, 1, − 1). By Definition 6.3.1 → , we see that the normal vector n is orthogonal to the vector P 0P for every point P in → the plane. Moreover, if → v is a normal vector of the plane, then k→ n also is a normal vector of the plane for every k ∈ ℝ with k ≠ 0. ut io n → The following result shows that the normal vector n is orthogonal to the vector PQ for any two different points P and Q in the plane. → D is tri b Theorem 6.3.2. Let → n be the normal vector of a plane that passes through a point P 0(x 0, y 0, z 0). Then → v is orthogonal to → any vector PQ, where P and Q are any two different points in the plane. Proof → → ,→ n ⋅ P 0P = 0 and → n ⋅ P 0Q = 0 . N ot fo r Let P and Q be any two different points in the plane. By Definition 6.3.1 → → → Because PQ = P 0Q − P 0P, we have → → → → → n ⋅ PQ = ( n ⋅ P 0Q) − ( n ⋅ P 0P) = 0. → . → Hence n is orthogonal to PQ. → Definition Processing math: 100% 6.3.2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/22 8/30/2020 6.3 Equations of planes in Math Equation → → A vector υ in ℝ 3 is said to be orthogonal to a plane if it is orthogonal to PQ for any two different points P → → 3 3 and Q in the plane. A vector υ in ℝ is said to be orthogonal to a line in ℝ if it is orthogonal to PQ for any two different points P and Q on the line. ut io n By Definition 6.3.2 , we see that a vector in ℝ 3 is orthogonal to a plane if and only if it is orthogonal to any line in the plane. Also, by Definitions 6.3.1 and 6.3.2 and Theorem 6.3.2 , we see that a normal vector of a plane is orthogonal to the plane as well as any line in the plane. Definition 6.3.3. By Definitions 6.2.1 the reader. and 6.3.3 D is tri b A vector in ℝ 3 is said to be parallel to a plane in ℝ 3 if there exists a line in the plane that is parallel to the vector. , we obtain the following results. Its proof is straightforward and is left to fo r Proposition 6.3.1. → N ot → 1. If P 1(x 1, y 1, z 1) and P 2(x 2, y 2, z 2) are two different points in a plane, then the vector P 1P 2 is parallel to the plane. 2. A vector → n is orthogonal to a plane if and only if it is orthogonal to any vector that is parallel to the plane. → → 3. If both → a and b are parallel to a plane and → a is not parallel to b, then the cross product → a × b is orthogonal to the plane. Now, we consider the following two planes linear system (6.3.3) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/22 8/30/2020 6.3 Equations of planes in Math Equation { a 1x + b 1y + c 1z = d 1 a 2x + b 2y + c 2z = d 2, which is a system of two linear equations. | , (6.3.3) | → has infinitely many solutions or no solutions because r(A b) ≤ 2 < 3 = n, → where A and (A b) are the coefficient matrix and the augmented matrix of (6.3.3) ut io n By Theorem 4.5.1 , respectively. Definition 6.3.4. D is tri b Geometrically, the two planes are the same or intersect in a line or are parallel. Hence, if they are not parallel, then they intersect in a straight line. Two planes are said to be parallel if their normal vectors are parallel, and to be orthogonal if their normal vectors are orthogonal. and Theorem 1.2.6 , we obtain the following results, (see Figure N ot Figure 6.4: (I), (II), (III) and 1.2.5 fo r By Definitions 1.1.6 6.4 (III)). Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/22 6.3 Equations of planes in Math Equation D is tri b ut io n 8/30/2020 Theorem 6.3.3. → → Let n 1 and n 2 be the normal vectors of two planes p 1 and p 2, respectively. Then the following assertions hold. Example 6.3.2. N ot fo r → → i. p 1ǁp 2 if and only if there exists k ∈ ℝ such that n 2 = kn 1. → → ii. p 1 ⊥ p 2 if and only if n 1 ⋅ n 2 = 0. For each pair of the following planes, determine whether it is parallel or orthogonal. a. − x − y + z = 2, 3x + 3y − 3z = 5; b. 2x − y + z = 1, 4x − 2y + 2z = 2; c. 2x − 2y + 3z = 1, x + y + z = 2; Processing math: 100% d. 2x − y + z = 2, x + 3y + z = 2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/22 8/30/2020 6.3 Equations of planes in Math Equation Solution → → → → →→ a. Let n 1 = ( − 1, − 1, 1) and n 2 = (4, − 2, 2). Then n 2 = − 3n 1 and n 1ǁn 2. The two planes are parallel. → → → → →→ b. Let n 1 = (2, − 1, 1) and n 2 = (4, − 2, 2). Then n 2 = 2n 1 and n 1ǁn 2. By Theorem 6.3.3 , the two As mentioned in Remark 4.3.1 the system d) in Example 4.3.1 D is tri b ut io n planes are parallel. → → → → c. Let n 1 = (2, − 2, 3) and n 2 = (1, 1, 1). Then n 2 ≠ kn 1 for any k ∈ ℝ and by Theorem 6.3.3 , the → → two planes are not parallel. Because n 1 × n 2 = 3 ≠ 0, the two planes are not orthogonal. → → → → d. Let n 1 = (2, − 1, 1) and n 2 = (1, 3, 1). Because n 1 × n 2 = 0, the two planes are orthogonal. , the two nonparallel planes x 1 + x 2 − 2x 3 = 1 and − 2x 1 − x 2 + x 3 = − 2 in have a line of intersection. fo r Example 6.3.3. Find the parametric equation of the line of intersection of the planes x − y − z = 1 and x + 2y − 4z = 2. N ot Analysis We need to find a point on the line and a vector parallel to the line as we mentioned above. The cross product of the normal vectors of the two planes is parallel to the line of intersection of the two planes, so we would choose it. But how to find a point on the line of intersection of the two planes? To find a point on the line of intersection, we must solve the system of the two equations. When we solve the system, we shall see that we automatically obtain the equation of the line of intersection. Hence, the methodology to find the equation of the line of intersection of the two planes is simply to solve the system of the two equations. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/22 8/30/2020 6.3 Equations of planes in Math Equation Solution Consider the following system (6.3.4) ( () ( 1 1 −1 −1 1 1 2 −4 4 ) ) R 1( − 1) + R 2→ 1 −1 −1 1 ( 1 −1 −1 1 0 3 −3 3 ) ut io n | → (A b) = x + 2y − 4z = 4. D is tri b { x−y−z=1 ( 1 0 −2 2 ) R2 3 → R (1) + R 1→ . 0 1 −1 1 2 0 1 −1 1 x − 2z = 2 y − z = 1, N ot { fo r The system corresponding to the last augmented matrix is where x, y are basic variables and z is a free variable. Let z = t. Then y = 1 + t and x = 2 + 2t. The parametric equation of the line of intersection of the planes is x = 2 + 2t, y = 1 + t, z = t. In Example 6.3.3 , we solve the system (6.3.4) to obtain the parametric equation of the line of intersection of the two planes. From this line, we can obtain a point, say, P 0(2, 1, 0) on the line, and a vector, → say, υ = (2, 1, 1) that is parallel to the line of intersection. In some cases, we do not want any points on the the line of intersection, but we only want a vector that is parallel to the line of intersection. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/22 8/30/2020 6.3 Equations of planes in Math Equation The following result provides such a vector that is parallel to the line of intersection of two planes (see Figure 6.4 (I)), where we do not need to solve the system of the two linear equations. Theorem 6.3.4. Assume that the two planes given in (6.3.3) → → are not parallel. Then n 1 × n 2 is parallel to the intersection ut io n line of the two planes. Proof Theorem 6.3.5. , we obtain the following result (see Figure 6.4 (II)). fo r By Theorem 6.3.4 D is tri b → We denote by p 1 and p 2 the two planes and by l the line of intersection of the two planes. Because n 1 is → perpendicular to the plane p 1, it is perpendicular to the line l. Similarly, n 2 is perpendicular to the line l. → → → → Hence, l is perpendicular to n 1 and n 2. This implies that l is parallel to n 1 × n 2. Proof N ot A vector is parallel to two nonparallel planes if and only if it is parallel to the intersection line of the two nonparallel planes. If a vector is parallel to two nonparallel planes, then it is orthogonal to the normal vectors of the two → → planes and thus, it is parallel to n 1 × n 2. It follows from Theorem 6.3.4 that the vector is parallel to the intersection line of the two planes. Conversely, if a vector is parallel to the intersection line of two planes, then by Definition 6.3.3 , the vector is parallel to each of the two planes. Processing math: 100% Example 6.3.4. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/22 8/30/2020 6.3 Equations of planes in Math Equation Find an equation of the line passing through P(1, 1, 1) that is parallel to the planes 2x + y − z = 1 and x − y + z = − 2. Analysis D is tri b ut io n Before we give the solution, we analyze the problem of finding the equation of the line. Recall the steps for finding an equation of a line given in Section 6.2 . To establish an equation of a line, we need to find a point on the line and a vector parallel to the line. Because the line passes through P(1, 1, 1), the point P(1, 1, 1) is on the line that we need. Next, we need to find a vector that is parallel to the line. Because the line we seek is assumed to be parallel to the two planes 2x + y − z = 1 and x − y + z = − 2, by Theorem 6.3.5 , the line is parallel to the intersection line of the two planes. By Theorem 6.3.4 , the cross product of the two normal vectors of the two planes is parallel to the intersection line of the two planes and thus, it is parallel to the line we seek. Hence, we can choose a vector that is parallel to the cross product of the two normal vectors. Solution N ot fo r → → → Let n 1 = (2, 1, − 1) and n 2 = (1, 1, − 1) be the normal vectors of the two planes. Because n 1 is not → → → parallel to n 2, the two planes intersect in a line. By Theorem 6.3.4 , n 1 × n 2 is parallel to the line of → → → intersection. By computation, we have n 1 × n 2 = (0, − 3, − 3) = − 3(0, 1, 1). Let υ = (0, 1, 1). Then → → → υǁn 1 × n 2. By (6.2.1) , x = 1, y = 1 + t, z = 1 + t is an equation of the line. Now, we study how to find an equation of a plane. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/22 8/30/2020 6.3 Equations of planes in Math Equation The steps to find an equation of a plane are 1. Step 1. Find a point P 0(x 0, y 0, z 0) in the plane. The point P 0(x 0, y 0, z 0) is often given directly or it is from a line in the plane. If a line is parallel to the → → ut io n plane, then we would not take any points from this line as the point P 0(x 0, y 0, z 0). 2. Step 2. Find a vector → n that is perpendicular to the plane, that is, find a normal vector of the plane. By Proposition 6.3.1 (3), to find a normal vector → n of the plane, we need to find two nonparallel vectors → a and b, which are parallel to the plane, and choose → n = k→ a × b for some k ≠ 0. In particular, → . D is tri b choose → n=→ a × b. 3. Step 3. Write the equation by using (6.3.1) Example 6.3.5. Find the equation of the plane passing through (1, − 1, − 2) and perpendicular to → n = (2, 1, − 1). Solution Example 6.3.6. N ot Find the equation of the plane passing through fo r 2x + y − z = (2, 1, − 1) ⋅ (1, − 1, − 2) = 3. P 1( − 1, 1, 1), P 2( − 2, 0, 1), P 3(0, 1, − 2). Solution 1. Step 1. We can choose one of the three points as a point in the plane, say, P 0(x 0, y 0, z 0) = P( − 1, 1, 1). → 2. Step 2. To find the normal vector → n, we need to find two nonparallel vectors → a and b, both of → Processing math: 100% which are parallel to the plane. By Proposition 6.3.1 (1), P 1P 2 = ( − 1, − 1, 0) and https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/22 8/30/2020 6.3 Equations of planes in Math Equation → → → P 1P 3 = (1, 0, − 3) are parallel to the plane. It is obvious that P 1P 2 is not parallel to P 1P 3 because ( − 1, − 1, 0) ≠ k(1, 0, − 3) for any k ∈ ℝ. By computation, we have | | → → j k (| 0 −3 −1 0 0 −3 | | , − = (3, − 3, 1). −1 0 1 −3 || , −1 −1 1 0 ut io n −1 −1 0 1 = → |) D is tri b → → P 1P 2 × P 1P 3 = i → → We choose n = P 1P 2 × P 1P 3 = (3, − 3, 1) as the normal vector of the plane. → 3. Step 3. By (6.3.1) and (6.3.2) , Example 6.3.7. (6.3.5) N ot Find an equation for the plane containing the line fo r 3x − 3y + z = (3, − 3, 1) ⋅ ( − 1, 1, 1) = − 5. x = 1 + t, y = − 1 − t, z = − 3 + 2t and that is parallel to the line of intersection of the planes x − y − z = 1 and x − 2y + z = 3. Analysis Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/22 8/30/2020 6.3 Equations of planes in Math Equation To establish an equation of a plane, we need to find a point in the plane and a normal vector of the plane. From the given line (6.3.5) , we can obtain two things: one is the point (1, − 1, − 3) when t ≠ 0 and → → another is the vector υ = (1, − 1, 2), which is parallel to the line, so υ = (1, − 1, 2), is parallel to the → → plane we seek. By Theorem 6.3.5 , the cross product n 1 × n 2 of the normal vectors of the two planes is ut io n parallel to the intersection line of the two planes. By the assumption, the plane we seek is parallel to the → → → → → line of intersection of the planes. Hence, n 1 × n 2 is parallel to the plane. The cross product υ × (n 1 × n 2) is the normal vector for the plane. Solution → → n1 × n2 = (| −1 −1 −2 1 D is tri b → → Let n 1 = (1, − 1, − 1) and n 2 = (1, − 2, 1). By computation we have | | | | |) , − 1 −1 1 1 , 1 −1 1 −2 = ( − 3, − 2, − 1). → → υ × (n 1 × n 2) = (| −1 2 −2 −1 | | , − N ot → fo r → Let υ = (1, − 1, 2). Then 1 2 −3 −1 || , 1 −1 −3 −2 |) = (5, − 5, − 5) = 5(1, − 1, − 1). Let P 0(x 0, y 0, z 0) = (1, − 1, − 3) and → n = (1, − 1, − 1). By (6.3.1) plane is and (6.3.2) , the equation of the x − y − z = (1, − 1, − 1) ⋅ (1, − 1, − 3) = 5. Processing math: 100% Example 6.3.8. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/22 8/30/2020 6.3 Equations of planes in Math Equation Find an equation for the plane through P( − 1, − 1, − 1) that contains the line of intersection of the planes x − y − z = 1 and x − 2y + z = 3. Analysis → From the line of intersection of the planes, we can obtain a point P 0(x 0, y 0, z 0) and a vector υ. Because the line of intersection of the planes is contained in the plane we seek, so P 0(x 0, y 0, z 0) is in the plane ut io n → solve the system of the two planes to get it. Solution We solve the following system ( ( ) 1 −1 −1 1 x − 2y + z = 3. N ot → (A | b) = fo r { x−y−z=1 R 1( − 1) + R 2→ 1 −2 1 3 1 −1 −1 1 0 1 −2 −2 ) D is tri b and υ is parallel to the plane. To find the normal vector, we need another vector that is is parallel to the → plane. Because P 0(x 0, y 0, z 0) and P( − 1, − 1, − 1) are in the plane, so P 0P is parallel to the plane. The → → → normal vector n = υ × P 0P. Because we need a point from the line of intersection of the planes, we must R 2(1) + R 1→ ( ( 1 −1 −1 1 0 −1 2 2 1 0 −3 −1 0 1 −2 −2 ) R 2( − 1)→ ) . The system corresponding to the last augmented matrix is Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/22 8/30/2020 6.3 Equations of planes in Math Equation { x − 3z = − 1 y − 2z = − 2, where x and y are basic variables and z is a free variable. Let z = t. Then → ut io n x = − 1 + 3z = − 1 + 3t and y = − 2 + 2z = − 2 + 2t. From the above equation of the line, we obtain P 0(x 0, y 0, z 0) = ( − 1, − 2, 0), υ = (3, 2, 1). D is tri b → P 0P = ( − 1, − 1, − 1) − ( − 1, − 2, 0) = (0, 1, − 1), and → υ × P 0P = → (| | | | | |) 2 1 1 −1 , − 3 1 0 −1 , 3 2 0 4 Let → n = (1, − 1, − 1). Then by (6.3.1) fo r = ( − 3, 3, 3) = − 3(1, − 1, − 1). and (6.3.2) , the equation of the plane is Example 6.3.9. N ot x − y − z = (1, − 1, − 1) ⋅ ( − 1, − 2, 0) = 1. Find the equation of the plane passing through the point P( − 1, 1, 1) and perpendicular to the planes 2x − y + z = 1 and x + y − z = 2. Solution Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/22 8/30/2020 6.3 Equations of planes in Math Equation → → n1 × n2 = (| −1 1 1 −1 | | | | |) , − 2 1 1 −1 , 2 −1 1 1 = (0, 3, 3) = 3(0, 1, 1). and (6.3.2) , the equation of the plane is D is tri b Let → n = (0, 1, 1). Then by (6.3.1) ut io n → → Let n 1 = (2, − 1, 1) and n 2 = (1, 1, − 1) be the normal vectors of the two given planes and → n be the → → → normal vector of the plane we must find. By Definition 6.3.4 , n is perpendicular to both n 1 and n 2 and → → thus, it is parallel to n 1 × n 2. By computation, we have y + z = (0, 1, 1) ⋅ ( − 1, 1, 1) = (0)( − 1) + (1)(1) + (1)(1) = 2. N ot Figure 6.5: (I), (II) fo r We end this section by deriving the formula for the (perpendicular) distance from a point to a plane (see Figure 6.5 (I)). Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/22 6.3 Equations of planes in Math Equation Definition 6.3.5. N ot The distance from a point Q(x 1, y 1, z 1) to a plane fo r D is tri b ut io n 8/30/2020 ax + by + cz = d is the distance between the point Q and the intersection point of the straight line passing through Q and perpendicular to the plane. Theorem 6.3.6. The distance D from a point Q(x 1, y 1, z 1) to the plane ax + by + cz = d is Processing math: 100% (6.3.6) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/22 8/30/2020 6.3 Equations of planes in Math Equation D= |ax1 + by1 + cz1 − d | √ a2 + b2 + c2 . ut io n Proof (6.3.7) D is tri b Let → n = (a, b, c) be the normal vector of the plane. Let P(x 0, y 0, z 0) be a point in the plane. Because → → n is parallel to the line passing through Q and perpendicular to the plane, the projection of PQ to → n → equals the projection of PQ to the line. The norms of the two projections are the same. Hence, we have | | → n ⋅ PQ → fo r → D = ǁProj nPQǁ = → , we obtain N ot Noting that ax 0 + by 0 + cz 0 = d, by (6.3.7) ǁ→ nǁ | | → n→ ⋅ PQ D = = Processing math: 100% ǁ n→ ǁ = | a ( x1 − x0 ) + b ( y1 − y0 ) + c ( z1 − z0 ) | √a 2 + b 2 + c 2 | ax1 + by1 + cz1 − ( ax0 + by0 + cz0 ) | √a 2 + b 2 + c 2 = | ax1 + by1 + cz1 − d | √a 2 + b 2 + c 2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/22 8/30/2020 6.3 Equations of planes in Math Equation and (6.3.6) holds. Example 6.3.10. Find the distance from the point Q(0, 0, 0) to the plane x + y + z = 1. , we have D= |ax0 + by0 + cz0 − d | √ a2 + b2 + c2 = |(1)(0) + (1)(0) + (1)(0) − 1| √ 12 + 12 + 12 = | − 1| √3 = √3 3 . D is tri b By (6.3.6) ut io n Solution The distance between two parallel planes is the distance from any point Q in one plane to the other (see Figure 6.5 (II)). Hence, to find the distance between two parallel planes, you need to find a point in one plane. Usually, one takes y = z = 0 and solves the equation of the plane for x. If the solution is denoted by x 0, Example 6.3.11. fo r then the point Q(x 0, 0, 0) is in the plane. N ot Find the distance between the following parallel planes x + y − z = 1 (1) 2x + 2y − 2z = 3. (2) Solution In equation (1), we let y = z = 0. Then x = 1 and Q(1, 0, 0) is a point in plane (1). The distance from Q(1, 0, 0) to plane (2) equals the distance between the two planes. By (6.3.6) , we have Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/22 8/30/2020 6.3 Equations of planes in Math Equation D= |ax0 + by0 + cz0 − d | √ a2 + b2 + c2 = |(2)(1) + (2(0) + ( − 2)(0) − 3| √ 2 2 + 2 2 + ( − 2) 2 = √3 6 . D is tri b 1. Find a normal vector for each of the following planes. a. x + 2y − z = 1, b. 4x − 2y − 3z = 5, c. − x − 6y + z − 1 = 0, d. 2x + 3z − 4 = 0. ut io n Exercises fo r 2. Determine whether the following planes are parallel or orthogonal. a. x − y + z = 2, 2x − 2y + 2z = 5, b. 2x − y + z = 1, x + y − z = 6, c. 2x − y + 3z = 2, x + 2y + z = 4, d. 4x − 2y + 6z = 3, − 2x + y − 3z = 5 N ot 3. For each pair of the following planes, find a parametric equation of the line of intersection of the two planes. a. x − y + z = 2, 2x − 3y + z = − 1; b. 3 x − y + 3z = − 1, − 4x + 2y − 4z = 2, c. x − 3y − 2z = 4, 3x + 2y − 4z = 6. 4. Find an equation for the line through P(2, − 3, 0) that is parallel to the planes 2x + 2y + z = 2 and x − 3y = 5. 5. Find an equation for the line through P(2, − 3, 0) that is perpendicular to the two lines Processing math: x 100% = − 1 + y, y = 2 − 2t, z = 1 − t and x = 1 − t, y = 1 − t, z = 2 + t. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/22 8/30/2020 6.3 Equations of planes in Math Equation 6. Find an equation of the plane passing through the point P and perpendicular to the vector → n. a. P(1, − 1, 0), → n = (1, 3, 5) → b. P(1, − 1, 3), n = ( − 1, 0, − 1) c. P(0, − 1, 2), → n = ( − 1, 1, − 1) d. P(1, 0, 0), → n = (1, 1, − 1) ut io n 7. Find an equation for the plane through P(2, 7, − 1) that is parallel to the plane 4x − y + 3z = 3. 8. Find an equation of the plane passing through the following given three points: a. P 1(1, 2, − 1), P 2(2, 3, 1), P 3(3, − 1, 2) D is tri b b. P 1(1, 0, 1) P 2(0, 1, − 1) P 3(1, 1, − 2) fo r 9. Find an equation for the plane containing the line x = 3 + 6t, y = 4, z = t and that is parallel to the line of intersection of the planes 2x + y + z = 1 and x − 2y + 3z = 2. 10. Find an equation for the plane through P(1, 4, 4) that contains the line of intersection of the planes x − y + 3z = 5 and 2x + 2y + 7z = 0. 11. Find an equation of the plane passing through the point Q(1, 2, − 1) and perpendicular to the planes x + y + z = 2 and − x + 2y + 3z = 5. 12. Find an equation for the plane through P(1, 1, 3) that is perpendicular to the line N ot x = 2 − 3t, y = 1 + t, z = 2t. 13. Find an equation for the plane that contains the line x = 3 + t, y = 5, z = 5 + 2t, and is perpendicular to the plane x + y + z = 4. 14. Find the distance from the point to the plane. a. P(1, − 2, − 1), x − y + 2z = − 1; b. P(0, 1, 2), 2x + y + 3z = 4; c. P(1, 0, 1), 2x − 5y + z = 3; d. P( − 1, − 1, 1), − x + 2y + 3z = 1. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 21/22 8/30/2020 6.3 Equations of planes in Math Equation N ot fo r D is tri b ut io n 15. Find the distance between the given parallel planes. a. x − y + 2z = − 3, 3x − 3y + 6z = 1 b. x + y − z = 1, 2x + 2y − 2z = − 5 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 22/22 8/30/2020 Chapter 7 Bases and dimensions ut io n Chapter 7 Bases and dimensions 7.1 Linear independence { be the same as in (1.4.2) and (4.1.5) standard matrix A given by → → a 1, ⋯, a n } → → and A = (a 1, ⋯, a n) , respectively. Let ℝ n → ℝ m be the linear transformation with fo r S= D is tri b Let N ot (7.1.1) → → T(X) = AX, where X = (x 1, x 2, ⋯, x n) T. By (2.2.3) → , we have (7.1.2) → → → → AX = x 1 a 1 + x 2 a 2 + ⋯ + x n a n . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/10 8/30/2020 Chapter 7 Bases and dimensions → → By Theorem 4.5.2 , we see that a homogeneous system AX = 0 has either only zero solution or infinitely many solutions. This, together with (7.1.2) , implies that the homogeneous system → → → → x 1a 1 + x 2a 2 + ⋯ + x na n = 0 ut io n (7.1.3) has either only zero solution or infinitely many solutions. Hence, either of the following assertions holds. 1. If (7.1.3) holds. D is tri b holds, then x 1 = x 2 = ⋯ = x n = 0. 2. There exists x 1, x 2, ⋯, x n ∈ ℝ not all zero such that (7.1.3) Hence, we introduce the following definition on linear dependence and independence. Definition 7.1.1. fo r If there exist x 1, x 2, ⋯, x n ∈ ℝ not all zero such that (7.1.4) N ot → → → → x 1a 1 + x 2a 2 + ⋯ + x na n = 0, → → → then S is said to be a linearly dependent set or the vectors a 1, a 2, ⋯, a n are said to be linearly dependent. If S is not linearly dependent, then S is said to be linearly independent. By Theorems 4.5.2 and 5.2.4 , we obtain the following criteria on linear independence. Theorem 7.1.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/10 8/30/2020 Chapter 7 Bases and dimensions The following assertions are equivalent. 1. S is linearly independent. → → ut io n 2. The homogeneous system AX = 0 has only zero solution. 3. The linear transformation T given in (7.1.1) is one to one. 4. r(A) = n. → → By Theorem 7.1.1 , we see that S is linearly dependent if and only if the homogeneous system AX = 0 has infinitely many solutions if and only if r(A) < n. Moreover, if n > m, then S is linearly dependent because r(A) ≤ min{m, n} = m < n. If S contains a zero vector, then S is linearly dependent because A contains a {a } is linearly dependent if a is a zero vector in ℝ m, → → independent if → a is a nonzero vector in ℝ m. Example 7.1.1. { } is linearly independent, where fo r 1. Determine whether S = → → → a 1, a 2, a 3 and linearly D is tri b zero column and r(A) < n. In particular, 2. Determine whether S = { N ot → → → a 1 = (1, 1, − 1), a 2 = (2, − 1, − 11), a 3 = ( − 1, 2, 10). → → → → a 1, a 2, a 3, a 4 } is linearly independent, where → → → → a 1 = (1, 0, − 1), a 2 = (2, − 1, − 10), a 3 = ( − 1, 2, 8), a 4 = (0, 2, − 5). Solution 1. Because https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/10 Chapter 7 Bases and dimensions A= ( 1 1 ) ( ) ( ) 2 −1 −1 2 − 1 − 11 10 R2 ( − 3 ) + R3 → R1 ( − 1 ) + R2 → R1 ( 1 ) + R3 1 2 −1 0 −3 3 0 −9 9 1 2 −1 0 −3 3 0 0 , 0 D is tri b r(A) = 2 < 3 = n. By Theorem 7.1.1 , S is linearly dependent. →→→→ 2. Let A = (a 1 a 2 a 3 a 4). By Theorem 2.5.2 , we have ut io n 8/30/2020 r(A) ≤ min{m, n} = m = 3 < 4 = n. By Theorem 7.1.1 , S is linearly dependent. fo r The following result gives the relation between linear dependence and linear combination. Theorem 7.1.2. the rest of vectors in S. Proof By Definition 7.1.1 such that (7.1.4) N ot → → S is linearly dependent if and only if there exists a vector a i in S such that a i is a linear combination of , we see that if S is linearly dependent, then there exist x 1, ⋯, x n, not all zero, holds. If we assume that xi is the nonzero number, then (7.1.4) implies that → → → → → x i a i = − x 1 a 1 − ⋯ − x i − 1a i − 1 − x i + 1a i + 1 − ⋯ − x n a n https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/10 8/30/2020 Chapter 7 Bases and dimensions and xn → a . xi n → → → → → Hence, a i is a linear combination of a 1, ⋯, a i − 1, a i + 1, ⋯, a n. . ut io n → x1 → xi − 1 → xi + 1 → ai = − a1 − ⋯ − a − a −⋯− xi xi i − 1 xi i + 1 D is tri b → → → → → Conversely, if a i is a linear combination of a 1, ⋯, a i − 1, a i + 1, ⋯, a n, then there exist y 1, ⋯, y i − 1, y i + 1, ⋯, y n such that → → → → → a i = y 1a 1 + ⋯ + y i − 1a i − 1 + y i + 1a i + 1 + ⋯ + y na n. If n > m, then by Theorem 7.1.1 , S is linearly dependent. If n ≤ m, then for each j ≠ i, we can use row operations R j( − y j) + R i on the transpose A T of A. Hence, the ith row of the resulting matrix denoted by B , fo r is a zero row. Moreover, A T and B are row equivalent. By Theorem 2.5.5 (i) and Theorem 2.5.4 T r(A) = r(A ) = r(B). Because B contains a zero row, r(B) < n. It follows that r(A) < n. By Theorem 7.1.1 , S is linearly dependent. Corollary 7.1.1. Let S = { } → → a 1, a 2 N ot By Theorem 7.1.2 , we obtain the following result, which shows that two nonzero vectors are linearly dependent if and only if they are parallel. → be a set of two nonzero vectors in ℝ m. Then S is linearly dependent if and only ifa 1 and → a 2 are parallel. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/10 8/30/2020 Chapter 7 Bases and dimensions Proof → → By Theorem 7.1.2 , S is linearly dependent if and only if there exists k ∈ ℝ such that a 1 = ka 2 or → → → → a 2 = ka 1. By Definition 1.1.6 , the latter holds if and only if a 1 and a 2 are parallel. ut io n The following result shows that if S is linearly dependent, then the set obtained by adding any finitely many vectors into S is linearly dependent. Corollary 7.1.2. { } ⊂ ℝ m. Then if S is linearly dependent, then is linearly dependent. } N ot Proof { → → → → a 1, ⋯, a n, b 1, ⋯, b k fo r S ∪ T: = D is tri b Let T = → → b 1, ⋯, b k → Because S is linearly dependent, one of vectors in S, say a i , is a linear combination of the rest of the → vectors in S. By Theorem 1.4.2 , a i is a linear combination of the rest of the vectors in S and all vectors in T. It follows that one of the vectors in S ∪ T is a linear combination of the rest of the vectors in S ∪ T. The result follows from Theorem 7.1.2 . Example 7.1.2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/10 8/30/2020 Chapter 7 Bases and dimensions Determine whether { → → → a 1, a 2, a 3 } is linearly dependent, where Solution { } → → a 1, a 3 is linearly dependent. By Corollary 7.1.2 dependent. By Theorems 7.1.2 and 5.2.2 , we obtain Theorem 7.1.3. then | → } is linearly → → → . Let b ∈ ℝ . If b is a linear combination of a 1, a 2, ⋯, a n, → m → → → → a 1, a 2, ⋯, a n, b N ot { { → fo r → → Let A = (a 1, ⋯, a n) be the same as in (4.1.5) , → → → a 1, a 2, a 3 D is tri b → → Because a 1 = 2a 3, ut io n → → → a 1 = (2, 4, 6), a 2 = (1, 0, 1), a 3 = (1, 2, 3). } is linear dependent and r(A) = r(A b). The following result shows that any nonempty subset of linearly independent vectors is linearly independent. Theorem 7.1.4. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/10 8/30/2020 Chapter 7 Bases and dimensions Let S 1 be a nonempty subset of S. If S is linearly independent, so is S 1. Proof The proof is by contradiction. Assume that S 1 is not linearly independent, that is, S 1 is linearly dependent. By Theorem 7.1.2 , one of the vectors in S 1 is a linear combination of the rest of the vectors in S 1. By ut io n Corollary 7.1.2 , S is linearly dependent, which contradicts the assumption that S is linearly independent. The contradiction shows that the assumption that S 1 is linearly dependent is false. Hence, S 1 is linearly independent. → → Let e 1 = (1, 0, 0) and e 3 = (0, 0, 1). Then { } → → e1, e3 D is tri b Example 7.1.3. is linearly independent. Solution { } is linearly independent. By Theorem 7.1.4 If n = m, then by Theorems 4.5.3 determine linear independence. N ot S= → → → e1, e2, e3 fo r →→→ → Let I 3 = ( e 1 e 2 e 3 ), where e 2 = (0, 1, 0). Then r(I 3) = 3. It follows from Theorem 7.1.1 and 7.1.1 , S1 = { } → → e1, e3 that is linearly independent. , we obtain the following result, which uses determinants to Theorem 7.1.5. Let A be an n × n matrix and let S be the set of column vectors of A. Then the following assertions hold. 1. S is linearly dependent if and only if |A | = 0 if and only if r(A) < n. 2. S is linearly independent if and only if |A | ≠ 0 if and only if r(A) = n. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/10 8/30/2020 Chapter 7 Bases and dimensions Example 7.1.4. → → → → Let e 1 , e 2 , e 2 , ⋯, e n be the standard vectors in ℝ n given in (1.1.1) . Show that S = { → → → e 1 , e 2 , ⋯, e n } is linearly independent. | | = | In | = 2 ≠ 0, it follows from Theorem 7.1.5 Because A ut io n Solution (2) that S is linearly independent. Determine whether S given in Example 7.1.1 is linearly independent by using Theorem 7.1.5 Solution Exercises 2 −1 1 −1 2 − 1 − 11 10 | | | R1 ( 1 ) + R2 ‗ 1 2 R1 ( − 1 ) + R2 0 −3 0 −9 . −1 3 = 0, 9 (1) that S is linearly dependent. N ot it follows from Theorem 7.1.5 | 1 fo r |A | = D is tri b Example 7.1.5. → → → 1. Let a 1 = (2, 4, 6) T, a 2 = (0, 0, 0) T, and a 3 = (1, 2, 3) T. Determine whether { → → → a 1, a 2, a 3 } is linearly dependent. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/10 8/30/2020 Chapter 7 Bases and dimensions } is linearly dependent, where () () () () 1 → a1 = 1 → , a2 = 2 1 0 3. Determine whether S = { } 0 → , a4 = 0 2 0 . 1 is linearly independent, where → , a2 = 3 } → 9 , a3 = 8 −1 → − 2 , a4 = 1 6 −2 . 5 fo r → → → a 1, a 2, a 3 7 is linearly independent, where N ot { 2 0 ( ) () ( ) ( ) 1 −4 4. Determine that S = 1 → , a3 = 0 → → → → a 1, a 2, a 3, a 4 → a1 = 0 0 ut io n { D is tri b 2. Determine whether → → → → a 1, a 2, a 3, a 4 → a1 = ( ) ( ) () −1 0 −1 → , a2 = 3 → −1 , a = 3 −6 1 2 . 3 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/10 8/30/2020 7.2 Bases of spanning and solution spaces 7.2 Bases of spanning and solution spaces ut io n Let (7.2.1) be a set of vectors in ℝ m. In Section 1.4 1.4.2 ) { , we introduced the following spanning space (see Definition { fo r (7.2.2) } → → → x 1a 1 + x 2a 2 + ⋯ + x na n : x i ∈ ℝ, i ∈ I n . N ot V : = span S = } D is tri b S= → → → a 1, a 2, ⋯, a n → → → In this spanning space, the vectors a 1, a 2, ⋯, a n are not required to be linearly independent. In some cases, we are interested in the case when these vectors are linearly independent. Hence, we introduce the notion of a basis. Definition 7.2.1. If S is linearly independent, then S is called a basis of V; n is called the dimension of V, denoted by dim V = n; and V is said to be an n dimensional subspace of ℝ m. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/20 8/30/2020 7.2 Bases of spanning and solution spaces Example 7.2.1. Let { be the set of standard vectors in ℝ n given in (1.1.1) assertions hold. or Example 7.1.4 1. ℝ n = span S. 2. S is linearly independent. 3. dim(ℝ n) = n. fo r Solution (1) follows from (1.4.9) } . Then the following D is tri b E: = → → → e 1 , e 2 , ⋯, e n ut io n (7.2.3) and (2) follows from Example 7.1.4 . By Definition 7.2.1 , the set E of the standard vectors constitutes a basis of ℝ n, dim(ℝ n) = n, and (3) holds. N ot The basis E in (7.2.3) is called the standard basis of ℝ n. Because the dimension of ℝ n is n, we call ℝ n an n-dimensional (Euclidean) space. Using vectors in S, we define a matrix as follows. (7.2.4) →→ → A = (a 1 a 2⋯ a n). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/20 8/30/2020 7.2 Bases of spanning and solution spaces By Definition 7.2.1 and Theorems 7.1.1 and 7.1.5 , we obtain the following result. Theorem 7.2.1. Let S and A be the same as in (7.2.1) and (7.2.4) . Then the following assertions hold. ut io n i. S is a basis of V if and only if r(A) = n. ii. If n = m, then S is a basis of V if and only if |A | ≠ 0 if and only if r(A) = n n if and only if A is invertible. Example 7.2.2. { } → → a 1, a 2 D is tri b → → T 1. Let a 1 = (1, 0, 1) and a 2 = (1, 1, 1) T. Show that S = dim(span S) = 2. is a basis of span S and → → → T T 2. Let a 1 = (1, 0) , a 2 = (1, 1) , and a 3 = (2, − 3) T. Determine whether S = N ot Solution →→ 1. Let A = (a 1 a 2). Then } is a basis fo r of span S. { → → → a 1, a 2, a 3 ( ) ( 1 0 1 R1 ( − 1 ) + R2 1 0 1 A = → 1 1 1 0 1 0 T ) and r(A T) = 2. By Theorem 2.5.5 , r(A) = r(A T) = 2 and by Theorem 7.2.1 span S. By Definition 7.2.1 , dim(span S) = 2. , S is a basis of https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/20 8/30/2020 7.2 Bases of spanning and solution spaces 2. Note that n = 3 and m = 2. Because n > m, it follows from Theorem 7.2.1 of span S. that S is not a basis Theorem 7.2.2. → ut io n Assume that S is a basis of span S. Then for each b ∈ span S, there exists a unique vector (x 1, x 2, ⋯, x n) such that (7.2.5) → → → b = x 1a 1 + x 2a 2 + ⋯x na n. D is tri b → Proof →→ → Let A = (a 1 a 2⋯ a n). By Theorem 7.2.1 → , r(A) = n ≤ m. Because b ∈ span S, by (7.2.2) has a solution. By Theorem 4.5.1 (7.2.5) has a unique solution. | N ot Note that the bases of span S are not unique. For example, if S = β i ≠ 0 for i = 1, 2, ⋯, n, then → (3), r(A) = r(A b) = n. By Theorem 4.5.1 fo r (7.2.5) T: = { { → → → β 1a 1, β 2a 2, ⋯, β na n → → → a 1, a 2, ⋯, a n } , the system (1), is a basis of span S and } is a basis of span S because T is linearly independent and span S = span T. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/20 8/30/2020 7.2 Bases of spanning and solution spaces Now, we assume that S is a basis and { → → → b 1, b 2, ⋯, b k } in span S is another basis such that { → → → span a 1, a 2, ⋯, a n } holds, is k equal to n? Equivalently, is the dimension of a spanning space D is tri b The question is that if (7.2.6) unique? } { → → → = span b 1, b 2, ⋯, b k . ut io n (7.2.6) In the following, we shall prove k = n, so we give a positive answer to the above question. Lemma 7.2.1. and let T = { → → → b 1, b 2, ⋯, b k } be a set of vectors in span S. Let fo r Let S be the same as in (7.2.1) N ot →→ → →→ → A m × n = (a 1 a 2⋯ a n) and B m × n = (b 1 b 2⋯ b n). Then the following assertions hold. i. There exists an n × k matrix C n × k such that B m × k = A m × nC n × k. ii. If k = n n and S and T are linearly independent, then there exists a unique invertible n × n matrix C n × n such that (7.2.7) B m × n = A m × nC n × n. Proof https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/20 7.2 Bases of spanning and solution spaces → i. Because b i ∈ span S for i ∈ I k, by (2.2.3) that (7.2.8) → , there exists a vector C i : = (c 1i, c 2i, ⋯, c ni) T such → → → → → b i = c 1ia 1 + c 2ia 2 + ⋯ + c nia n = AC i . →→ → Let C n × k = (C 1C 2⋯C k). By (2.2.6) and (7.2.8) , we have ut io n 8/30/2020 D is tri b →→ → → → → B m × k = (b 1 b 2⋯ b k ) = (A m × kC 1A m × kC 2⋯A m × kC k) = A m × nC n × k. fo r ii. Because S is linearly independent, by Theorem 7.2.1 , there exists a unique vector → C i : = (c 1i, c 2i, ⋯, c ni) T such that (7.2.8) holds for i ∈ I n and (7.2.7) with k = n holds. We will → prove that C n × n is invertible by contradiction. If not, then there exists a nonzero vector X 0 ∈ ℝ n → → such that C n × nX 0 = 0. Because B m × n = A m × nC n × n, it follows that N ot → → → → B m × nX 0 = A m × nC n × nX 0 = A 0 = 0. By Theorem 7.1.1 , T is linearly dependent, which contradicts the hypothesis that T is linearly independent. Hence, C n × n is invertible. Lemma 7.2.2. Let S and T be the same as in Lemma 7.2.1 following assertions hold. . Assume that S is linearly independent. Then the https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/20 8/30/2020 7.2 Bases of spanning and solution spaces i. If k > n, then T is linearly dependent. ii. If T is linearly independent, then k ≤ n. Proof →→ → →→ → A m × n = (a 1 a 2⋯ a n) and B m × k = (b 1 b 2⋯ b k ). (i), there exists an n × k matrix C n × k such that B m × k = A m × nC n × k. Because k > n, it → → follows from Theorem 4.5.2 (2) that the homogeneous system C n × kX = 0 has infinitely many solutions. This implies that the homogeneous system → → D is tri b By Lemma 7.2.1 ut io n Note that (i) and (ii) are equivalent, so we only prove (i). Let → → B m × kX = A m × nC n × kX = A m × n 0 = 0 has infinitely many solutions. By Theorem 7.1.1 , T is linearly dependent. fo r By Lemma 7.2.2 (ii), we see that the number of linearly independent vectors in span S is less than or equal to the number of vectors in S. Theorem 7.2.3. N ot The following result shows that if T is linearly independent and span T = span S, then k = n. Let S and T be the same as in Lemma 7.2.1 k = n if and only if . Assume that S and T are linearly independent. Then (7.2.9) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/20 8/30/2020 7.2 Bases of spanning and solution spaces { → → → span a 1, a 2, ⋯, a n } { } → → → = span b 1, b 2, ⋯, b k . Proof (ii), n ≤ k. Hence, k = n. Now, we assume that k = n. In this case, T : = → → → b 1, b 2, ⋯, b n } is a set of → holds. Because b i ∈ span S for each D is tri b vectors in the spanning space span S. We prove that (7.2.9) → { ut io n Assume that (7.2.9) holds. We will prove k = n. By (7.2.9) , T is the set of vectors in span S. By Lemma 7.2.2 (ii), k ≤ n. Similarly, by (7.2.9) , S is the setofvectorsinspan T and by Lemma 7.2.2 → i ∈ {1, 2, ⋯, n}, every vector b ∈ span T implies that b ∈ span S. We will prove that for → → b ∈ span S, b ∈ span T. Let → fo r →→ → →→ → A m × n = (a 1 a 2⋯ a n) and B m × n = (b 1 b 2⋯ b n). → → −1 → Let X = C n × n Y. Then N ot By Lemma 7.2.1 → → , there exists a vector Y : = (y 1, y 2, ⋯, y n) ∈ ℝ n such that A m × nY = b. −1 (ii), there exists a unique invertible n × n matrix C n × n such that A m × n = B m × nC n × n . For each b ∈ span S, by (7.2.2) BX = B m × nC n−×1n Y = A m × nY = b. → → → → → This implies b ∈ span T. Hence, span T = span S. By Theorem 7.2.3 , we see that the dimension of span S is independent of the choices of linearly independent vectors whose spanning space is span S and thus, the dimension of a spanning space is unique. Moreover, the spanning space of any n linearly independent vectors in span S is equal to span S. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/20 8/30/2020 7.2 Bases of spanning and solution spaces Theorem 7.2.3 requires all vectors of T to be in span S. If this condition is not satisfied, then span T might not be equal to span S. Example 7.2.3. {( )} 0 and T = {( )} 0 1 . Show that span T ≠ span S. Solution Note that () 0 1 0 :t ∈ ℝ and span T = ∉ span S and thus, span T ≠ span S. {( ) } 0 s :s ∈ ℝ . fo r It is obvious that {( ) } t D is tri b span S = ut io n Let S = 1 Corollary 7.2.1. Let S = { → → → a 1, a 2, ⋯, a n } and T = { N ot However, in Theorem 7.2.3 , if m = n, then we do not need to verify the condition that all vectors of T are in span S because this condition is satisfied automatically. → → → b 1, b 2, ⋯, b n } be sets of linearly independent vectors in ℝ n. Then span T = span S = ℝ n. Proof https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/20 8/30/2020 7.2 Bases of spanning and solution spaces By Theorem 7.2.3 } , span S = span E = ℝ n. Similarly, we can prove that span T = ℝ n. Hence, span E. By Theorem 7.2.3 T = span S. , we see that any set of n linearly independent vectors in ℝ n constitutes a basis of ℝ n. ut io n By Corollary 7.2.1 { → → → (1), ℝ = span e 1 , e 2 , ⋯, e n . Because all vectors of S are in ℝ n, they are in span n Theorem 7.2.4. { → → → a 1, a 2, ⋯, a n } be a basis of span S and assume that S ≠ ℝ m. Then S 1 = → linearly independent for each b ∉ span S. Proof Let A be the same as in (7.2.4) { → → → → a 1, a 2, ⋯, a n, b D is tri b Let S = . Then by Theorems 7.2.1 is , fo r (7.2.10) and 2.5.2 } → By (1.4.9) { N ot n = r(A) ≤ r(A | b) ≤ n + 1. } → → → , ℝ m = span e 1 , e 2 , ⋯, e m . Because span S ≠ ℝ m, by Theorem 7.2.3 , n < m and thus → n + 1 ≤ m. Let b ∉ span S. Then the system (7.2.11) → → → b = x 1a 1 + x 2a 2 + ⋯x na n → https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/20 8/30/2020 7.2 Bases of spanning and solution spaces → → has no solutions. By Theorem 4.5.1 (3), r(A) < r(A | b). By (7.2.10) 7.1.1 (i), S 1 is linearly independent. S= { → → → a 1, a 2, ⋯, a n , if n < m, then we can extend a set of linearly independent vectors } toabasisof ℝ m. →→ → Let A = (a 1 a 2⋯ a n) be the same as in (7.2.4) . We use methods given ut io n By Theorem 7.2.4 , r(A | b) = n + 1. By Theorem → → → m in Sections 4.7 or 4.8 to find a vector b 1 ∈ ℝ such that the system AX = b 1 has no solutions (Section 4.7) or , S1 : = { }{ → S, b 1 = → → → → a 1, a 2, ⋯, a n, b 1 D is tri b → such that b 1 ∈ span S (Section 4.8). By Theorem 7.2.4 } is linearly independent. If n + 1 = m, then S 1 isabasisof ℝ m. If n + 1 < m, then repeating the process and using S 1, we → → obtain a vector b 2 such that b 2 ∉ span S 1 and S 2 = { } → S 1, b 2 is linearly independent. We repeat the process until we find S m − n. fo r Example 7.2.4. basis of span S. Solution N ot → → → T Let a 1 = (1, 0, 0, 1) and a 2 = (1, − 1, 1, − 1) T. Find a vector b ∈ ℝ 4 such that S = { → → → a 1, a 2, b } is a →→ → Let A = (a 1 a 2) and b = (b 1, b 2, b 3, b 4) T. Then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/20 7.2 Bases of spanning and solution spaces ( ) ( ) ( ) 1 1 b1 1 1 0 − 1 b2 R1 ( − 1 ) + R4 0 − 1 → 0 1 b3 0 1 → (A | b) = 1 − 1 b4 → R2 ( 1 ) + R3 b2 b3 0 − 2 b4 − b1 1 1 b1 0 −1 b2 0 0 b3 + b2 = (B | → c). D is tri b R2 ( − 2 ) + R4 b1 ut io n 8/30/2020 0 0 b 4 − b 1 − 2b 2 → → If either b 3 + b 2 ≠ 0, or b 4 − b 1 − 2b 2 ≠ 0, then r(A) = 2 and r(A | b) ≥ 3. Hence, r(A) < r(A | b). It follows from Theorem 4.5.1 (3) that if either b 3 + b 2 ≠ 0, or b 4 − b 1 − 2b 2 ≠ 0, then the system has no → By Theorem 7.2.4 Theorem 7.2.5. ,S= { → → → a 1, a 2, b } is a basis of span S. N ot Theorem 7.2.4 fo r solutions. Hence we can choose b 1 = b 4 = 0, b 2 = b 3 = 1. Then b 3 + b 2 ≠ 0. Let b = (0, 1, 1, 0) T. By , we prove the following result. Let S and A be the same as in (7.2.1) and (7.2.4) . Let T = { → → → b n , b n , ⋯, b n 1 2 r } be a set of r vectors in span S. Assume that T is linearly independent and r(A) = r. Then the following assertions hold. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/20 8/30/2020 7.2 Bases of spanning and solution spaces i. span T = span S. ii. Each of the vectors in S but not in T is a linear combination of vectors in T. Proof i. It is obvious that every vector in span T is in span S. We will prove that every vector in span S → → } D is tri b { → → → → b n , b n , ⋯, b n , b 1 2 r ut io n must belong to span T. In fact, if not, then there exists a vector b ∈ span S and b ∈ span T. Because T is linearly independent, by Theorem 7.2.4 , the following set is linearly independent. Hence, we have r(A) ≥ r + 1, which contradicts the condition r(A) = r. → → → ii. For every vector b in S, because span T = span S, b ∈ span T. Hence, b is a linear combination of vectors in T. In particular, result (ii) holds. { } Theorem 7.2.6. Let { → → → a 1, a 2, ⋯, a n } N ot fo r → → → The following result provides a method to choose basis vectors from span a 1, a 2, ⋯, a n . be a set of vectors in ℝ m and let A be the same as in (7.2.4) r(A) = r ≥ 1. Then there exists a subset { → → → a n , a n , ⋯, a n 1 2 r }{ of . Assume that } → → → a 1, a 2, ⋯, a n . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/20 8/30/2020 7.2 Bases of spanning and solution spaces { → → → span a 1, a 2, ⋯, a n } { } → → → = span a n , a n , ⋯, a n . 1 2 r Proof . Because r(A) = r ≥ 1, there is at least one nonzero vector in S. We → choose one nonzero vector in S, denoted by a n . Let I n : = 1, 2, ⋯, n and we use the notation: { 1 {} → → 1. If a i ∈ span a n 1 means that i ∈ I n with i ≠ n i. D is tri b { } i ∈ I n\ n 1 } ut io n Let S be the same as in (7.2.1) { } for i ∈ I n\ n 1 , then there exists y i ∈ ℝ such that (7.2.12) {} → a n . Then span S = span T. Indeed, it is obvious that span T is a subset of span S. We N ot Let T = fo r → → y ia n = a i . 1 1 → prove that span S is a subset of span T. Let b ∈ span S. Then there exist x 1, x 2, ⋯, x n ∈ ℝ such that → b = x 1a 1 + x 2a 2 + ⋯ + x n a n + ⋯ + x na n 1 1 = x 1y 1a n + x 2y 2a n + ⋯ + x n a n + ⋯ + x ny na n 1 2 1 1 1 = (x 1y 1 + x 2y 2 + ⋯ + x n + ⋯ + x ny n)a n . 1 1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/20 8/30/2020 7.2 Bases of spanning and solution spaces → This shows b ∈ span T and span S = span T. Now, we prove that r(A) = 1. By (7.2.12) 1 { } → → an , a1 1 is linearly dependent and →→ → r((a n a 1)) = r((a n ) = 1. 1 1 7.1.3 , { → → → a n , a 1, a 2 1 } → , a 2 is a linear combination of is linearly dependent and { } → → a n , a 1 . By Theorem 1 D is tri b → → Because y 2a n = a 2, by Theorem 1.4.2 1 ut io n → linear combination of a n . By Theorem 7.1.3 → , a 1 is a fo r → → → →→ → r((a n , a 1, a 2)) = r((a n a 1)) = r((a n ) = 1. 1 1 1 Repeating the process implies N ot → →→ → → → r((a n a 1 a 2⋯a j − 1a j + 1⋯ a n)) = 1. 1 → → → → → → Let B = (a n , a 1, a 2, ⋯a j − 1, a j + 1, ⋯, a l ). By the third row operations, it is easy to prove 1 r(B T) = r(A T). By Theorem 2.5.5 , r(B) = r(A) = 1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/20 8/30/2020 7.2 Bases of spanning and solution spaces {} → → 2. In the above, we have proved that if a i ∈ span a n 1 for each i ∈ I n : = {} Theorem 7.2.4 , { } → → an , an 1 2 {} → such that a n ∉ span a n . By 2 1 is linearly independent and by Theorem 7.1.1 { } } { } 1 2 for each D is tri b → → → Without loss of generalization, we assume that n 1 < n 2. If a i ∈ span a n , a n 1 2 { → → , r((a n , a n )) = 2. ut io n → then r(A) = 1. Hence, if r(A) > 1, there exists a n ∈ S\ a n 2 1 {1, 2, ⋯, n } with i ≠ n1, i ∈ I n\ n 1, n 2 , then for each i ∈ I n\ n 1, n 2 , there exists y 1, y i2 ∈ ℝ such that (7.2.13) → { } → → a n , a n . Then span S = span T. We prove that span S is a subset of span T. Let 1 2 N ot Let T = fo r → → → y i1a n + y i2a n = a i . 1 2 b ∈ span S. Then there exist x 1, x 2, ⋯, x n ∈ ℝ such that https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/20 8/30/2020 7.2 Bases of spanning and solution spaces → b = x 1a 1 + x 2a 2 + ⋯ + x n a n + ⋯ + x n a n + ⋯ + x na n 1 1 2 2 n i = 1 , i ≠ n1 , n2 n = ∑ i = 1 , i ≠ n1 , n 2 = [ x ia i + x n a n + x n a n 1 1 2 2 → → x i(y i1a n + y i2a n ) + x n a n + x n a n 1 2 1 1 2 2 n (∑ i = 1 , i ≠ n1 , n 2 x iy i1) + x n 1 ] [ n → an + ( ∑ 1 → i = 1 , i ≠ n1 , n 2 ut io n ∑ x iy i2) + x n 2 ] an . 2 D is tri b = → , a 1 is a This shows b ∈ span T and span S = span T. Now, we prove that r(A) = 2. By (7.2.13) linear combination of { } → → a n , a n . By Theorem 7.1.3 1 1 , { → → → an , an , a1 1 2 } is linearly dependent and fo r → →→ →→ r((a n a n a 1)) = r((a n a n ) = 2. 1 2 1 2 Theorem 7.1.3 , { N ot → → → Because y 21a n + y 22a n = a 2, by Theorem 1.4.2 1 2 → → → → a n , a n , a 1, a 2 1 2 } → → → → , a 2 is a linear combination of a n , a n , a 1. By 1 2 is linearly dependent and → → →→ → →→ →→ r((a n a n a 1 a 2)) = r((a n a n a 1)) = r((a n a n ) = 2. 1 2 1 2 1 2 Repeating the process implies https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/20 8/30/2020 7.2 Bases of spanning and solution spaces → →→ → → → → → → r((a n a n a 1, a 2, ⋯a n − 1, a n + 1⋯a n − 1, a n + 1⋯ a n)) = 2. 1 2 1 1 2 2 → →→ → → → → → → Let B = (a n a n a 1, a 2, ⋯a n − 1, a n + 1⋯a n − 1, a n + 1⋯ a n). Then, by the third row operations, it 1 2 1 1 2 2 , r(B) = r(A) = 2. ut io n is easy to prove r(B T) = r(A T). By Theorem 2.5.5 Finally, we apply linear independence to study bases of the solution spaces of the homogenous system → → AX = 0 . D is tri b Theorem 7.2.7. → → → Let A be the same as in (7.2.4) . Assume that null(A) = k. Then the basis vectors υ 1 , υ 2 , ⋯, υ k given in Definition 4.3.1 are linearly independent. Proof (7.2.14) fo r Assume that there exist x 1, ⋯, x k ∈ ℝ such that N ot → → → → x 1 υ 1 + x 2 υ 2 + ⋯ + x k υ k = 0. → → → Let υ = x 1 υ 1 + x 2 υ 2 + ⋯ + x k υ k . By Theorem 4.3.1 → (P 1), the components on n 1, n 2⋯, n k(1 ≤ n 1 < n 2 < ⋯ < n k ≤ n) are x 1, x 2, ⋯, x k, respectively. By (7.2.14) → → → x 1 = x 2 = ⋯ = x k = 0 and thus, υ 1 , υ 2 , ⋯, υ k are linearly independent. → → , υ = 0 implies that https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/20 8/30/2020 7.2 Bases of spanning and solution spaces → → → , we see that the set of vectors υ 1 , υ 2 , ⋯, υ k given in is a basis of the solution space N A. As mentioned above, the bases of a spanning space By Theorem 7.2.7 and Definition 7.2.1 Definition 4.3.1 are not unique; hence, the solution space N A has many other bases. Theorem 7.2.8. → → → 1. u 1, u 2, ⋯, u l are linearly independent. { } N ot Proof fo r → → → 2. N A = span u 1, u 2, ⋯, u l . Then l = k. → → → and let u 1, u 2, ⋯, u l be vectors D is tri b → → → Let υ 1 , υ 2 , ⋯, υ k be the basis vectors of N A given in Definition 4.3.1 in N A. Assume that the following conditions hold. ut io n By Theorems 7.2.3 and 7.2.7 , we will prove the following result, which shows that the number of vectors in all bases of N A is equal to the nullity null(A) of the matrix A. → → → By Theorem 7.2.7 , the basis vectors υ 1 , υ 2 , ⋯, υ k are linearly independent. By conditions (1) and (2) and Theorem 7.2.3 , l = k. Theorem 7.2.8 { → → → shows that any set of k-linearly independent vectors in N A = span υ 1 , υ 2 , ⋯, υ k constitutes a basis of N A. We shall use Theorem 7.2.8 in the proof of Theorem 8.2.4 } . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/20 8/30/2020 7.2 Bases of spanning and solution spaces Exercises → → 1. Let a 1 = (1, − 1, 1, 0), and a 2 = (1, 0, 1, 1). Determine whether S = { } → → a 1, a 2 is a basis of span S. If 2. Determine whether S = { → → → a 1, a 2, a 3 } is a basis of span S, where D is tri b → → → i. a 1 = (1, 0, − 1, 1), a 2 = (0, 0, 1, 1) and a 3 = ( − 2, 1, 0, 1). → → → ii. a 1 = (1, 0, 1, 1), a 2 = (0, 0, 1, − 1) and a 3 = (1, 0, 2, 0). ut io n so, find dim(span S). → → → → 3. Let a 1 = (1, − 1), a 2 = (0, 1), a 3 = ( − 3, 2) and a 4 = (1, 0). Determine whether S = fo r is a basis of span S. → → 4. Let a 1 = (1, 0, 1) and a 2 = (1, 1, 1). Find a vector S = to prove Theorem 7.2.2 to prove Theorem 7.2.4 } } is a basis of span S. . . N ot 5. Use Theorem 7.1.1 6. Use Theorem 7.1.1 { → → → a 1, a 2, b { → → → → a 1, a 2, a 3, a 4 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/20 8/30/2020 7.3 Methods of finding bases of spanning spaces 7.3 Methods of finding bases of spanning spaces and (7.2.4) Definition 7.3.1. and B = AT . Let T be the linear transformation from fo rD is tri bu tio n Let S and A be the same as in (7.2.1) n R to Rm with standard matrix A. The spanning space span S is said to be the column space of A, and to be the row space of B. We denote by CA and RB the column space of A and the row space of B, respectively. Then (7.3.1) n CA = RB = span S = A(R ≠ RA If the set S given in (7.2.1) . linearly independent vectors in Rm , say T is, (7.3.2) → → → = { b1 , b2 , → span{ b1 , b2 , By Definition 7.2.1 ). is not linearly independent, then it is not a basis of span S. Our purpose is to find → ⋯ , br } N ot Note that in general, CA n ) = T (R in Rm , such that span T → → → ⋯ , br } = span{a1 , a2 , = span S, that − → ⋯ , an }. , T is a basis of span S. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/14 8/30/2020 7.3 Methods of finding bases of spanning spaces Usually, we seek bases with all the vectors satisfying one of the following requirements: 1. All the basis vectors in T belong to S. 2. The basis vectors in T may or may not be in S. fo rD is tri bu tio n We first prove two results, which provide the methods to find basis vectors that satisfy the above requirement (1) or (2). We start with bases that satisfy the above requirement (1). Let 7.3.3 c11 c12 ⋯ c1n ⎜ c21 ⎜ C = ⎜ ⎜ ⎜ ⋮ c22 ⋯ ⋮ ⋱ c2n ⎟ ⎟ ⎟ ⎟ ⎟ ⋮ cm2 ⋯ ⎛ ⎝ − → cm1 cmm ⎞ ⎠ be a row echelon matrix and Ci the ith column vector of C. N ot By Theorems 7.1.1 and 7.2.5 , we prove the following result, which shows that the column vectors with the leading entries of a row echelon matrix C form a basis for the column space of C. Lemma 7.3.1. − − → Let C be a row echelon matrix and let Cn , entries. Then the following assertions hold. 1 i. − − → − − → − − → {Cn 1 , Cn 2 ⋯ , Cn r } − − → − − → Cn 2 ⋯ , Cn r be the column vectors of C containing leading is linearly independent. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/14 8/30/2020 7.3 Methods of finding bases of spanning spaces − − → ii. span{Cn − − → − − → − → − → − → , Cn 2 ⋯ , Cn r } = span{C1 , C2 ⋯ , Cn } 1 . Proof 7.1.1 − − →− − → − − → = (Cn 1 Cn 2 ⋯ Cn r ). By Theorem 2.5.1 − − → (1), implies that {Cn − − → (4), r(M ) = r. This, together with Theorem − − → is linearly independent. , Cn 2 ⋯ , Cn r } 1 ii. By the result (i) and Theorem 7.2.5 fo rD is tri bu tio n i. Let M , the result (ii)holds. By Lemma 7.3.1 , we prove the following result, which shows that if C is a row echelon matrix of A and a set of column vectors of C formsabasisfor the column space of C, then the corresponding column vectors of A form a basis for the column space of A, which is equal to span S. Theorem 7.3.1. Let S be the same as in (7.2.1) − →− → − → C = (C1 C2 ⋯ Cn ) →→ and A = (a1 a2 − → ⋯ an ) be the same as in (7.2.4) − − → be a row echelon matrix of A. Let Cn 1 − − → , Cn 2 , − − → ⋯ , Cn r be the column vectors of C − → − → , an 2 , 1 containing leading entries. Then the set of the pivot columns of A, that is, {an Proof − → ⋯ , an r } is a basis N ot of span S. . Let Because C is a row echelon matrix of A and there are r columns vectors of C that contain the leading entries of C, by Theorem 2.5.1 , r(A) = r(C) = r. By Lemma 7.3.1 , we have (7.3.4) − − → − − → span{Cn 1 , Cn 2 , − − → − → − → ⋯ , Cn r } = span{C1 , C2 , − → ⋯ , Cn }. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/14 8/30/2020 7.3 Methods of finding bases of spanning spaces It follows from Theorem 2.7.3 (2.2.6) , we have that there exists an invertible matrix F such that A = F C. Hence, by − →− → − → − → − → − → →→ − → A = (a1 a2 ⋯ an ) = F (C1 C2 ⋯ Cn ) = F (F C1 F C2 ⋯ F Cn ) (7.3.5) fo rD is tri bu tio n and thus, − → → ai = F Ci Let A∗ ∗ A − →− → − → = (an 1 an 2 ⋯ an r ) and C ∗ − − → − − →− − → − − → = (Cn 1 Cn 2 ⋯ Cn r ). − − → − − → for i ∈ In . By (7.3.5) − − →− − → and (2.2.6) − − → , − →− → − → ∗ = (an 1 an 2 ⋯ an r ) = (F Cn 1 F Cn 2 ⋯ F Cn r ) = F (Cn 1 Cn 2 ⋯ F Cn r ) = F C . invertible, by Corollary 2.7.3 , ∗ r(A ) = r(C By (7.3.5) (1), S ∗ and (7.3.4) − → − → = {an 1 , an 2 , , we have ) = r(C) = r(A) = r. − → ⋯ , an r } is linearly independent. N ot By Theorem 7.1.1 ∗ → → span S = span{a1 , a2 , − → − → − → − → ⋯ , an } = span{F C1 , F C2 , − → = F span{C1 , C2 , − − → − → − − → − → ⋯ , F Cn } − − → ⋯ , Cn } = F span{Cn 1 , Cn 2 , − − → = span{F Cn 1 , F Cn 2 , Because F is − − → ⋯ , F Cn r } = span S ∗ − → . an 2 , − − → ⋯ , E Cn r } − → ⋯ , an r }. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/14 8/30/2020 7.3 Methods of finding bases of spanning spaces By Definition 7.2.1 , S ∗ is a basis of span S ∗ and thus a basis of span S. Next, we work with bases that satisfy the above requirement (2). We first give the following result, which shows that if B and D are row equivalent (see Definition 2.5.2 ), then all the row vectors of D are contained in the spanning space of the row vectors of B. fo rD is tri bu tio n Lemma 7.3.2. Assume that B and D are row equivalent. Then, each row of D is a linear combination of the row vectors of B. Proof Because B and D are row equivalent, by Theorem 2.7.3 D = F B. By Theorem 2.2.3 , we have D where A is the same as in (7.2.4) . → − → be the ith row vector of D for i ∈ In and Xi i ∈ In . By (2.2.6) , we have for i ∈ In , di →T di and the result follows. T = B F = (x1i , T = AF T ⋯ xni ) , T be the ith column vector of F T for N ot Let T , there exists an invertible matrix F such that T − → = AX i → → − → = x1i a1 + x2i a2 + ⋯ + xni an Theorem 7.3.2. Assume that D is a row-echelon matrix of B. Then the row vectors with the leading entries of D form a basis for RB . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/14 8/30/2020 7.3 Methods of finding bases of spanning spaces Proof Let → → → − → d1 , d2 , ⋯ , dn → d1 , d2 , → ⋯ , dr be the row vectors of D. Because D is a row-echelon matrix, we assume that − − → − → are the row vectors of D with the leading entries. Then other rows from dr+1 to dn are →→ → , the set D∗ is fo rD is tri bu tio n zero rows if r < n. Let D∗ = ( d1 d2 ⋯ dr ). Because r(D∗ ) = r, by Theorem 7.1.1 linearly independent. Then by Theorem 2.5.4 and Theorem 2.5.1 (3), ∗ r(B) = r(D) = r(D ) = r. Because D is a row-echelon matrix of B, it follows from Lemma 7.3.2 → → → di ∈ span{a1 , a2 , where S is the same as in (7.2.1) . By Theorem 7.2.5 → → span{ d1 , d2 , → d2 , By Theorems 7.3.1 → ⋯ , dr } is a basis of RB . and 7.3.2 for i = 1, 2, ⋯ , r, , → → → ⋯ , dr } = span{a1 , a2 , − → ⋯ , an } N ot → and { d1 , − → ⋯ , an } = span S that , we see that there are two methods to find the bases. Method 1. Finding bases of span S satisfying the above requirement (1) is equivalent to finding the pivot columns of A (see Section 2.9). Hence, we use row operations to change A to a row echelon matrix − →− → − → C = (C1 C2 ⋯ Cn ). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/14 8/30/2020 7.3 Methods of finding bases of spanning spaces − − → − − → − − → Let Cn , Cn , ⋯ , Cn be the column vectors of C containing leading entries. We shall prove that the corresponding column vectors (pivot columns) in A 1 2 r − → − → {an 1 , an 2 , − → ⋯ , an r } fo rD is tri bu tio n are a basis of span S. Method 2. To find bases of span S that satisfy the above requirement (2), we use row operations to change B − − → to a row echelon matrix D. Let dm1 , Then, we shall prove that − − → dm2 , − − → ⋯ , dmr be the column vectors of D containing leading entries. − − → − − → − − → {dm1 , dm2 , is the basis vector of span S. Example 7.3.1. Find a basis for the column space of C, where 1 1 N ot ⎛ ⋯ , dmr } C = ⎜0 ⎝ 0 0 0 2 1 ⎞ −1 −3 ⎟ . 0 8 ⎠ Solution − → We denote the column vectors of C by Ci , i = 1, 2, 3, 4, that is, − →− →− →− → C = (C1 C2 C3 C4 ). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/14 8/30/2020 7.3 Methods of finding bases of spanning spaces − → The column vectors of C containing the leading entries are C1 , − → − → − → {C1 , C3 , C4 } − → − → C3 , C4 , and by Theorem 7.3.1 , is a basis for the column space of C. Example 7.3.2. fo rD is tri bu tio n Find a set of column vectors of A that forms a basis for CA and dim(CA ), where 1 1 2 A = ⎜1 1 1 −2 ⎟ . 2 1 ⎛ ⎝ Solution We denote the column vectors of A by 2 → ai , i = 1, 2, 3, 4. 1 1 ⎞ ⎠ Then →→→→ A = (a1 a2 a3 a4 ). ⎛ 1 A = ⎜1 2 1 2 2 1 1 −−−− → ⎜0 1 −2 ⎟ − 0 −1 −3 ⎟ 0 −3 −1 1 1 ⎞ ⎠ 1 R1 (−1)+R2 R1 (−2)+R3 N ot ⎝ 1 1 1 − −−−− → ⎜0 0 R2 (−3)+R3 ⎛ ⎝ where C is the same as in Example 7.3.1 − → − → − → C1 , C3 , C4 0 0 2 1 ⎛ ⎝ 0 2 1 ⎞ ⎠ ⎞ −1 −3 ⎟ = C, 0 8 ⎠ . From the matrix C, we see that the column vectors → contain leading entries of C. The pivot column vectors in A are a1 , → → a3 , a4 , that is, https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/14 8/30/2020 7.3 Methods of finding bases of spanning spaces 1 2 1 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ → → → a1 = ⎜ 1 ⎟ , a3 = ⎜ 1 ⎟ , a4 = ⎜ −2 ⎟ . ⎝ By Theorem 7.3.1 2 → , the column vectors a1 , ⎠ ⎝ → → a3 , a4 1 ⎠ ⎝ 1 ⎠ of A formabasisof CA and fo rD is tri bu tio n → → → → → → → span{a1 , a3 , a4 = span{a1 , a2 , a3 , a4 }. It is obvious that dim(CA ) = 3 . Example 7.3.3. Find a basis for the row space of the following matrix 1 0 0 4 ⎜0 D = ⎜ ⎜0 1 0 7 0 ⎟ ⎟. 0 −1 ⎟ 0 0 0 ⎛ ⎝ N ot Solution Note that D is a row echelon matrix. By Theorem 7.3.2 form a basis for RD . Let → → → d1 , d2 , d3 → d1 = (1, 0, 0, 4) form a basis of RD . 0 T ⎞ ⎠ , the row vectors with the leading entries in D → , d2 = (0, 1, 0, 7) T , and → d3 = (0, 0, 0, − 1) T . Then Example 7.3.4. Find the set of basis vectors for RB , where https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/14 8/30/2020 7.3 Methods of finding bases of spanning spaces ⎛ B = ⎜ ⎝ 1 −1 3 2 −1 0 ⎞ 4⎟. −3 1 ⎠ Solution 1 ⎛ B = ⎜ ⎝ ⎛ 2 0 −1 −3 1 −1 ⎜0 ⎝ Let → → d1 0 → d2 ⎞ 1 ⎠ ⎞ ⎝ R1 (1)+R3 0 0 ⎠ ⎝ 0 Then by Theorem 7.3.2 −1 ⎞ 1 2 −3 R2 (2)+R3 − − − − → −2 ⎟ − 2 −4 , 3 → 4 → d1 , d2 → ⎠ are basis vectors for RB . → d2 } does not satisfy the N ot where ⎛ → a1 = ⎜ 1 is not a row vector of B, so the basis { d1 , Example 7.3.5. → → → → = {a1 , a2 , a3 , a4 }, ⎛ −−−− → ⎜0 4⎟ − requirement (1). Let S R1 (−2)+R2 2 ⎟ = D. − 1, 3), d2 = (0, 2, 2). is a row vector of B, but 3 3 2 → d1 = (1, Note that −1 fo rD is tri bu tio n We use row operations to change B to a row echelon matrix. ⎞ ⎛ → ⎟ , a2 = ⎜ ⎠ ⎝ −2 0 4 ⎞ ⎛ → ⎟ , a3 = ⎜ ⎠ ⎝ 0 4 −2 −2 ⎛ ⎞ → ⎟ , a4 = ⎜ −4 ⎟ . ⎞ ⎠ ⎝ 6 ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/14 8/30/2020 7.3 Methods of finding bases of spanning spaces a. Use Theorem 7.3.1 b. Use Theorem 7.3.2 to find a subset of S that forms a basis for span S. to find another basis for span S. Solution fo rD is tri bu tio n a. Because we seek a subset of S that forms a basis for span S, we need to put the vectors in S as the column vectors of a matrix A, that is, 1 −2 0 −2 2 0 4 −4 ⎟ . −3 4 −2 ⎛ A = ⎜ ⎝ 6 ⎞ ⎠ We use row operations to change A into a row echelon matrix. ⎛ A = ⎜ ⎝ R2 ( 1 2 1 2 −3 )+R3 −2 0 4 4 ⎛ 0 −2 1 The column vectors C1 , vectors of A are → → a1 , a2 . − → C2 R1 (−2)+R2 R1 (3)+R3 ⎠ 6 −2 ⎛ 1 4 4 0 0 0 0 ⎝ 0 −2 0 −2 4 4 0 −2 −2 0 ⎞ ⎟ ⎠ ⎞ ⎟ = C. ⎠ N ot − → 0 ⎞ −−−− → ⎜0 −4 ⎟ − −2 0 − − − − − → ⎜0 ⎝ −2 of C contain the leading entries of C. The corresponding column By Theorem 7.3.1 → , {a1 , → a2 } forms a basis of span S. b. We define a matrix B by putting the vectors in S as its row vectors ⎛ 1 ⎜ −2 B = ⎜ ⎜ 0 ⎝ −2 2 −3 0 4 4 −4 ⎞ ⎟ ⎟. −2 ⎟ 6 ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/14 8/30/2020 7.3 Methods of finding bases of spanning spaces We use row operations to change B to a row echelon matrix D. ⎛ 1 ⎜ −2 B = ⎜ ⎜ 0 ⎝ 2 −3 0 R1 (2)+R2 ⎟ ⎜0 − − − − →⎜ ⎟ − −2 ⎟ R1 (2)+R4 ⎜ 0 4 −2 ⎟ R2 (−1)+R3 −−−− → ⎟ − −2 ⎟ 4 6 ⎠ −3 ⎜0 ⎜ ⎜0 4 −2 ⎟ ⎟ = D. 0 ⎟ 0 0 0 0 ⎠ → = (1, 2, 4 −4 ⎞ − 3), d2 = (0, 4, − 2). forms a basis for span S. Exercises 1. Find a basis for the column space of C, where ⎝ 1 ⎜0 C = ⎜ ⎜0 ⎝ 0 4 0 0 ⎞ ⎠ It follows from Theorem 7.3.2 −3 4 N ot ⎛ ⎛ fo rD is tri bu tio n d1 1 ⎞ 2 ⎝ → −3 1 ⎛ Let −2 2 0 −2 0 1 3 0 0 0 0 0 0 5 4 → that { d1 , → d2 } ⎞ −2 −6 ⎟ ⎟. 1 5 ⎟ 0 0 ⎠ 2. For each of the following matrices, find a set of its column vectors that forms a basis for its column space. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/14 7.3 Methods of finding bases of spanning spaces ⎛ 0 A1 = ⎜ 1 ⎝ ⎛ ⎜ A3 = ⎜ ⎜ ⎝ 2 −1 −1 3 0 1 2 1 3 0 2 0 ⎜1 A2 = ⎜ ⎜2 1 1 0 4 1 −1 0⎟ ⎟ 0⎟ 0 0 1 ⎛ ⎞ ⎟ ⎠ ⎝ 1 −3 4 −2 5 4 2 −6 9 −1 8 2 2 −6 −1 3 0 3 ⎞ ⎠ ⎞ ⎟ ⎟. ⎟ fo rD is tri bu tio n 8/30/2020 9 −1 9 7 −4 2 −5 −4 ⎠ 3. For each of the following row echelon matrices, find the basis and dimension of its row space and use the basis vectors to denote the row space. 1 0 D1 = ⎜ 0 1 0⎟ 0 0 ⎛ ⎝ 0 0 1 ⎜0 D2 = ⎜ ⎜0 0 0 1 0 0 0 3⎟ ⎟ 0⎟ 0 0 0 0 ⎛ 1 ⎞ ⎠ ⎝ 0 −2 0 1 1 2 5 0 3 ⎜0 D3 = ⎜ ⎜0 1 3 0 0 0 1 0⎟ ⎟ 0⎟ 0 0 0 0 ⎞ ⎛ ⎠ ⎝ 0 ⎞ ⎠ 4. Find a set of basis vectors for RB and dim RB , where 1 N ot ⎛ B = ⎜ ⎝ 5. Let S → → → → → = {a1 , a2 , a3 , a4 , a5 }, 2 −1 −1 3 0 ⎞ 4⎟. −3 1 ⎠ where → a1 = (1, − 2, 0, 3) → a4 = (2, − 1, 4, T → , a2 = (2, − 7) T − 5, → , a5 = (5, − 3, 6) − 8, 1, 2) T T → T , a3 = (0, 1, 3, 0) , . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/14 8/30/2020 7.3 Methods of finding bases of spanning spaces N ot fo rD is tri bu tio n a. Find a subset of S that forms a basis for the spanning space span S. b. Find another basis for span S. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/14 8/30/2020 7.4 Coordinates and orthonormal bases 7.4 Coordinates and orthonormal bases In Section 7.2 , we see that ℝ n has a standard basis E given in (7.2.3) n and Theorem 7.2.1 (ii) n set of n linearly dependent vectors in ℝ n constitutes a basis of ℝ n. { } be a basis of ℝ n. Then D is tri b Let S = → → → a 1, a 2, ⋯, a n ut io n provides a necessary and sufficient condition for a set of n vectors in ℝ to be a basis of ℝ . Moreover, any { } → → → ℝ = span a 1, a 2, ⋯, a n . n with n = m, we obtain the following result. fo r By Theorem 7.2.2 If S = { → → → a 1, a 2, ⋯, a n such that } N ot Theorem 7.4.1. is a basis of ℝ n, then for each b ∈ ℝ n, there exists a unique vector (x 1, x 2, ⋯, x n) → (7.4.1) → → → b = x 1a 1 + x 2a 2 + ⋯x na n. → Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/21 8/30/2020 7.4 Coordinates and orthonormal bases Definition 7.4.1. The numbers x 1, x 2, ⋯, x n in (7.4.1) S. We write → are said to be the coordinates of the vector b relative to the basis → ( b) S = (x 1, x 2, ⋯, x n). → → ut io n (7.4.2) → (7.4.3) → → b = ( b) E. D is tri b If b = (b 1, b 2, ⋯, b n) ∈ ℝ n, then the coordinates b 1, b 2, ⋯, b n of b are the coordinates of b relative to the standard basis E. Hence, according to (7.4.2) , we have as fo r → → → Let A be the matrix whose column vectors are a 1, a 2, ⋯, a n. Then we can rewrite (7.4.1) (7.4.4) → N ot → ( b) E = A( b) S. By Theorem 7.2.1 (ii), A is invertible. Hence, we obtain (7.4.5) → → → → ( b) S = A − 1( b) E = A − 1 b for b ∈ ℝ n. Processing math: 100% Example 7.4.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/21 8/30/2020 7.4 Coordinates and orthonormal bases Let S = {( ) ( )} 1 1 −1 , 1 be a set of vectors in ℝ 2. i. Show that S is a basis of ℝ 2. → → ut io n ii. Let b = (2, − 4). Find ( b) S. Solution ( ) 1 −1 1 1 . Then |A | = 1 + 1 = 2 ≠ 0. By Theorem 7.2.1 ii. By Theorem 2.7.5 ( ) 1 1 1 , A −1 = 2 . By (7.4.5) −1 1 → → ( b) S = A − 1 b = , ( )( ) ( ) 1 1 2 −1 1 2 −4 = −1 −3 , → fo r → 1 (ii), S is a basis of ℝ 2. D is tri b i. Let A = that is, ( b) S = ( − 1, − 3) are the coordinates of the vector b = (2, − 4) relative to the basis S. → , we used the inverse matrix method to find the coordinates ( b) S. In fact, all the other methods given in Chapter 4 N ot In Example 7.4.1 , including Gaussian elimination, Gauss-Jordan elimination, and Cramer’s → → rule, can be employed to solve the system AX = b . Example 7.4.2. → → → Let a 1 = (1, 0, 1), a 2 = (0, 1, 1), and a 3 = (1, 1, 0). Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/21 8/30/2020 7.4 Coordinates and orthonormal bases i. Show that S = { → → → a 1, a 2, a 3 → } is a basis of ℝ 3. → ii. Let b = (1, 1, 1) T. Find ( b) S. | | 1 0 1 i. Because |A | = 0 1 1 = − 2 ≠ 0, by Theorem 7.2.1 1 1 0 → (A | b) = } is a basis of ℝ 3. ( ) ( ) 1 0 1 1 R ( −1) +R 1 0 1 1 R ( −1) +R 1 3 2 3 0 1 1 1 0 1 1 1 → → 1 1 0 1 1 0 1 1 1 ) R3 0 0 −2 −1 ( ) 1 −2 ( ) 1 0 1 1 R3 ( − 1 ) + R2 fo r 1 0 1 0 1 −1 0 → 0 1 1 1 1 0 0 1 2 → R3 ( − 1 ) + R1 N ot ( { → → → a 1, a 2, a 3 D is tri b ii. (ii), ut io n Solution ( ) 1 1 0 0 2 1 0 1 0 2 . 1 Processing math: 100% 0 0 1 2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/21 8/30/2020 7.4 Coordinates and orthonormal bases In Example 7.4.2 ( 1 1 1 ) , , . 2 2 2 → , the coordinates of b = (1, 1, 1) relative to the basis S are coordinates relative to the standard basis E are (1, 1, 1). Equation (7.4.5) standard basis E. → ( 1 2 1 1 , 2, 2 ) while its ut io n → Hence, ( b) S = gives the relation between the coordinates of a vector b relative to a basis S and the → D is tri b Now, we study the relations between the coordinates of a vector b relative to two bases of ℝ n, that is, we shall generalize the relation given in (7.4.5) by replacing the standard basis E by any basis of ℝ n. By Theorem 7.2.1 (ii) and Lemma 7.2.1 the change of bases. (ii) with m = n = k, we obtain the following result, which gives { } and T = { → → → b 1, b 2, ⋯, b n } be two bases of ℝ n. Let N ot Let S = → → → a 1, a 2, ⋯, a n fo r Theorem 7.4.2. →→ → →→ → A = (a 1 a 2⋯ a n), B = (b 1 b 2⋯ b n). Then A and B are invertible and there exists a unique invertible n × n matrix (7.4.6) Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/21 8/30/2020 7.4 Coordinates and orthonormal bases ( ) c 11 c 12 ⋯ c 1n C= c 21 c 22 ⋯ c 2n ⋮ ⋮ ⋱ ⋮ ut io n c n c n ⋯ c nn 1 2 such that C = A − 1B and Definition 7.4.2. Let S = { → → → a 1, a 2, ⋯, a n } and T = { → → → → b 1 = c 11a 1 + c 21a 2 + ⋯ + c n a n 1 fo r → → → → b 2 = c 12a 1 + c 22a 2 + ⋯ + c n a n 2 ⋮ → → → → b n = c 1na 1 + c 2na 2 + ⋯ + c nna n. N ot { D is tri b (7.4.7) → → → b 1, b 2, ⋯, b n } be two bases of ℝ n. The matrix C given in (7.4.6) is said to be the transition matrix from the basis T to the basis S. Processing math: 100% The following result gives the change of coordinates under two bases. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/21 8/30/2020 7.4 Coordinates and orthonormal bases Theorem 7.4.3. Let S = { → → → a 1, a 2, ⋯, a n } and T = { → → → b 1, b 2, ⋯, b n } → be two bases of ℝ n. Then for each vector b ∈ ℝ n, → ut io n (7.4.8) → ( b) S = C( b) T. Proof → D is tri b . Because S and T are bases of ℝ n, by (7.4.4) Let A and B be the same as in Theorem 7.4.2 n have for each vector b ∈ ℝ , (7.4.9) → → , we By Theorem 7.4.2 and (7.4.9) , we have → → fo r → ( b) E = A( b) S and (→ b) E = B( b) T. → → → and (7.4.8) N ot ( b) S = A − 1( b) E = A − 1(B( b) T) = (A − 1B)( b) T = C( b) T holds. Example 7.4.3. Let S = {( ) ( )} 2 1 , 1 −2 and T = {( ) ( )} 1 2 , −1 1 → . Assume that ( b) T = ( ) 5 − 10 → . Find ( b) S. Processing math: 100% Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/21 7.4 Coordinates and orthonormal bases Let A = ( ) 2 1 1 −2 and B = ( ) 1 −1 2 1 . By Theorem 2.7.5 A −1 = ( , ( ) ) 1 −2 −1 = −5 −1 2 2 1 5 5 1 5 2 . ut io n 8/30/2020 −5 C = A − 1B = ( ) ( ) 2 1 5 5 1 ( ) 1 −1 2 1 2 −5 4 = 1 −5 5 3 . 3 −5 −5 fo r 5 , we have ( ) N ot By Theorem 7.4.3 D is tri b Hence, 4 → → ( b) S = C( b) T = 1 −5 5 3 3 −5 −5 ( ) () 5 − 10 = 6 3 . Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/21 8/30/2020 7.4 Coordinates and orthonormal bases Let S = { → → → a 1, a 2, ⋯, a n } be a basis of ℝ n. In general, these vectors may not be orthogonal. In some cases, we are interested in those bases whose basis vectors are orthogonal. Let S = { → → → a 1, a 2, ⋯, a n } ut io n Definition 7.4.3. be a basis of ℝ n. D is tri b i. If S satisfies the following condition: → → a i ⊥ a j for i ≠ j and i, j ∈ {1, 2, ⋯, n}, || | | = 1 for i = 1, 2, ⋯, n, N ot → ai fo r then S is called an orthogonal basis of ℝ n. ii. If S is an orthogonal basis and satisfies then S is called an orthonormal basis. Example 7.4.4. Determine which of the following bases are orthogonal bases and orthonormal bases. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/21 7.4 Coordinates and orthonormal bases S1 = S3 = {( ) ( )} −1 1 {( ) ( )} S4 = {( ) ( ) ( )} −1 {( ) ( )} 1 1 √2 √2 1 S2 = 1 , , − √2 1 √2 1 1 −1 , 1 1 0 0 0 , 0 1 , 0 0 S 1 is not an orthogonal basis because ( − 1, 1) ⋅ S 2 is an orthogonal basis because (1, 1) ⋅ ( ) 1 −1 ( ) 1 −1 D is tri b Solution 1 ut io n 8/30/2020 = − 2 ≠ 0. = 0, but S 2 is not an orthonormal basis because fo r | | (1, 1) | | = √2 ≠ 1. () S 3 is an orthonormal basis because ǁ Processing math: 100% N ot 1 ( ) 1 1 , √2 √2 ( ) 1 , 1 √2 √2 . √2 − 1 = 0 and √2 ǁ = 1 and ǁ ( 1 √2 , − 1 √2 ) ǁ = 1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/21 8/30/2020 7.4 Coordinates and orthonormal bases → → → → → S 4 is an orthonormal basis. In fact, because it is easy to verify that e 1 ⋅ e 2 = 0, e 3 = 0 and e 2 ⋅ e 3 = 0. 3 Hence, S is an orthogonal basis of ℝ . Because || | | | | | | → e1 → e2 = 1, = 1 and || | | → e3 = 1, S is an orthonormal basis of ℝ 3. ut io n , the basis S 4 is the standard basis of ℝ 3. In fact, the standard basis E given in (7.2.1) In Example 7.4.4 is an orthonormal basis of ℝ n. The coordinates of a vector b in ℝ n relative to an orthogonal basis or an orthonormal basis can be obtained by using the methods given in Section 8.4. However, because the basis vectors in an orthogonal basis or an orthonormal basis are orthogonal, we can provide formulas for their coordinates. D is tri b → Theorem 7.4.4. { } is an orthogonal basis of ℝ n, then (7.4.10) (|| | fo r i. If S = → → → a 1, a 2, ⋯, a n → ( b) S = ii. If S = Processing math: 100% { → → → a 1, a 2, ⋯, a n } → b ⋅ a2 N ot → b ⋅ a1 → → a1 |2 → , || | ) || | → b ⋅ an → , ⋯, → a2 |2 . → an |2 is an orthonormal basis of ℝ n, then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/21 8/30/2020 7.4 Coordinates and orthonormal bases (7.4.11) → → → → → ( b) S = ( b ⋅ a 1, b ⋅ a 2, ⋯, b ⋅ a n). → → Because S = { → → → a 1, a 2, ⋯, a n } ut io n Proof → is a basis of ℝ n, by Theorem 7.4.1 , for each b ∈ ℝ n, there exists a unique solution (x 1, x 2, ⋯, x n) such that D is tri b → → → → b = x 1a 1 + x 2a 2 + ⋯ + x i a i + ⋯ + x na n. → This implies that for each i ∈ I n, (7.4.12) fo r → → → → → → → → → ( b ⋅ a i ) = x 1(a 1 ⋅ a i ) + x 2(a 2 ⋅ a i ) + ⋯ + x i( a i ⋅ a i ) + ⋯ + x n(a n ⋅ a i ). → N ot i. Because S is an orthogonal basis of ℝ n, → → a j ⋅ a i = 0 for j ≠ i and i, j ∈ I n. This, together with (7.4.12) , implies that for i ∈ I n, → → → b ⋅ a i = x i( a i ⋅ a j ) = x i → Processing math: 100% || | → ai |2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/21 8/30/2020 7.4 Coordinates and orthonormal bases and → xi = → b ⋅ ai for i ∈ I n. || | → Hence, (7.4.10) holds. ii. If S is an orthonormal basis, then || | | → ai = 1 for i ∈ I n. This, together with (7.4.10) holds. Example 7.4.5. → → → Let b = (0, 1) T. Find ( b) S and ( b) T, where Solution By Example 7.4.4 1 , 1 {( ) ( )} 1 1 √2 √2 fo r {( ) ( )} −1 and T = N ot S= 1 , implies D is tri b that (7.4.11) ut io n ai |2 1 √2 , − 1 . √2 () → 1 , S is an orthogonal basis but not an orthonormal basis. Let a 1 = and 1 ( ) → −1 a2 = Processing math: 100%1 . Then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/21 8/30/2020 7.4 Coordinates and orthonormal bases ( ) || | → 1 b ⋅ a 1 = (0, 1) = 1, 1 → → a 1 | 2 = 2. and ( ) || | → , we obtain ( b) S = By Example 7.4.4 ( ) 1 1 , . 2 2 , T is an orthonormal basis. Let () ( ) 1 √2 1 → and a 2 = √2 − 1 Then () 1 → √2 b ⋅ a 1 = (0, 1) 1 → √2 Processing math: 100% . √2 N ot √2 1 fo r → a1 = D is tri b By (7.4.10) → a 2 | 2 = 2. ut io n → −1 b ⋅ a 2 = (0, 1) = 1, 1 → () 1 = 1 √2 → and b ⋅ a 2 = (0, 1) → √2 − 1 √2 = − 1 √2 . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/21 8/30/2020 7.4 Coordinates and orthonormal bases By (7.4.11) → , ( b) S = ( 1 1 ) , −2 . 2 The following result is the well-known Gram-Schmidt process, which shows that all bases of ℝ n can be changed to orthogonal bases and orthonormal bases of ℝ n. We omit its proof. { } be a basis of ℝ n. Let D is tri b Let S = → → → a 1, a 2, ⋯, a n ut io n Theorem 7.4.5 (Gram-Schmidt process) → → b 1 = a 1, → → → → ( a2 ⋅ b1 ) → b2 = a2 − b 1, || | → b1 |2 → → || | || | → → → → || | || | fo r → → → → ( a3 ⋅ b1 ) → ( a3 ⋅ b1 ) → b3 = a3 − b1 − b 2, → ⋮ → b2 |2 N ot b1 |2 → → → → ( an ⋅ b1 ) → ( an ⋅ b2 ) → ( bn ⋅ bn − 1 ) → bn = an − b1 − b2 − ⋯ − b n − 1. → b1 |2 → b2 |2 || | → bn − 1 |2 Then Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/21 8/30/2020 7.4 Coordinates and orthonormal bases i. { ii. { | | | | | | | | | | | |} } is an orthogonal basis of ℝ n. → → → b1 b2 bn , , ⋯, → → → b1 b2 bn is an orthonormal basis of ℝ n. ut io n → → → b 1, b 2, ⋯, b n Example 7.4.6. 1. Find an orthogonal basis S 1 = → { → → → b 1, b 2, b 3 } D is tri b → → → Let a 1 = (1, 0, 1), a 2 = (0, 1, 1), and a 3 = (1, 1, 0) be the same as in Example 7.4.2 by using the Gram-Schmidt process. Moreover, if → b = (1, 1, 1) T, find ( b) S . 1 { } fo r 2. Find an orthonormal basis S 2 : = . → → → q 1, q 2, q 3 → by normalizing S 1. Moreover, if b = (1, 1, 1) T, find → N ot ( b) S . 2 Solution By Example 7.4.2 (i), S is a basis of ℝ 3. We use Theorem 7.4.5 to find S 1 and S 2. → → 1. Let b 1 = a 1 = (1, 0, 1) and let Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/21 8/30/2020 7.4 Coordinates and orthonormal bases → → → → (a 2 ⋅ b 1) → b2 = a2 − b 1. → b1 |2 || | || | → b 1 | 2 = 2, we have D is tri b and || | → → → 2 b 1 | . Because a 2 ⋅ b 1 = (0, 1, 1) ⋅ (1, 0, 1) = 0 ut io n → → → To find b 2, we need to compute a 2 ⋅ b 1 and → 1 b 2 = (0, 1, 1) − (1, 0, 1) = (0, 1, 1) − 2 Let ( 1 2 , 0, 1 2 ) ( ) 1 1 = − , 1, . 2 2 fo r → → → → → → (a 3 ⋅ b 1) → (a 3 ⋅ b 2) → b3 = a3 − b1 − b 2. → → b1 |2 b2 |2 N ot || | || | → → → → → To find b 3, we need to compute a 3 ⋅ b 1, a 3 ⋅ b 2 and and Processing math: 100% || | → → → → → 1 2 b 2 | . Because a 3 ⋅ b 1 = 1, a 3 ⋅ b 2 = 2 , || | → 3 b 2 | 2 = 2 , we obtain https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/21 7.4 Coordinates and orthonormal bases 1 ( → 1 1 1 2 b 3 = (1, 1, 0) − (1, 0, 1) − 3 − , 1, 2 2 2 2 ) ( || | | √2, = → 2 b ⋅ b 3 = 3 . By Theorem 7.4.4 → || || → b2 → 1 ( || | → b ⋅ b1 = 2 and → b1 |2 → b ⋅ b2 → , || | → b2 |2 ) form an orthogonal basis of ℝ 3. || | | → b3 → → → = 3 , ( b ⋅ b 1) = 2, b ⋅ b 2 = 1, and 2 √3 → ) ( ) || | → b ⋅ b3 → , 2 → b3 |2 = 2 2 2 3 3 , , 2 4 4 3 3 ( = 1, 2 3 , 1 2 ) . N ot 2. By computation, we obtain √6 , → ( b) S = 2 , , −3 3 3 fo r By computation, → b1 2 ) 2 2 2 , , − . 3 3 3 D is tri b ( → → → 1 1 Hence, b 1 = (1, 0, 1), b 2 = − 2 , 1, 2 , b 3 = ) ( = ut io n 8/30/2020 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/21 7.4 Coordinates and orthonormal bases → q1 : = → b1 || || → √2 = 2 (1, 0, 1) = b1 → b2 || || → √6 ( √3 ( = 3 b2 → q3 : = → b3 || || → b3 Hence, S 2 = { → → → q 1, q 2, q 3 } = 2 1 1 − 2 , 1, 2 2 3 2 2 ) , 0, 2 , ) ( √6 √6 √6 ) = − 6 , 3 , 6 , 2 , 3, − 3 ) ( = √3 √3 3 √3 ) , 3 , − 3 . forms an orthonormal basis of ℝ 3 and ( √2, √6 √3 3 , 3 ) . N ot → fo r → → → → → ( b) S = ( b ⋅ q 1, b ⋅ q 2, b ⋅ q 3) = 2 → Exercises √2 D is tri b → q2 : = ( √2 ut io n 8/30/2020 1. For each of the following pairs of vectors, verify that it is a basis of ℝ 2 and find the coordinates of the → vector b = (x, y) T. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/21 8/30/2020 7.4 Coordinates and orthonormal bases {( ) ( )} {( ) ( )} {( ) ( )} {( ) ( )} −1 S1 = 1 5 S3 = −6 1 , 0 4 , −3 S2 = −1 2 3 S4 = −2 { → } is a basis of ℝ 3. → ii. Let b = (1, 1, 1) T. Find ( b) S. , −2 5 D is tri b i. Show that S = 3 ut io n → → → 2. Let a 1 = (1, 0, 1) T, a 2 = (0, 1, 1) T, and a 3 = (1, 1, 0) T. → → → a 1, a 2, a 3 2 , → → → T T 3. Let a 1 = (1, 0, 0) , a 2 = (2, 4, 0) , and a 3 = (3, − 1, 5) T. { } → is a basis of ℝ 3. fo r i. Show that S = → → → a 1, a 2, a 3 → ii. Let b = (2, − 1, − 2). Find ( b) S. {( ) ( )} −1 1 , 1 0 and S 2 = → → {( ) ( )} −3 2 N ot 4. Let S 1 = 2 , 3 . i. If ( b) S = (2, − 1), find ( b) S . 2 1 → → ii. If ( b) S = (13, − 26), find ( b) S . 1 → 2 → 5. If ( b) T = (1, 2, 3), find ( b) S, where Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/21 7.4 Coordinates and orthonormal bases {( ) ( ) ( )} {( ) ( ) ( )} T= ) { } 0 0 0 0 , 1 1 0 1 0 , 0 1 1 S= ( 1 → → → 4 3 6. Let b 1 = (0, 1, 0), b 2 = − 5 , 0, 5 , and b 3 = i. Show that S : = → → → → b 1, b 2, b 3 ( , 1 1 , −1 3 , . 1 4 ) , 0, 5 . 5 ut io n −1 D is tri b 8/30/2020 is an orthonormal basis of ℝ 3. → ii. Let b = (1, − 1, 1). Find ( b) S. → ( b) S . 1 { → → → b 1, b 2, b 3 } of ℝ 3 by using the Gram-Schmidt process and N ot 1. Find an orthogonal basis S 1 : = fo r → → → → 7. Let a 1 = (1, 1, 1), a 2 = (0, 1, 1), a 3 = (0, 0, 1), and b = ( − 1, 1, − 1) T. 2. Find an orthonormal basis S 2 : = { → → → q 1, q 2, q 3 } → of ℝ 3 by normalizing S 1 and ( b) S . 2 Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 21/21 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability fo rD is tri bu tio n Chapter 8 Eigenvalues and Diagonalizability 8.1 Eigenvalues and eigenvectors Let A be an n × n matrix given in (3.2.1) , that is, (8.1.1) a11 a12 ⋯ a1n ⎜ a21 ⎜ A = ⎜ ⎜ ⎜ ⋮ a22 ⋯ ⋮ ⋱ a2n ⎟ ⎟ ⎟. ⎟ ⋮ ⎟ ⎛ Definition 8.1.1. an 1 an 2 ⋯ ann N ot ⎝ ⎞ ⎠ A real number λ is called an eigenvalue of A if there exists a nonzero vector → X in Rn such that (8.1.2) → → AX = λX . The vector → X is said to be an eigenvector of the matrix A corresponding to the eigenvalue λ . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability Note that we restrict our study on eigenvalues in R, the set of real numbers. In fact, eigenvalues could be complex numbers, which we shall study in Chapter 10 . Therefore, eigenvalues in the field of complex numbers can be further studied after Chapter 10 . An eigenvalue could be positive, zero, or negative. But, according to Definition 8.1.1 , an eigenvector must be a nonzero vector. In other words, the zero vector is not an eigenvector although it is a solution of (8.1.2) . and Theorems 4.5.2 and 4.5.3 → , we see that A has a zero eigenvalue if and only if the fo rD is tri bu tio n From (8.1.2) homogeneous system A X = 0 has a nonzero solution if and only if r(A) < n if and only if |A| = 0 if and only if A is not invertible. Hence, A is invertible if and only if all the eigenvalues of A are nonzero, see Corollary 8.1.3 for another proof. By (8.1.2) , we see that λ ∈ R is an eigenvalue of A if and only if (8.1.3) → (λI − A) X = has a nonzero solution in Rn . Theorem 8.1.1. λ ∈ R 0 , we obtain the following result, which shows that we can use determinants to find the N ot By Theorem 4.5.3 eigenvalues. → is an eigenvalue of A if and only if (8.1.4) p(λ) := |λI − A| = 0. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability Note that the determinant ∣ λ − a11 ∣ ∣ −a21 −a12 ⋯ −a1n λ − a22 ⋯ −a2n ∣ ∣ ∣ p(λ) = ∣ ∣ ∣ ⋮ ⋮ ⋮ −an 1 −an 2 ∣ ⋮ ∣ ⋯ λ − ann ∣ fo rD is tri bu tio n ∣ ∣ is a polynomial of degree n, which is called the characteristic polynomial of A. The equation p(λ) = 0 is called the characteristic equation of A. The following result can be proved by using the fundamental theorem of algebra from complex analysis. Lemma 8.1.1. Any polynomial p(z) = a0 + a1 z + a2 z 2 + ⋯ + an z n (an ≠ 0) of degree (n ≥ 1) can be expressed as a product of the form N ot p(z) = c(z − z 1 )(z − z 2 ) ⋯ (z − z n ), where c and zj (j ∈ In ) are complex numbers (see Chapter 10 Note that some of these complex numbers z1 , be rewritten as follows. z2 , ⋯ , zn for the study of complex numbers). may be the same. By Lemma 8.1.1 , p(λ) can (8.1.5) p(λ) = (λ − λ1 ) r1 (λ − λ2 ) r2 ⋯ (λ − λk ) rk , https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability where λi is a real or complex number, λi satisfying ≠ λj for i ≠ j, and ri is a positive integer for i, j ∈ {1, ⋯ , k} (8.1.6) r1 + r2 + ⋯ + rk = n. fo rD is tri bu tio n Hence, the characteristic equation p(λ) = 0 has n (real or complex) roots including repeated roots, which are those roots λi with ri > 1 . Definition 8.1.2. The number ri is called the algebraic multiplicity of the eigenvalue λi for each i = 1, 2, ⋯, k . In Definition 8.1.1 , if we allow the eigenvalues to be complex numbers, then the eigenvectors would take complex numbers as their components, so complex vector space Cn , complex matrices, complex determinants, etc., are involved. But from now on, we restrict our discussion of eigenvalues and eigenvectors to real numbers. Hence, we shall only consider matrices A such that all of λ1 , ⋯ , λk in (8.1.5) are real. But we give one question involving complex eigenvalues in question 8 of Exercise 9.2 . For each λi , let N ot (8.1.7) → n Eλi := N (λi I − A) = { X ∈ R → : (λi I − A) X = → 0 }. The set Eλ is called the eigenspace of A. It is obvious that the eigenspace Eλ of A is exactly the solution i i → → space of (λi I − A) X = 0 or the nullspace of λi I − A, see (4.1.13) . Hence, the eigenspace Eλ contains all eigenvectors corresponding to the eigenvalue λi as well as the zero vector in Rn , which is not an eigenvector (as mentioned above). i https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability Definition 8.1.3. The dimension dim(Eλ ) of the eigenspace Eλ is called the geometric multiplicity of the eigenvalue λi for each i = 1, 2, ⋯ , k . i , the eigenspace Eλ contains a nonzero vector, so the geometric multiplicity is greater i fo rD is tri bu tio n By Definition 8.1.1 than or equal to 1. i We emphasize that there are two numbers, one being the algebraic multiplicity ri given in (8.1.5) another the geometric multiplicity dim(Eλ ), corresponding to each eigenvalue λi . i and The following result shows that the geometric multiplicity is less than or equal to the algebraic multiplicity. The proof is omitted because it involves additional properties of determinants. Theorem 8.1.2. Let λi and ri be the same as in (8.1.5) for i ∈ {1, 2, ⋯ , k}. Then 1 ≤ dim(Eλi ) ≤ ri . Corollary 8.1.1. Let λi and ri bethesameasin (8.1.5) j ∈ {1, 2, ⋯ , k}, then N ot As a special case of Theorem 8.1.2 , we obtain the following result, which shows that if the algebraic multiplicity of an eigenvalue is 1, then its geometric multiplicity is 1. for i ∈ {1, 2, ⋯ , k}, Then if rj = 1 for some dim(Eλi ) = rj = 1. By Theorem 8.1.2 and (8.1.6) , we obtain the following result, which shows that the sum of the geometric multiplicities is less than or equal to n. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability Corollary 8.1.2. Let λi and ri bethesameasin (8.1.5) . Then k k ≤ ∑ dim(Eλi ) ≤ n. fo rD is tri bu tio n i=1 The following result gives the eigenvalues of triangular (upper, lower, or diagonal) matrices. Theorem 8.1.3. Let A be an n × n triangular (upper, lower, or diagonal) matrix. Then the eigenvalues of A are all the entries on the main diagonal of A. Proof We prove this only when A is an upper triangular matrix. Let ⎛ ⎜ ⎜ A = ⎜ ⎜ ⎜ be an upper triangular matrix. Then a12 ⋯ a1n 0 a22 ⋯ ⋮ ⋮ ⋱ a2n ⎟ ⎟ ⎟ ⎟ ⋮ ⎟ 0 0 ⋯ ann N ot ⎝ a11 ∣ λ − a11 ∣ ∣ |λI − A| = 0 ⎞ ⎠ −a12 ⋯ −a1n λ − a22 ⋯ −a2n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ⋮ ⋮ ⋱ 0 0 ⋯ ⋮ ∣ ∣ λ − ann ∣ = (λ − a11 )(λ − a22 ) ⋯ (λ − ann ). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability Solving |λI − A| = 0 implies λ1 = a11 , λ2 = a22 , ⋯ , λn = ann . Example 8.1.1. For each of the following matrices, find its eigenvalues and determine its algebraic multiplicity. A = ⎜ ⎝ −1 −1 0 2 0 0 ⎞ ⎛ 3⎟, 3 Solution By Theorem 8.1.3 of each of λ1 , λ2 , 0 , the eigenvalues of A are λ1 λ3 is 1 because 4 B = ⎜0 −1 0 4 3 0 −3 ⎞ ⎟. fo rD is tri bu tio n ⎛ ⎠ ⎝ = −1, λ2 = 2, 0 and λ3 ⎠ = 3. The algebraic multiplicity |λI − A| = (λ − λ1 )(λ − λ2 )(λ − λ3 ). All the eigenvalues of B are λ1 = λ2 = 4 and λ3 = −3. The algebraic multiplicity for the eigenvalue 4 is 2 and the algebraic multiplicity for the eigenvalue 3 is 1 because ∣∣λI − A∣∣ = (λ − 4) 2 (λ + 3) . The procedure for finding eigenvalues and eigenvectors is given below. N ot 1. Step 1. Find p(λ) = |λI − A| . 2. Step 2. Find all roots λ1 , λ2 , ⋯ , λn of p(λ) = 0. Some of these eigenvalues may be the same. 3. Step 3. For each eigenvalue λi , solve the corresponding homogenous system → (λi I − A) X = and find the basis vectors by using the methods in Section 4.4 4. Step 4. Give the eigenspace Eλ using the basis vectors. → 0 . i https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability Example 8.1.2. Let A = ( 2 0 0 2 ). Solution ∣λ−2 ∣ 2 ∣ = (λ − 2) = 0, λ − 2∣ 0 1. Because |λI − A| = ∣ ∣ λ1 = λ2 = 2 For λ1 0 . → → = λ2 = 2, (λ1 I − A) X = 0 ( = (1, 0) T → and υ2 0 0 0 + 0x2 = 0, x ) = ( y → 0 )( t s = (0, 1) T x1 ) = ( x2 0 ). 0 where x1 and x2 are free variables. Let x1 N ot The corresponding system is 0x1 x2 = s . Then we obtain a repeated eigenvalue becomes ( where υ1 fo rD is tri bu tio n 1. Find the eigenvalues and eigenspaces of A. 2. Find the algebraic and geometric multiplicities for each eigenvalue. ) = t( 1 ) + s( 0 is a solution of 0x1 0 1 = t and → → ) = tυ1 + sυ2 , + 0x2 = 0. Moreover, → → Eλ1 = Eλ2 = span{υ1 , υ2 }. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability 2. The algebraic multiplicity for the eigenvalue λ1 λ−2 = λ2 = 2 is 2 because 2 is the power of the term in the characteristic polynomial ∣∣λI − A∣∣ = (λ − 2) basis vectors 2. → υ1 and → υ2 , dim(Eλ1 ) = 2. 2 Because the eigenspace Eλ has two . 1 Hence, the geometric multiplicity of the eigenvalue 2 is ⎛ 0 Let A = ⎜ 1 ⎝ 0 0 −1 −2 1 0 3 ⎞ ⎟. fo rD is tri bu tio n Example 8.1.3. Find the eigenvalues and eigenspaces of A. ⎠ Solution ∣ λ p(λ) = |λI − A| = ∣ ∣ 1 −1 λ+2 −1 0 0 ∣ Solving p(λ) = λ(λ + 2)(λ − 3) = 0 gets λ1 = 0 , ⎛ 0 λ1 I − A = ⎜ −1 ⎝ ∣ ∣ = λ(λ + 2)(λ − 3). λ−3∣ = 0, λ2 = −2, N ot For λ1 ∣ 0 0 0 and λ3 1 = 3 . ⎞ 2 −1 ⎟ . 0 −3 ⎠ We solve the following system. ⎛ 0 ⎜ −1 ⎝ 0 0 1 ⎞⎛ x1 ⎞ ⎛ 0 ⎞ 2 −1 ⎟ ⎜ x2 ⎟ = ⎜ 0 ⎟ . 0 −3 ⎠⎝ x3 ⎠ ⎝ 0 ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/17 Chapter 8 Eigenvalues and Diagonalizability ⎛ ∣→ (A∣ 0 ) = ⎜ ⎜ ∣ ⎝ 0 0 1 0 ⎞ R1 , 2 ⎛ −1 2 −1 0 ⎞ 0 ⎟ ⎟ 0 ⎠ −1 2 −1 0 ⎟ − → ⎜ ⎟ − ⎜ 0 0 0 0 −3 0 ⎠ 0 0 −3 −1 2 0 0 ⎞ 0 0 1 0 ⎟. ⎟ 0 0 0 0 R2 (3)+R3 ⎛ − − − − − →⎜ ⎜ R2 (1)+R1 ⎝ ⎝ 1 ⎠ fo rD is tri bu tio n 8/30/2020 The system corresponding to the last augmented matrix is { −x1 + 2x2 = 0 x3 = 0, where x1 and x3 are basic variables and x2 is a free variable. Let x2 → T υ1 = (2, 1, 0) . For λ2 Then Eλ 1 → = span{υ1 } . = −2, ⎛ −2 λ2 I − A = ⎜ −1 We solve the following system. N ot ⎝ ⎛ −2 ⎜ −1 ⎝ 0 0 1 ⎞⎛ 0 x1 0 = t. 1 Then x2 = 2t. Let ⎞ 0 −1 ⎟ . 0 −5 ⎞ ⎠ ⎛ 0 ⎞ 0 −1 ⎟ ⎜ x2 ⎟ = ⎜ 0 ⎟ . 0 −5 ⎠⎝ x3 ⎠ ⎝ 0 ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability −2 0 1 0 ⎞ R1, ⎛ −1 0 −1 0 ⎞ −2 0 0 ⎟ ⎟ 0 ⎠ 2 −1 0 −1 0 ⎟− ⎜ ⎟ →⎜ 0 0 −5 0 ⎠ R1 (−2)+R2 ⎛ −1 − −−−− → ⎜ ⎜ ⎝ R2 (5)+R3 ⎛ ⎝ 0 −1 0 0 0 0 −5 −1 3 0 −1 − − − − − →⎜ ⎜ 0 0 ⎝ 0 0 0 ⎞ 0 R1 ( 1 0 −5 1 3 ) ⎛ ⎟ − − − → ⎜ ⎟ ⎜ 0 ⎠ 0 ⎝ −1 0 −1 0 ⎞ 0 0 0 ⎟ ⎟ 0 0 −5 1 0 ⎠ fo rD is tri bu tio n ⎛ ∣→ (A∣ 0 ) = ⎜ ⎜ ∣ ⎝ 0 ⎞ ⎛ R2 (1)+R1 −1 0 0 0 ⎞ 1 0 ⎟ − − − − →⎜ ⎟ − ⎜ 0 0 1 0 ⎟. ⎟ 0 0 ⎠ 0 0 0 0 ⎠ ⎝ The system corresponding to the last augmented matrix is { x1 = 0 x3 = 0, Eλ2 → = span{υ2 } For λ3 . = 3, N ot where x1 and x3 are basic variables and x2 is a free variable. Let x2 ⎛ 3 λ3 I − A = ⎜ −1 ⎝ 0 0 = t 1 → and υ2 = (0, 1, 0) T . Then ⎞ 5 −1 ⎟ . 0 0 ⎠ We solve the following system. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/17 Chapter 8 Eigenvalues and Diagonalizability 3 ⎛ 0 ⎛ ∣→ (A∣ 0 ) = ⎜ ⎜ ∣ ⎝ 0 3 0 R1 (3)+R2 0 0 −1 ⎛ ⎜ ⎜ ⎝ 0 0 5 0 0 0 0 −1 0 0 −1 1 − 0 0 5 −1 ⎝ −1 0 1 − − − − − →⎜ ⎜ ⎛ x1 ⎞⎛ ⎞ ⎛ 0 ⎞ 5 −1 ⎟ ⎜ x2 ⎟ = ⎜ 0 ⎟ . ⎜ −1 ⎝ 1 2 15 0 ⎠⎝ ⎞ x3 R1, ⎠ ⎛ ⎝ −1 0 ⎠ 5 −1 0 ⎞ 0 0 ⎟ ⎟ 2 ⎟− ⎜ ⎟ →⎜ 3 1 fo rD is tri bu tio n 8/30/2020 5 ⎠ ⎝ −1 0 ⎞ 15 −2 0 0 0 0 0 0 R2 ( 0 1 15 0 ⎠ 0 ) ⎟ − −− → ⎟ ⎠ ⎛ ⎞ −1 0 − R2 (−5)+R1 0 ⎜ ⎟ − −−−− → ⎜ ⎟ ⎜ 0 ⎠ ⎝ 0 1 − 0 0 1 3 2 15 0 0 ⎞ 0 ⎟ ⎟. ⎟ 0 ⎠ The system corresponding to the last augmented matrix is −x1 − { 3 2 N ot x2 − 1 15 x3 = 0 x3 = 0, where x1 and x2 are basic variables, and x3 is a free variable. Let x3 → Eλ3 = span{υ3 } . = 15t → and υ3 = (−5, 2, 15) T . Then We shall give more examples of finding eigenvalues and eigenspaces in the next section. Now, we turn our attention to the relations between A and its eigenvalues. Recall that the trace of A is given by tr(A) = a11 + ⋯ + ann , https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability see (2.3.2) in Section 2.2 . The matrix A and its eigenvalues have the following relationships, which show that the sum of all the eigenvalues equals the trace of A and the product of all the eigenvalues equals the determinant of A. Theorem 8.1.4. ⋯ , λn 1. λ1 2. λ1 be all the eigenvalues of the matrix A given in (8.1.1) + ⋯ +λn = tr(A). × ⋯ × λn = tr(A). Proof We only prove the case when n = 2. Let A = ( ∣ λ − a11 p(λ) = ∣ ∣ −a21 It follows that (8.1.8) λ − a22 a11 a12 a21 a22 ). Then ∣ ∣ = (λ − a11 )(λ − a22 ) − a12 a21 ∣ 2 − (a11 + a22 )λ + (a11 a22 − a12 a21 ) = λ − tr(A)λ + ∣ ∣A∣ ∣. N ot 2 = λ −a12 . Then the following assertions fo rD is tri bu tio n Let λ1 , hold. 2 p(λ) = λ − tr(A)λ + ∣ ∣A∣ ∣. On the other hand, assume that λ1 and λ2 are solutions of p(λ) = 0. Then (8.1.9) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability 2 p(λ) = (λ − λ1 )(λ − λ2 ) = λ By (8.1.8) and (8.1.9) , we see that λ1 + λ2 = tr(A) − (λ1 + λ2 )λ + λ1 λ2 . and λ1 λ2 = |A| . Example 8.1.4. . Find tr(A) and |A| using Theorem 8.1.4 Solution By Example 8.1.3 , λ1 = 0, λ2 = −2, tr(A) = λ1 + λ2 + λ3 = 1 and λ3 and |A| = λ1 λ2 λ3 . fo rD is tri bu tio n Let A be the same as in Example 8.1.3 = 3. = 0 By Theorem 8.1.4 , . By Theorem 8.1.4 (2), we obtain the following result, which provides a criterion for a matrix to be invertible (see Corollary 2.7.1 and Theorem 3.4.1 for other criteria). Corollary 8.1.3. A square matrix A is invertible if and only if all its eigenvalues are nonzero. By Corollary 8.1.3 Example 8.1.5. or equation (8.1.4) , we see that if zero is an eigenvalue of A, then A is not invertible. Solution N ot Assume that A is a 2 × 2 matrix and 2I − A and 3I − A are not invertible. Find |A| and tr(A) . Because 2I − A and 3I − A are not invertible, |2I − A| = 0 and |3I − A| = 0. Hence, 2 and 3 are the eigenvalues of A. By Theorem 8.1.4 , |A| = 2 × 3 = 6 and tr(A) = 2 + 3 = 5 . Theorem 8.1.5. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability 1. Let λ be an eigenvalue of A and let → X be an eigenvector of A. Then for each integer k, λk is an → eigenvalue of Ak and X is an eigenvalue of Ak . 2. Let ϕ(x) = a0 + a1 x + ⋯ + am xm be a polynomial of degree m and λ be an eigenvalue of A. Then ϕ(λ) is an eigenvalue of ϕ(A) . Proof → 1. Because A X → = λX , 2 A → → 1 λ is an eigenvalue of A−1 . fo rD is tri bu tio n 3. Assume that A is invertible and λ is an eigenvalue of A. Then → → → 2 X = A(A X ) = A(λ X ) = λA X = λ(λ X ) = λ → → X . → Repeating the process implies that Ak X = λk X . The result follows. 2. Because λ is an eigenvalue of A, then for each positive integer k, k λ → k X = A Hence, → → X for some vector X ≠ m N ot ϕ(λ)I − ϕ(A) = (a0 + a1 λ + ⋯ + am λ 2 → 0 . m )I − (a0 I + a1 A + ⋯ + am A 2 m = a1 (λI − A) + a2 (λ I − A ) + ⋯ + am (λ It follows that → → 2 2 → m I −A m (ϕ(λ)I − ϕ(A)) X = a1 (λI − A) X + a2 (λ I − A ) X + ⋯ + am (λ ) ). m I −A → )X = → 0 and ϕ(λ) is an eigenvalue of ϕ(A) . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability → 3. If λ is an eigenvalue of A, then A X −1 A exists and → = λX This implies that → −1 X = λA → (X ) Example 8.1.6. → −1 (A X ) = A and 1 λ → −1 X = A → → X ≠ → 0 . −1 (λ X ) = λA → ( X ). Hence, 1 λ Because A is invertible, → ( X ). is an eigen-value of A−1 . fo rD is tri bu tio n −1 A for some vector Let A be a 3 × 3 matrix. Assume that 1, 2, and 3 are the eigenvalues of A. 1. Find all eigenvalues of 2A + 3A2 . 2. Find all eigenvalues of A−1 . Solution Let ϕ(x) = 2x + 3x2 . Then ϕ(A) = 2A + 3A2 . Because 1, 2, 3 are the eigenvalues of A, it follows from Theorem 8.1.5 that ϕ(1), ϕ(2), and ϕ(3) are the eigenvalues of ϕ(A). By computation, we have 2 ϕ(1) = 2 + 3 = 5, ϕ(2) = 2(2) + 3(2) 2 = 16 and ϕ(3) = 2(3) + 3(3) = 33. (2) By Theorem 8.1.5 N ot Hence, all the eigenvalues of 2A + 3A2 are 5, 16, and 33. (3), all the eigenvalues of A−1 are 1, 1 2 , and 1 3 . Exercises 1. For each of the following matrices, find its eigenvalues and determine which eigenvalues are repeated eigenvalues. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/17 8/30/2020 Chapter 8 Eigenvalues and Diagonalizability A = ( 1 0 2 1 2 0 0 B = ⎜1 1 0 ⎛ ) ⎝ 2 −3 −4 ⎛ ⎞ ⎟ ⎠ −1 0 0 0 0 2 0 0 0 −1 0⎟ ⎟ 0⎟ 0 0 0 ⎜ C = ⎜ ⎜ ⎝ 0 ⎞ ⎠ A = ( 5 7 3 1 ) B = ( 3. Let A be a 3 × 3 matrix. Assume that 2I − A, 1. Find |A| and tr(A). 2. Prove that I + 5A is invertible. fo rD is tri bu tio n 2. For each of the following matrices, find its eigenvalues and eigenspaces. 3 −1 1 5 ) C = ( 3I − A, 3 0 0 3 1 2 3 ) D = ⎜2 1 3⎟ 1 2 ⎛ ⎝ 1 ⎞ ⎠ and 4I − A are not invertible. 2 − 2A + 3A + I ∣ ∣ . N ot 4. Assume that the eigenvalues of a 3 × 3 matrix A are 1, 2, 3. Compute ∣∣A3 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/17 8/30/2020 8.2 Diagonalizable matrices 8.2 Diagonalizable matrices By (2.3.7) , we see that it is easy to compute D k, where D is a diagonal matrix. For a nondiagonal matrix A, ut io n is it possible to find a suitable diagonal matrix D such that A k can be computed by D k ? The answer is yes if A is diagonalizable. Definition 8.2.1. By Definition 8.2.1 and Theorem 3.4.1 D is tri b An n × n matrix A is said to be diagonalizable if there exists an n × n invertible matrix P such that P − 1AP is a diagonal matrix. P is said to diagonalize A. (i), we obtain the following result. Theorem 8.2.1. N ot (8.2.1) fo r An n × n matrix A is diagonalizable if and only if there exist an n × n matrix P with |P | ≠ 0 and a diagonal matrix D such that AP = PD. The following result shows that if A is diagonalizable, then (8.2.2) Processing math: 77% → → D = diag(λ 1, λ 2, ⋯, λ n) and P = (p 1⋯ p n), https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/28 8/30/2020 8.2 Diagonalizable matrices → where λ i is the eigenvalue of A and the column vector p i of P is the eigen-vector of A corresponding to λ i for i ∈ I n. Theorem 8.2.2. Proof D is tri b → → Assume that A has n linearly independent eigenvectors: p 1, ⋯, p n such that ut io n An n × n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. In this case, the matrix P in (8.2.2) diagonalizes A and the diagonalizable matrix is the matrix D in (8.2.2) . → → A p i = λ i p i for i = 1, ⋯, n. Let D and P be the same as in (8.2.2) . Then P is invertible and N ot fo r → → → → → → AP = A(p 1⋯ p n) = (Ap 1⋯Ap n) = (λ 1p 1⋯λ np n) → → = (p 1⋯ p n)diag(λ 1, λ 2, ⋯, λ n) = PD. By Theorem 8.2.1 , A is diagonalizable. Conversely, if A is diagonalizable, then AP = PD, where P and D are given by (8.2.2) and P is invertible. Because → → → → → → AP = A(p 1⋯ p n) = (Ap 1⋯Ap n) and PD = (λ 1p 1⋯λ np n), it follows that Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/28 8/30/2020 8.2 Diagonalizable matrices → → A p i = λ i p i for i = 1, ⋯, n. and 3.4.1 , { } is linearly independent. ut io n Because P is invertible, by Theorems 7.1.5 → → p 1, ⋯, p n Remark 8.2.1. fo r D is tri b 1. By (8.2.2) and the proof of Theorem 8.2.2 , we see that the matrices D and P depend on the order of eigenvalues and both follow the same order of the eigenvalues. Hence, different orders of eigenvalues give different matrices D and P. Hence, the choices of D and P are not unique (see Remark 8.2.3 ). 2. If there is more than one linearly independent eigenvector corresponding to one eigenvalue, then different orders of these eigenvectors give different matrices P, but the diagonal matrix D remains unchanged. N ot By Theorem 8.2.2 , we see that, to determine whether A is diagonalizable, we need linearly independent eigenvectors of A and the number of linearly independent eigenvectors. The following result provides the linear independence of the eigenvectors of a matrix. Theorem 8.2.3. Let λ i and r i be the same as in (8.1.5) → → → for i ∈ {1, 2, ⋯, k}. Assume that pi1, pi2, ⋯, pir i are linearly independent eigenvectors corresponding to λ r for i ∈ I n. Then the set of all eigenvectors i Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/28 8/30/2020 8.2 Diagonalizable matrices → → → → → → → → → p11, p12, ⋯, p1r 1; p21, p22, ⋯, p2r 2; ⋯; pk1, pk2, ⋯, pkr k is linearly independent. Assume that there exist x i1, x i2, ⋯xir i ∈ ℝ for i ∈ I k such that (8.2.3) ( ) ri ∑ → ∑ x ijpi j = →0. i=1 j=1 We prove below that x ij = 0 for j ∈ {1, 2, ⋯, ri } and i ∈ Ik. Indeed, for i ∈ Ik, let fo r (8.2.4) becomes ∑ k i=1 ri → ∑ x ijp ij. N ot → qi = Then (8.2.3) D is tri b k ut io n Proof j=1 → → q i = 0. Let i ∈ I k be fixed. Then (8.2.5) Processing math: 77% → → → → q 1 + ⋯ + q i + ⋯ + q k = 0. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/28 8/30/2020 8.2 Diagonalizable matrices → → Because Ap ij = λ ip ij for j ∈ {1, 2, ⋯, ri }, we have for i ∈ In, → A qi = A [ ] ri → ∑ x ijp ij = j=1 ri r j=1 j=1 i → → → ∑ x ijAp ij = λ i ∑ x ijp ij = λ i q i and This, together with (8.2.5) D is tri b → → → → → → A(q 1 + ⋯ + q i + ⋯ + q k ) = λ 1q 1 + ⋯ + λ i q i + ⋯ + λ k q k . ut io n (8.2.6) , implies fo r (8.2.7) Multiplying (8.2.5) by λ k implies N ot → → → → λ 1q 1 + ⋯ + λ i q i + ⋯ + λ k q k = 0. → → → → λ kq 1 + ⋯ + λ k q i + ⋯ + λ k q k = 0. This, together with (8.2.7) implies that (8.2.8) Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/28 8/30/2020 8.2 Diagonalizable matrices → → → → (λ k − λ 1)q 1 + ⋯ + (λ k − λ i) q i + ⋯ + (λ k − λ k − 1)q k − 1 = 0. By (8.2.6) and (8.2.7) , we obtain (8.2.9) Multiplying (8.2.8) ut io n → → → → (λ k − λ 1)λ 1q 1 + ⋯ + (λ k − λ i)λ i q i + ⋯ + (λ k − λ k − 1)λ k − 1q k − 1 = 0. by λ k − 1 implies This, together with (8.2.9) D is tri b → → → → (λ k − λ 1)λ k − 1q 1 + ⋯ + (λ k − λ i)λ k − 1 q i + ⋯ + (λ k − λ k − 1)λ k − 1q k − 1 = 0. implies fo r → → (λ k − λ 1)(λ k − 1 − λ 1)q 1 + ⋯ + (λ k − λ i)(λ k − 1 − λ i) q i + ⋯ N ot → → + (λ k − λ k − 1)(λ k − 1 − λ k − 2)q k − 2 = 0. Repeating the above process by eliminating the following terms in order: → → → → → → q k − 2, q k − 3, ⋯, q i + 1, q i − 1, ⋯, q 2, q 1, we obtain Processing math: 77% → → (λ k − λ i)(λ k − 1 − λ i)⋯(λ i + 1 − λ i)(λ i − 1 − λ i)⋯(λ 2 − λ i)(λ 1 − λ i) q i = 0. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/28 8.2 Diagonalizable matrices → → Because λ j ≠ λ i for j ≠ i, we have q i = 0. By (8.2.4) linearly independent, x ij = 0 for j ∈ → , ∑ rj =i 1x ijp ij = 0. → → → → Because pi1, pi2, ⋯, pir i are {1, 2, ⋯, ri } and i ∈ Ik. The result follows. Theorem 8.2.3 ut io n 8/30/2020 shows that for each eigenvalue λ i, we can find linearly independent eigenvectors. For example, for each eigenvalue λ i, we can take all the basis vectors of E λ and put them together, then the set of i D is tri b all the vectors is linearly independent. The following result gives the largest number of linearly independent eigenvectors of A. Theorem 8.2.4. Let λ i and r i be the same as in (8.1.5) for i ∈ {1, 2, ⋯, k}. Then the number fo r number of linearly independent eigenvectors of A. ∑ ki= 1dim(E λi) is the largest Solution { N ot → → → Let p 1, p 2, ⋯, p r be linearly independent eigenvectors of A. We regroup the eigenvectors corresponding to λ 1, λ 2, ⋯, λ k as }{ → → → p11, p12, ⋯, p1s 1 ; By Theorem 7.2.8 Processing math: 77% }{ → → → p21, p22, ⋯, p2s 2 ; → → → pk1, pk2, ⋯, pks r k } . , we have It follows that . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/28 8/30/2020 8.2 Diagonalizable matrices s i ≤ dimE λ It follows that i for i ∈ {1, 2, ⋯, k}. ∑ ki= 1s i ≤ ∑ ki= 1dim(E λi). ut io n Combining Corollary 8.1.2 , Theorems 8.2.2 , and 8.2.3 , we obtain the following result, which provides a necessary and sufficient condition for a matrix to be diagonalizable. Theorem 8.2.5. ∑ ki= 1dim(E λi) = n, k ii. If ∑ i = 1 dim(E λ ) < n, i i. If i ∈ {1, 2, ⋯, k}. Then the following assertions hold. then A is diagonalizable. then A is not diagonalizable. Proof ,k≤ ∑ ki= 1dim(E λi) ≤ n and for each eigenvalue λ i, independent eigenvectors. By Theorem 8.2.3 ∑ ki= 1dim(E λi) = n, 8.2.2 ii. If ∑ , the set of all these eigenvectors is linearly independent. then A has n linearly independent eigenvectors. It follows from Theorem that A is diagonalizable. k dim(E λ ) i=1 i 8.2.2 i N ot i. If there are dim(E λ ) linearly fo r By Corollary 8.1.2 D is tri b Let λ i and r i be the same as in (8.1.5) < n, then A does not have n linearly independent eigenvectors and by Theorem , A is not diagonalizable. The following result provides the methods used to determine whether an n × n matrix A is diagonalizable. Corollary 8.2.1. Processing math: Let77% A be an n × n matrix and let λ i and r i be the same as in (8.1.5) for i ∈ {1, 2, ⋯, k}. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/28 8/30/2020 8.2 Diagonalizable matrices 1. If A has n distinct eigenvalues, then A is diagonalizable. 2. If A has repeated eigenvalues: λ n , λ n , ⋯λ nl with algebraic multiplicities, r n , r n , ⋯r nl, respectively, 1 2 1 2 then the following assertions hold. i. If dim(E λ ) = r n for i ∈ {1, 2, ⋯, l}, then A is diagonalizable. i i {1, 2, ⋯, l } such that dim(Eλ ) < rn , then A is not diagonalizable. i0 i0 ut io n ii. If there exists i 0 ∈ Proof 1. Let λ 1, λ 2, ⋯, λ n be distinct eigenvalues of A. By Corollary 8.1.1 ∑ ki= 1dim(E λi) = n. The result follows from Theorem 8.2.5 (i). D is tri b This implies that , dim(E λ ) = 1 1 for each i ∈ I n. i 2. Note that if r j = 1 for j ∈ {1, 2, ⋯, k}, then by Corollary 8.1.1 , dim(E λ ) = 1. This, together with i dim(E λ ) = r n for i ∈ {1, 2, ⋯, l}, implies dim(E λ ) = r i for all i ∈ {1, 2, ⋯, k}. By (8.1.6) i i i k , i=1 , A is diagonalizable and the result (i) holds. If there exists i 0 ∈ such that dim(E λ ) < r n , then ni i0 0 {1, 2, ⋯, l } N ot By Theorem 8.2.5 fo r ∑ dim(E λi) = r 1 + r 2 + ⋯ + r k = n. k ∑ dim(E λi) < r 1 + r 2 + ⋯ + r k = n. i=1 By Theorem 8.2.5 , A is not diagonalizable and the result (ii)holds. Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/28 8/30/2020 8.2 Diagonalizable matrices By Corollary 8.2.1 , we see that if each of the algebraic multiplicities is equal to the corresponding geometric multiplicity, then A is diagonalizable (see (1) and (i) in Corollary 8.2.1 ) and if there exists a geometric multiplicity, which is less than the corresponding algebraic multiplicity, then A is not diagonalizable (see (ii) in Corollary 8.2.1 ). Example 8.2.1. ( ) ( ) ( ) 0 1 0 0 0 1 B= 4 − 17 8 3 2 4 1 0 0 2 0 2 C= 4 2 3 1 2 0 Solution | 0 λ 0 −1 − 4 17 λ − 8 | = λ 2(λ − 8) − 4 + 17λ fo r p(λ) = |λI − A | = λ −1 −3 5 2 D is tri b A= ut io n For each of the following matrices, determine whether it is diagonalizable. = λ 2 − 8λ 2 + 17λ − 4 = (λ − 4)(λ 2 − 4λ + 1) 2 N ot = (λ − 4)[(λ − 2) 2 − 3] = (λ − 4)[(λ − 2) 2 − (√3) ] = (λ − 4)(λ − 2 − √3)(λ − 2 + √3). Solving p(λ) = 0 implies that λ 1 = 4, λ 2 = 2 + √3, and λ 3 = 2 − √3. Hence, A has three distinct eigenvalues. By Corollary 8.2.1 (1), A is diagonalizable. Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/28 8/30/2020 8.2 Diagonalizable matrices p(λ) = |λI − B | = | λ − 3 −2 −4 −2 λ −2 −4 −2 λ − 3 | = [λ(λ − 3) − 16 − 16] − [16λ + 4(λ − 3) + 4(λ − 3)] = [λ 3 − 6λ 2 + 9λ − 32] − [24λ − 24] = λ 3 − 6λ 2 − 15λ − 8 ut io n = (λ − 8)(λ 2 + 2λ + 1) = (λ + 1) 2(λ − 8). Solving p(λ) = 0, we have λ 1 = − 1 and λ 2 = 8. Moreover, λ 1 = − 1 has algebraic multiplicity 2 and λ 2 = 8 −2 −1 −2 −4 −2 −4 We solve the above system. → ( x2 0 = x3 ( ) ) ( ( ) 1 − 4 − 2 − 4 0 R1 − 2 N ot | (B 0) = )( ) ( ) x1 0 . 0 fo r ( −4 −2 −4 D is tri b has algebraic multiplicity 1. We find dim(E λ ). For λ 1 = 1, the system λ 1I − B = 0 becomes i −2 −1 −2 0 → −4 −2 −4 0 R1 ( 2 ) + R2 → R1 ( 4 ) + R3 2 1 2 0 −2 −1 −2 0 −4 −2 −4 0 ) 2 1 2 0 0 0 0 0 . 0 0 0 0 Processing math: 77% The system corresponding to the last augmented matrix is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/28 8/30/2020 8.2 Diagonalizable matrices 2x + y + 2z = 0 and has two free variables. Thus, dim(E λ ) = 2 (algebraic multiplicity). By Corollary 8.2.1 1 (i), B is diagonalizable. | 0 0 −1 λ − 2 3 0 −5 λ − 2 | = (λ − 1)(λ − 2) 2 = 0. D is tri b p(λ) = | λI − C | = λ−1 ut io n Because λ 1 = 1 has algebraic multiplicity 2 and λ 2 = 2 has algebraic multiplicity 2. We find dim(E λ ). For λ 2 = 2, the 2 system λ 2I − C = 0 becomes ( )( ) ( ) x1 −1 0 0 x2 We solve the above system. = 0 . 0 x3 N ot 3 −5 0 0 fo r 0 0 1 Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/28 8/30/2020 8.2 Diagonalizable matrices ( ) ( ) ( ) 1 (C 0) = 0 0 0 R (1) +R 1 2 −1 0 0 0 → 3 −5 0 0 R2 , R1 ( − 3 ) + R3 1 0 0 0 0 0 0 0 0 −5 0 0 1 0 0 0 3 0 −5 0 0 . 0 0 0 0 → ut io n | → The system corresponding to the last augmented matrix has one free variable and thus, dim(E λ ) = 1 < 2 2 (ii), C is not diagonalizable. Example 8.2.2. Find x ∈ ℝ such that A is diagonalizable, where ( ) 2 0 1 fo r 2 1 x . 0 0 1 N ot A= D is tri b (algebraic multiplicity). By Corollary 8.2.1 Solution Because | || λI − A = λ−2 0 −2 λ − 1 0 0 −1 −x λ−1 | = (λ − 1) 2(λ − 2) = 0, Processing math: 77% λ 1 = 1 (with algebraic multiplicity 2), λ 2 = 2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 13/28 8/30/2020 8.2 Diagonalizable matrices We find dim(E λ ). For λ 1 = 1, the system λ 1I − A = 0 becomes 1 ( )( ) ( ) x1 −2 0 −x x2 0 0 0 x3 0 0 . 0 = We solve the above system. (A | 0) = ( ) ( ) −1 0 −1 0 R ( −2) +R −1 0 −1 0 1 2 −2 0 −x 0 0 0 −x + 2 0 . → 0 0 0 0 0 0 0 0 D is tri b → ut io n −1 0 −4 If x = 2, then the system corresponding to the last augmented matrix has two free variables and thus, dim(E λ ) = 2 (algebraic multiplicity). By Corollary 8.2.1 (i), A is diagonalizable. fo r 1 N ot In the above examples, we used eigenvalues or eigenvectors of a matrix to determine whether the matrix is diagonalizable. Now, we use (8.2.2) and Theorem 8.2.2 to find the diagonalizable matrix D and the invertible matrix P that diagonalizes A. Example 8.2.3. Let A = ( ) 4 2 3 3 . 1. Find the eigenvalues and eigenspace of A. 2. Find an inverse matrix P and a diagonal matrix D such that Processing math: 77% A = PDP − 1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 14/28 8/30/2020 8.2 Diagonalizable matrices Solution | | | 1. p(λ) = λI − A = λ−4 −2 −3 λ−3 | = (λ − 4)(λ − 3) − 6 Solving p(λ) = (λ − 1)(λ − 6) = 0 implies λ 1 = 1 and λ 2 = 6. → → For λ 1 = 1, (λ 1I − A)X = 0 becomes −2 −3 1−3 )( ) ( x y = −3 −2 −3 −2 The above system is equivalent to the equation )( ) ( ) x y = 0 0 . D is tri b ( 1−4 ut io n = λ 2 − 7λ + 12 − 6 = λ 2 − 7λ + 6 = (λ − 1)(λ − 6). 3x + 2y = 0, fo r where x is a basic variable and y is a free variable. Let y = 3t. Then x = − 2t. Hence, () ( ) ( ) x = =t 3t −2 3 → = tυ1, N ot y − 2t → where υ 1 = ( − 2, 1) T. Then E λ = 1 → { } → tυ1 : t ∈ ℝ . → For λ 2 = 6, (λ 2I − A)X = 0 becomes Processing math: 77% ( 6−4 −2 −3 6−3 )( ) ( x y = 2 −2 −3 3 )( ) ( ) x y = 0 0 . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 15/28 8/30/2020 8.2 Diagonalizable matrices The above system is equivalent to x − y = 0, where x is a basic variable and y is a free variable. Let y = t. Then x = t and y { =t t } ( , A = P − 1DP. ( −2 1 3 1 )( ) ( 1 0 0 6 = ( )( −2 6 3 6 4 2 −2 1 3 3 3 1 ) 1 1 ) . to obtain A = P − 1DP. Infact,we cancheck AP = PD. N ot In the above solution, we use Theorem 8.2.2 Indeed, and PD = = tυ 2, fo r Remark 8.2.2. AP = 1 → tυ2 : t ∈ ℝ . →→ −2 2. Let D = diag(λ 1, λ 1) = diag(1, 6) and P = ( υ 1 υ 2 ) = 3 Then by Theorem 8.2.2 1 D is tri b → where υ 2 = (1, 1) T. Then E λ = 2 t = ut io n () () ( ) x ) ( = −2 6 3 6 ) . Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 16/28 8/30/2020 8.2 Diagonalizable matrices Remark 8.2.3. , the choice of D and P is not unique. In Example 8.2.1 ( ) 6 D 1 = diag(λ 2, λ 1) = diag(6, 1)= 0 and | | ≠ 0. 3 ) 1 . fo r By direct verification, we have AP 1 = P 1D 1 and P 1 −2 0 D is tri b ( →→ 1 P1 = (υ2 υ1) = 1 , we ut io n As we mentioned in Remark 8.2.1 can make the following choice: By Theorem 8.2.2 N ot Note that both D and P must follow the same order of the choices of eigenvalues. For example, we cannot choose either D and P 1 or D 1 and P because you can check AP 1 ≠ P 1D and AP ≠ PD 1. , we can choose P as follows: → → P = (α υ 1 β υ 2 ), where α ≠ 0 and β ≠ 0. Inmath: Example Processing 77% 8.2.3 , the geometric multiplicity of each eigenvalue is 1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 17/28 8/30/2020 8.2 Diagonalizable matrices The following example deals with the case when geometric multiplicities of eigenvalues are greater than 1. The matrix below is taken from Example 8.2.2 with x = 2. Example 8.2.4. Let ( ) 0 1 2 1 0 0 2 . 1 ut io n A= 2 D is tri b 1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1. 1. | λ−2 0 1 −2 λ−1 −2 0 λ−1 0 N ot p(λ) = | λI − A | = | fo r Solution = (λ − 1) 2(λ − 2). Solving p(λ) = (λ − 1) 2(λ − 2) = 0 implies λ 1 = 1 and λ 2 = 2. The algebraic multiplicities of λ 1 and λ 2 are 2 and 1, respectively. For λ 1 = 1, the system λ 1I − A = 0 becomes Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 18/28 8.2 Diagonalizable matrices ( −1 0 −1 −2 0 −2 0 0 0 )( ) ( ) x1 0 x2 0 . 0 = x3 We solve the above system. ( 0 −1 −2 0 −2 0 0 0 ) ( 0 R ( −2) +R −1 1 2 0 0 → 0 −1 0 0 0 0 0 0 0 ) 0 . 0 D is tri b → (A | 0) = −1 ut io n 8/30/2020 The system with the last augmented matrix is equivalent to the equation x + 0y + z = 0, fo r where x is a basic variable and y and z are free variables. Let y = s and z = t. Then x = − t. Hence, () ( ) () ( ) x −t y s =s 1 0 −1 +t N ot = 0 z t → → T where υ 1 = (0, 1, 0) and υ 2 = ( − 1, 0, 1) T. Then E λ = 1 0 → → = tυ1 + sυ2, 1 { } → → s υ 1 + t υ 2 : s, t ∈ ℝ . For λ 2 = 2, the system λ 2I − A = 0 becomes Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 19/28 8.2 Diagonalizable matrices ( 0 0 −1 −2 1 −2 0 0 1 )( ) ( ) x1 0 x2 0 . 0 = x3 We solve the above system. → = ( 0 −1 −2 1 −2 0 0 1 R2 ( 1 ) + R3 → R2 ( − 2 ) + R1 )( 0 R −2 1, 2 0 → 0 0 0 ( −2 0 0 1 −2 0 −1 0 1 0 0 0 ) D is tri b | (A 0) 0 ut io n 8/30/2020 1 0 0 −1 0 0 0 ) 0 . 0 fo r The system with the last augmented matrix is equivalent to the equation − z = 0, N ot { − 2x + y = 0 → where x and z are basic variables and y is a free variable. Let y = 2t. Then x = t. Let υ 3 = (1, 2, 0) T. Then E λ = 2 { } → tυ3 : t ∈ ℝ . 2. Let D = diag(λ 1, λ 2, λ 3) = diag(1, 1, 2) and Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 20/28 8/30/2020 8.2 Diagonalizable matrices →→→ P = (υ1 υ2 υ3) = −1 1 0 0 1 1 ) 2 . 0 , A = PDP − 1. ut io n Then by Theorem 8.2.2 ( 0 Remark 8.2.4. ( −1 0 0 1 1 0 1 ) 2 , 0 N ot fo r →→→ P1 = (υ2 υ1 υ3) = D is tri b → → In Example 8.2.4 , there are two basis eigenvectors υ 1 and υ 2 corresponding to the eigenvalue → → → λ 1 = 1. We choose the column vectors of P in the order υ 1 , υ 2 , υ 3 . By Theorem 8.2.2 , we can → → → choose the following order υ 1 , υ 2 , υ 3 . If we let −1 then it is easily verified that A = P 1DP 1 . Finally, we give the methods used to compute A k if A is diagonalizable. Theorem 8.2.6. Processing math: 77% −1 If P AP = D, then A k = PD kP − 1 for each k ∈ ℕ. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 21/28 8/30/2020 8.2 Diagonalizable matrices Proof Because P − 1AP = D, AP = PD and A = PDP − 1. A 2 = (PDP − 1)(PDP − 1) = PD(P − 1P)DP − 1 = PD 2P − 1. Example 8.2.5. Let ( −1 3 2 2 1 4 ) D is tri b A= 1 ut io n Repeating the process implies that the result holds. −1 . −1 fo r 1. Find the eigenvalues and eigenspace of A. 2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1. Solution 1. N ot 3. Compute A 3 by using the result (2). Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 22/28 8/30/2020 8.2 Diagonalizable matrices p(λ) = | || λI − A = λ−1 1 −4 −3 λ−2 1 −2 −1 λ+1 | = (λ − 1)(λ − 2)(λ + 1) − 12 − 2 − 8(λ − 2) + (λ − 1) + 3(λ + 1) = (λ 2 − 1)(λ − 2) − 14 − 8λ + 16 + λ − 1 + 3λ + 3 ut io n = λ 3 − 2λ 2 − 5λ + 6 = (λ − 1)(λ 2 − λ − 6) = (λ − 1)(λ − 3)(λ + 2). Solving p(λ) = 0, we get three eigenvalues → λ 2 = 1, and λ 3 = 3. → For λ 1 = − 2, (λ 1I − A)X = 0 becomes ( 1 −4 −3 −4 1 −2 −1 −1 )( ) ( ) x1 x2 x3 fo r → BX : = −3 D is tri b λ 1 = − 2, 0 = 0 . 0 N ot Now we use row operations to solve the above system. Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 23/28 8/30/2020 8.2 Diagonalizable matrices ( 1 −4 −3 −4 −2 −1 R1 ( − 2 ) + R3 → R1 ( − 3 ) + R2 R2 ( − 1 ) + R3 → ( ( ) ( ( ) ) ( )( ) ( ) 2 −3 0 1 0 R ( −1) +R −1 3 1 0 −3 → −4 1 0 1 0 −1 −1 0 −1 2 −3 0 0 1 −1 0 0 1 −1 0 −1 2 0 − 10 0 −5 −2 −1 2 −3 0 1 −1 0 0 0 1 −3 0 R3 − 5 10 0 5 0 → R2 − 1 10 0 R ( −2) +R −1 2 1 0 0 → 0 0 0 1 0 ) ut io n (B | 0) = −3 ) −1 0 −1 0 . 0 D is tri b → 0 The system corresponding to the last augmented matrix is =0 fo r { x1 + x3 x2 − x3 = 0, N ot where x 1 and x 2 are basic variables and x 3 is a free variable. Let x 3 = t. Then x 2 = t and x 1 = − t. → Let υ 1 = ( − 1, 1, 1) T. Then Eλ = 1 Processing math: 77% { } → tυ1 : t ∈ ℝ . For λ2=1, (λ2I−A)X→=0→ becomes (01−4−3−11−2−12)(x1x2x3)=(000). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 24/28 8/30/2020 8.2 Diagonalizable matrices Now we use row operations to solve the above system. (B|0→)= (01−40−3−110−2−120)→R1, 2(−3−11001−40−2−120)→R3(−1)+R1(−10−1001−40−2−120)→R1(−1) (101001−40−2−120)→R1(2)+R3(101001−400−140)→R2(1)+R3(101001−400000). { x1+x3=0x2−4x3=0, ut io n Hence, we obtain D is tri b where x1 and x2 are basic variables and x3 is a free variable. Let x3=t. Then x2=4t and x1=−t. Let υ2→=(−1, 4, 1)T. Then Eλ2={tυ2→:t∈ℝ}. For λ3=3, (λ3I−A)X→=0→ becomes (21−4−311−2−14)(x1x2x3)=(000). This implies N ot fo r Now we use row operations to solve the above system. (B|0→)(21−40−3110−2−140)→R2(1)+R1(−12−30−3110−2−140)→R1(−1) (1−230−3110−2−140)→R1(2)+R3R1(3)+R2(1−2300−51000−5100)→R2(−1)+R3(1−2300−51000000)→R2(−15) (1−23001−200000)→R2(2)+R1(10−1001−200000). { x1−x3=0x2−2x3=0, where x1 and x2 are basic variables and x3 is a free variable. Let x3=t. Then x2=2t and x1=t. Let υ3→=(1, 2, 1)T. Then Eλ3={tυ3→:t∈ℝ}. 2. Let D=diag(λ1, λ2, λ3)=diag(−2, 1, 3) and Processing math: 77% P=(υ1→υ2→υ3→)=(−1111−421−11). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 25/28 8/30/2020 8.2 Diagonalizable matrices Then by Theorem 8.2.2 , A=PDP−1. 3. By computation, we have D3=(−200010003)3=(−8000100027). Now, we find P−1. ut io n (P|I3)= (−1−11100142010111001)→R1(1)+R2R1(1)+R3(−1−11100033110111001)→R2(13),R3(12)R1(−1) (11−1−1000111313000112012)→R3(−1)+R2R3(1)+R1(110−12012010−1613−1200112012)→R3(−1)+R1 (100−13−131010−1613−1200112012)=(I3|P−1). D is tri b By Theorem 8.2.6 , we obtain A3=PD3P−1=(−1−11142111)(−8000100027)(−13−131−1613−1212012)=(8−127−8454−8127) (−13−131−1613−1212012)=(11−322294171635). Exercises fo r 1. For each of the following matrices, determine whether it is diagonalizable. A=(−124003170058000−2)B=(−1430−110−43). N ot 2. Find x∈ℝ such that A is diagonalizable, where A=(201−41x102). 3. Assume that υ→=(1, 1, 0)T is an eigenvector of the matrix A=(1−113x2y01). i. Find x, y∈ℝ R and the eigenvalue corresponding to υ→. ii. Is A diagonalizable? Processing math: 77% 4. Let https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 26/28 8/30/2020 8.2 Diagonalizable matrices A=(1−1−41). 1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A=PDP−1. A=(324202423). D is tri b 1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A=PDP−1. ut io n 5. Let 3. Compute A2 by using the result (2). 6. Let A=(00−2121103). 7. Let N ot fo r 1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A=PDP−1. 3. Compute A2 by using the result (2). A=(1−113−32001). 1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A=PDP−1. 3. Compute A3 by using the result (2). Processing math: 77% 8. Let https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 27/28 8/30/2020 8.2 Diagonalizable matrices A=(01−10). 1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A=PDP−1. 3. Compute A2 by using the result (2). fo r D is tri b Show that A is similar to B. 10. Assume that A~B. Show that the following assertions hold. i. |λI−A|=|λI−B|. ii. A and B have the same eigenvalues. iii. tr(A)=tr(B). iv. |A|=|B|. v. r(A)=r(B). ut io n 9. Two n×n matrices A and B are said to be similar (or A is similar to B) if there exists an invertible n×n matrix P such that B=P−1AP. We write A~B. Let A=(1−121),B=(1−211), and P=(01−10). N ot 11. Assume that A, B, C are n×n matrices. Show that the following assertions hold i. A~A. ii. If A~B, then B~A. iii. If A~B and B~C, then A~C. 12. Assume that A and B are n×n matrices and A~B. Prove that the following assertions hold. a. Ak~Bk for every positive integer k. b. Let ϕ(λ)=amλm+am−1λm−1+⋯+a1λ+a0. Then ϕ(A)~ϕ(B). Processing math: 77% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 28/28 8/30/2020 Chapter 9 Vector spaces ut io n Chapter 9 Vector spaces 9.1 Subspaces of ℝ n D is tri b Definition 9.1.1. Let W be a nonempty subset of ℝ n. W is called a subspace of ℝ n if it satisfies the following two conditions: → i. If → a = (x 1, x 2, ⋯, x n) ∈ W, and b = (y 1, y 2, ⋯, y n) ∈ W, then → a + b = (x 1 + y 1, x 2 + y 2, ⋯, x n + y n) ∈ W. fo r → N ot ii. If → a = (x 1, x 2, ⋯, x n) ∈ W and k ∈ ℝ is a real number, then k→ a = (kx 1, kx 2, ⋯, kx n) ∈ W. Note that the above conditions (i) and (ii) are equivalent to the following condition: for → → a = (x 1, x 2, ⋯, x n) ∈ W, b = (y 1, y 2, ⋯, y n) ∈ W, α ∈ ℝ and β ∈ ℝ, (9.1.1) → α→ a + β b ∈ W. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/12 8/30/2020 Chapter 9 Vector spaces By Definition 9.1.1 , we see that ℝ n is a subspace of itself, and the set containing only the origin W : {(0, 0, ⋯, 0)︸n} is a subspace of ℝ n. In Definition 9.1.1 → (ii), if k = 0 and → a ∈ W, then k→ a = 0→ a = 0 ∈ W. . This shows that every subspace of ℝ n ut io n → contains the zero vector 0, Hence, we have the following remark. Remark 9.1.1. D is tri b Any nonempty subsets of ℝ n that do not contain the origin of ℝ n are not subspaces of ℝ n. The following result shows that the solution space of a homogeneous system (see (4.1.13) of ℝ n. Theorem 9.1.1. NA = Proof → {X ∈ ℝ : AX = 0 } → n → → N ot is a subspace of ℝ n. fo r Let A be an m × n matrix. Then the solution space ) is a subspace → → → → → → Note that X ∈ N A if and only if AX = 0. Let → a ∈ N A, b ∈ N A, and k ∈ ℝ. Then A→ a = 0 and A b = 0. Hence, by Theorem 2.2.1 , → → → → → → A(→ a + b) = A→ a + A b = 0 + 0 and A(k→ a) = k 0 = 0. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/12 8/30/2020 Chapter 9 Vector spaces → This implies → a + b ∈ N A and k→ a ∈ N A. By Definition 9.1.1 , N A is a subspace of ℝ n. → → The following result shows that the set of solutions of a nonhomogeneous system AX = b is not a subspace of ℝ n because it doesn’t contain the zero vector in ℝ n. → Let A be an m × n matrix and b ∈ ℝ m and let SA = → → → { X ∈ ℝ : AX = b } . → n → → → Proof → → → → D is tri b If AX = b is consistent and b ≠ 0, then S A is not a subspace of ℝ n. ut io n Theorem 9.1.2. Because AX = b is consistent, AX = b has at least one solution and the set S A is a nonempty set. → → → → → → , S A is not a fo r Because b ≠ 0, 0 is not a solution of AX = b. Hence, 0 ∉ S A. By Theorem 9.1.2 subspace of ℝ n. What types of subsets in ℝ n are subspaces of ℝ n ? In the following, we find all the subspaces in ℝ 2 and ℝ 3. Example 9.1.1. N ot The same approach can be used to deal with ℝ n with n ≥ 4. First, we consider subspaces in ℝ 2. Show that W is a subspace of ℝ 2, where W= {(x, y) ∈ ℝ : x + 2y = 0}. 2 Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/12 8/30/2020 Chapter 9 Vector spaces → Let A = (1, 2) and X = (x, y) T. Then W= It follows from Theorem 9.1.1 {(x, y) ∈ ℝ : x + 2y = 0 } = {(x, y) ∈ ℝ : AX = 0 }. 2 2 → → that W is a subspace of ℝ 2. ut io n Note that x + 2y = 0 represents the equation of the line passing through the origin (0, 0). By Theorem 9.1.1 , one can show that any line passing through the origin (0, 0) is a subspace of ℝ 2. The equation of a line passing through the origin (0, 0) is of the form D is tri b ax + by = 0, where one of a and b is not zero. Conclusion: All the subspaces in ℝ 2 are the origin {(0, 0)}, the entire space ℝ 2, and all the lines passing through the origin (0, 0). Any other subsets of ℝ 2 are not subspaces of ℝ 2. We do not prove the result here, Example 9.1.2. N ot Show that W is not a subspace of ℝ 2, where fo r but we give two examples that are not subspaces of ℝ 2. W= {(x, y) ∈ ℝ : x + 2y = 1}. 2 Solution Because (x, y) = (0, 0) is not a solution of x + 2y = 1, (0, 0) ∉ W. By Remark 9.1.1 , W is not a subspace of ℝ 2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/12 8/30/2020 Chapter 9 Vector spaces Note that x + 2y = 1 represents the equation of the line that does not pass through the origin (0, 0). By Theorem 9.1.2 , one can see that any line that does not pass through the origin (0, 0) is not a subspace of ℝ 2. The equation of a line that does not pass through the origin (0, 0) is of the form ax + by = c, Example 9.1.3. Show that W is not a subspace of ℝ 2, where {(x, y) ∈ ℝ : y = x }. 2 Solution Let (x 1, y 1) = (1, 1) and k = 2. Then (x 1, y 1) ∈ W and D is tri b W= ut io n where one of a and b is not zero, and c ≠ 0. 2 fo r k(x 1, y 1) = 2(1, 1) = (2, 2) ∉ W Example 9.1.4. and W is N ot 2 because 2 = y 1 ≠ x 1 = 2 2 = 4. Hence, W does not satisfy the condition (ii) of Definition 9.1.1 not a subspace of ℝ 2. Show that W is a subspace of ℝ 3, where W= {(x, y, z) ∈ ℝ : x + 2y + 3z = 0 }. 3 Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/12 8/30/2020 Chapter 9 Vector spaces → Let A = (1, 2, 3) and X = (x, y, z) T. Then W= {(x, y, z) ∈ ℝ : x + 2y + 3z = 0 } = {(x, y, z) ∈ ℝ : AX = 0 }. It follows from Theorem 9.1.1 3 3 → → that W is a subspace of ℝ 2. ut io n Note that x + 2y + 3z = 0 represents the equation of the plane passing through the origin (0, 0, 0). By Definition 9.1.1 or Theorem 9.1.1 , one can show that any plane passing through the origin (0, 0, 0) is a subspace of ℝ 3. ax + by + cz = 0, where at least one of a, b,and c is not zero. Example 9.1.5. { fo r Show that W is a subspace of ℝ 3, where Let A = ( 3 −2 3 1 −3 5 ) → { 3x − 2y + 3z = 0, x − 3y + 5z = 0 N ot W = (x, y, z) ∈ ℝ 3 : Solution D is tri b The equation of a plane passing through the origin (0, 0, 0) is of the form } . → , X = (x, y, z) T, and 0 = (0, 0, 0) T. Then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/12 8/30/2020 Chapter 9 Vector spaces { W = (x, y, z) ∈ ℝ 3 : By Theorem 9.1.1 { 3x − 2y + 3z = 0, x − 3y + 5z = 0 } = {(x, y, z) ∈ ℝ : AX = 0 }. 3 → → , W is a subspace of ℝ 3. Show that W is not a subspace of ℝ 3, where {(x, y, z) ∈ ℝ : x + 2y + 3z = 1 }. 3 D is tri b W= ut io n Example 9.1.6. Solution Note that (x, y, z) ∈ W if and only if x + 2y + 3z = 1. Because 0 + 2(0) + 3(0) ≠ 1, so (0, 0, 0) ∉ W and W is not a subspace of ℝ 3. Note that x + 2y + 3z = 1 represents the equation of the plane that does not pass through the origin (0, 0). In fo r general, one can show that any plane that does not pass through the origin (0, 0, 0) is not a subspace of ℝ 3. The equation of a plane that does not pass through the origin (0, 0) is of the form N ot ax + by + cz = d, where at least one of a, b, and c is not zero, and d ≠ 0. Recall that the equation of a line in ℝ 3 is of the form (9.1.2) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/12 8/30/2020 Chapter 9 Vector spaces { −∞ < t < ∞ z = z 0 + tc, → → → . Let X = (x, y, z), P 0 = (x 0, y 0, z 0), and υ = (a, b, c). Then (9.1.2) as → → X = P 0 + t υ, → can be written D is tri b − ∞ < t < ∞. ut io n see Theorem 6.2.1 x = x 0 + ta y = y 0 + tb → If the line does not pass through the origin, then P 0 ≠ (0, 0, 0). If the line passes through the origin, then → P 0 = (0, 0, 0) and the equation of the line becomes → − ∞ < t < ∞. fo r → X = t υ, Example 9.1.7. → N ot The following example shows that any lines passing through the origin (0, 0, 0) are subspaces of ℝ 3. Let υ ∈ ℝ 3 be a given nonzero vector. Show that → W = span υ = {t υ : t ∈ ℝ } → is a subspace of ℝ 3. Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/12 8/30/2020 Chapter 9 Vector spaces → → Note that → a ∈ W if and only if there exists t ∈ ℝ such that → a = t υ. Let → a ∈ W, b ∈ W, and k ∈ ℝ. Then → → → there exist t 1, t 2 ∈ ℝ such that → a = t 1 υ and b = t 2 υ. Hence, → → → → → a + b = t 1 υ + t 2 υ = (t 1 + t 2) υ ∈ W → → k→ a = k(t 1 υ) = (kt 1) υ ∈ W. By Definition 9.1.1 , W is a subspace of ℝ 3. ut io n and ℝ 3 are not subspaces of ℝ 3. In Example 9.1.7 , we note that W = D is tri b Similar to ℝ 2, all the subspaces of ℝ 3 contain the origin {(0, 0, 0)}, the entire space ℝ 3, all the planes passing through the origin (0, 0, 0), and all the lines passing through the origin (0, 0, 0). Any other subsets of {t υ : t ∈ ℝ } = span {υ }. Hence, by Example 9.1.7 → → → the spanning Theorem 9.1.3. { fo r space of υ is a subspace of ℝ 3. The following result shows that spanning spaces in ℝ n are subspaces of ℝ n. } N ot → → → → → → Let a 1, a 2, ⋯, a n ∈ ℝ m. Show that span a 1, a 2, ⋯, a n Proof { is a subspace of ℝ m. } → → → Let W = span a 1, a 2, ⋯, a n . Note that → u ∈ W if and only if there exist x 1, x 2, ⋯, x n ∈ ℝ such that https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/12 8/30/2020 Chapter 9 Vector spaces → → → u = x 1a 1 + x 2a 2 + ⋯ + x na n. → → Let → u ∈ W, υ ∈ W, and k ∈ ℝ. Then there exist x 1, x 2, ⋯, x n ∈ ℝ and y 1, y 2, ⋯, y n ∈ ℝ such that → → → → → → → u = x 1a 1 + x 2a 2 + ⋯ + x na n and υ = y 1a 1 + y 2a 2 + ⋯ + y na n . ut io n → Hence, → → → → → → u + υ = x 1a 1 + x 2a 2 + ⋯ + x na n + y 1a 1 + y 2a 2 + ⋯ + y na n → D is tri b → → → → = (x 1 + y 1)a 1 + (x 2 + y 2)a 2 + ⋯ + (x n + y n)a n ∈ W and → → → k u = (kx 1)a 1 + (kx 2)a 2 + ⋯ + (kx n)a n ∈ W. Exercises N ot This shows that W is a subspace of ℝ n. fo r → {(x, y) ∈ ℝ : x − y = 0 }. Show that W is a subspace of ℝ . 2. Let W = {(x, y) ∈ ℝ : x + y ≤ 1 }. Show that W is not a subspace of ℝ . 1. Let W = 2 2 + 2 2 3. Let https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/12 8/30/2020 Chapter 9 Vector spaces 2 ℝ+ = {(x, y) : x ≥ 0 } and y ≥ 0 . i. Show that if → a ∈ ℝ 2+ and b ∈ ℝ 2+ , then → a + b ∈ ℝ 2+ . → → 2 ii. Show that ℝ + is not a subspace of ℝ 2 by using Definition 9.1.1 . 2 ut io n 4. Let 2 W = ℝ + ∪ ( − ℝ + ). D is tri b i. Show that if → a ∈ W and k is a real number, then k→ a ∈ W. ii. Show that W is not a subspace of ℝ 2 by using definition 9.1.1 5. Let W = . {(x, y, z) ∈ ℝ : x − 2y − z = 0}. Show that W is a subspace of ℝ 3 fo r Theorem 9.1.1 , respectively. 6. Show that W is a subspace of ℝ 4, where { { x − 2y + 2z − w = 0, − x + 3y + 4z + 2w = 0 N ot W = (x, y, z, w) ∈ ℝ 4 : 3 by Definition 9.1.1 } and . 7. Show that W is a subspace of ℝ 3, where W= { (x, y, z) ∈ ℝ 3 : { x + 2y + 3z = 0, − x + y − 4z = 0, x − 2y + 5z = 0 } . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/12 8/30/2020 Chapter 9 Vector spaces 8. Let W = {(x, y, z) ∈ ℝ : x − 2y − z = 3}. Show that W is not a subspace of ℝ 3 3 by Definition 9.1.1 . 9. Let W = [0, 1]. Show that W is not a subspace of R. 10. Show that all the subspaces of ℝ are {0} and ℝ. 11. Let W = {(0, 0, ⋯, 0) } ∈ ℝ . Show that W is a subspace of ℝ . n n → W= → . N ot fo r D is tri b is a subspace of ℝ n by Definition 9.1.1 {t υ : t ∈ ℝ } ut io n 12. Let υ ∈ ℝ n be a given nonzero vector. Show that https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 12/12 8/30/2020 9.2 Vector Spaces 9.2 Vector Spaces fo rD is tri bu tio n Before we give the definition of a vector space, we introduce the notion of a set. Definition 9.2.1. A set is a collection of objects. We always use capital letters such as A, B, C, V, W, X, Y, Z to denote sets. Each object in a set is called an element. We denote an element by small letters such as a, b, c, v, w, x, y, z. We use the symbol x ∈ A to denote “x belongs to A” or ”x is in A” and the symbol x ∉ A to denote “x does not belong to A”or”x is not in A.” When a set A is a collection of objects that satisfies some property P, we write A = {x : x has the property P }. Example 9.2.1. p 5. Q = { q : p, q ∈ Z, q ≠ 0} , 6. A = {(x, 2 y) : x +y 2 N ot 1. N = {1, 2, ⋯}, the set of natural numbers. 2. Z = {⋯ , − 2, − 1, 0, 1, 2, ⋯}, the set of integers. 3. R = (−∞, ∞), the set of real numbers. 4. Rn = {(x1 , x2 , ⋯ , xn ) : xi ∈ R, i = 1, 2, ⋯ , n}. the set of rational numbers; = 1}. 7. V = {Amn : Amn is an m × n matrix}, the set of all m × n matrices. 8. C[a, b] = {f : f : [a, b] → R is a continuous function}, the set of all the continuous functions defined on [a, b]. 9. V = {f : f : R → Ris a function}, the set of all the functions defined on R. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/11 8/30/2020 9.2 Vector Spaces We use B ⊂ A to denote that A contains B or B is a subset of A, see Figure 9.1 N ⊂ Z ⊂ Q ⊂ R. . It is easy to see that fo rD is tri bu tio n Figure 9.1: Subset A set is a collection of objects in which there may be no operations. If we introduce suitable operations in a set, then such a set is more useful. For example, we introduce addition + and scalar multiplication ⋅ together with dot product into Rn . We use the addition + and scalar multiplication ⋅ in Rn to study the linear combination of vectors, spanning space, linear independence, bases, linear transformations, and eigenvalues. Recall some sets in which we introduce suitable addition and scalar multiplication. → → 1. → a + 2. → k ⋅ a = (ka1 , b = (a1 + b1 , → N ot 1. In Rn : for a = (a1 , ⋯ , an ) and multiplication ⋅ are defined by b = (b1 , ⋯ , bn ), and k ∈ R, the addition + and scalar ⋯ , an + b1 ), ⋯ , kan ). 2. For m × n matrices, the addition + and scalar multiplication ⋅ are defined as follows. Let A = (aij ), B = (bij ), and k ∈ R, 1. A + B = (aij + bij ), 2. k ⋅ A = (kaij ). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/11 8/30/2020 9.2 Vector Spaces 3. In the set of all the functions defined on R: V = {f : f : R → R is a function}, fo rD is tri bu tio n the addition and scalar multiplication in V are defined as follows: 1. For f, g ∈ V , (f + g)(x) = f(x) + g(x) and x ∈ R . 2. For k ∈ R and f ∈ V , (k ⋅ f) = kf(x) for x ∈ R . Let V be a nonempty set with addition + and scalar multiplication ⋅. Each element in V is called a vector. We denote a vector by small letters in bold like u. When V Definition 9.2.2. n = R , then u = → u . Let V be a nonempty set with addition + and scalar multiplication ⋅. . The set V is called a vector space if the addition and scalar multiplication satisfy the following ten axioms: 1. If u, v ∈ V, then u + v [closure under addition] ∈ V u + v = v + u [commutative law of vector addition] 3. u + (v + w) = (u + v) + w N ot 2. 4. There is a vector 0 for each u 5. If u ∈ V, ∈ V ∈ V such that 0 + u [associative law of vector addition] [0 is called the additive identity] = u + 0 = u . there is a −u ∈ V, such that u + (−u) = 0 . [−u ∈ V is called the additive https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/11 8/30/2020 9.2 Vector Spaces inverse of u] ∈ R and u ∈ V, then ku 7. k(u + v) = ku + kv 8. (k + l)u = ku + lu 9. k(lu) = (kl)u ∈ V . [closure under scalar multiplication] [first distributive law] fo rD is tri bu tio n 6. If k [second distributive law] [associative law of scalar multiplication] 10. [the scalar 1 is called a 1u = u multiplicative identity] Remark 9.2.1. When we say that V is a vector space, we mean: N ot 1. V is a set. 2. In V, there are two operations: addition and scalar multiplication. 3. The two operations satisfy the ten axioms. Remark 9.2.2. Let V be a vector space, u ∈ V , and k ∈ R. Then a. 0u = 0; b. k0 = 0; https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/11 8/30/2020 9.2 Vector Spaces c. ku = 0, then k = 0 or u = 0 . Now, we provide several examples of vector spaces. The first one is the n-dimensional Euclidean space. Example 9.2.2. Solution Let → u = (u1 , u2 , n ⋯ , un ) ∈ R , fo rD is tri bu tio n The n-dimensional Euclidean space Rn is a vector space with the standard addition + and scalar multiplication ⋅ given in Definition 1.1.5 . → υ = (υ1 , υ2 , n ⋯ , υn ) ∈ R → → u + υ = (u1 + υ1 , u2 + υ2 , and → k u = (ku1 , ku2 , and k ∈ R. Then n ⋯ , un + υn ) ∈ R n ⋯ , kun ) ∈ R . Example 9.2.3. Let N ot Hence, the axioms 1 and 6 hold. It is easy to verify that the other eight axioms given in Definition 9.2.2 hold. Hence, Rn is a vector space with the standard operations: + and ⋅. V = {M22 : M22 is a 2 × 2 matrix}. Show that V is a vector space with the standard matrix addition and matrix scalar multiplication given in Definition 2.1.4 . Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/11 8/30/2020 9.2 Vector Spaces u = ( u11 u12 u21 u22 ) ∈ V, v = ( υ11 υ12 υ21 υ22 We verify that V satisfies the ten axioms given in Definition 9.2.2 2. u + v = ( u11 + υ11 u12 + υ12 u21 + υ21 u22 + υ22 and k, l ∈ R. . ) ∈ V. u11 + υ11 u12 + υ12 u21 + υ21 u22 + υ22 fo rD is tri bu tio n 1. u + v = ( ) ∈ V, ) = ( υ11 + υ11 υ12 + υ12 υ21 + υ21 υ22 + υ22 ) = v + u. 3. If w ∈ V , then u + (v + w) = (u + v) + w. 4. Let 0 = ( 0 0 0 0 ). Then 0 0 0+u = ( )+( 0 u+0 = ( 6. ku = ( −u11 −u12 −u21 −u22 ku11 ku12 ku21 ku22 u11 u12 u21 u22 ) ∈ V. ) ∈ V. k(u + v) = k ( 7. = ( u12 u21 u22 0 ) = ( 0 u11 + υ11 u12 + υ12 u21 + υ21 u22 + υ22 ku11 ku12 ku21 ku22 )+( ) = ( 0 )+( N ot 5. −u = ( 0 u11 0 ) = ( kυ11 kυ12 kυ21 kυ22 u11 u12 u21 u22 u11 u12 u21 u22 ) = u ) = u. ku11 + kυ11 ku12 + kυ12 ku21 + kυ21 ku22 + kυ22 ) ) = ku + kv. 8. (k + l)u = ku + lu. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/11 8/30/2020 9.2 Vector Spaces 9. k(lu) = (kl)u. 10. 1u = u. Hence, V is a vector space. , one can prove the following result. Example 9.2.4. Let V be the set of all m × n matrices, that is, fo rD is tri bu tio n By a method similar to Example 9.2.3 V = {Mmn : Mmn is an m × n matrix} with the standard matrix addition and matrix scalar multiplication given in Definition 2.1.4 vector space. Example 9.2.5. Let V = {f follows: : f : R → R is a function}. . Then V is a The addition and scalar multiplication in V are defined as N ot i. For f, g ∈ V , (f + g)(x) = f(x) + g(x) for each x ∈ R . ii. For k ∈ R and f ∈ V , (k ⋅ f)(x) = kf(x) for each x ∈ R . Then V is a vector space. Solution It is obvious that if f, g ∈ V, then f + g ∈ V , and if k ∈ R and f ∈ V, then kf ∈ V. Hence, the ˆ axioms (1) and (6) hold. Let ˆ 0 denote the zero function defined on R, that is, 0(x) = 0 for each x ∈ R. Then 0̂ + f = f + 0̂ = f for each f ∈ V and the axiom (4) holds. It is easy to see that other axioms hold. Example 9.2.6. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/11 8/30/2020 9.2 Vector Spaces In R2 , we introduce the following addition and scalar multiplication: 1. For → → → → u = (u1 , u2 ), υ = (υ1 , υ2 ), u + υ = (u1 + υ1 , u2 + υ2 ). 2. For k ∈ R and → → u = (u1 , u2 ), k ⋅ u = (ku1 , 0). Solution fo rD is tri bu tio n Then V is not a vector space under the above two operations. The addition given in (1) is the standard addition, but the scalar multiplication is not the standard scalar multiplication. The axiom (10) does not hold for the scalar multiplication given in (2). For example, let → u = (2, 3), then by the scalar multiplication given in (2), we have → → 1 ⋅ u = (1(2), 0) = (2, 0) ≠ (2, 3) = u . Hence, V is not a vector space under the two operations (1) and (2). N ot In a vector space V, there are two operations: addition and scalar multiplication, which satisfy 10 axioms. Hence, similar to Section 1.4 , we can introduce linear combinations, spanning spaces, and similar to Sections 7.1, 7.2 and 7.4, we can introduce linear independence, bases, coordinates. Also, we can introduce linear transformations from one vector space to another, and eigenvalues and eigenvectors of linear transformations. We can generalize the dot product of two vectors and the norm of a vector from Rn to vector spaces and study some related problems such as those in Sections 1.7, 1.8, and 8.5. We do not discuss these further here in this book. We only generalize the notion of subspaces in Rn in Section 10.1 to vector spaces. Definition 9.2.3. Let V be a vector space and let W ⊂ V be a nonempty subset of V. W is called a subspace of V if W is itself a vector space under the addition and multiplication defined on V. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/11 8/30/2020 9.2 Vector Spaces To show that W is a subspace of V, according to Definition 9.2.3 , we would verify that the addition and multiplication defined on V satisfy the ten axioms given in Definition 9.2.2 , where V is replaced by W. However, in the following, we show that it suffices to verify that the axioms (1) and (6) hold. Theorem 9.2.1. fo rD is tri bu tio n Checking the axioms (1) to (10), we see that each of the six axioms (2), (3), (7)-(10) only depends on the addition and multiplication while the other four axioms (1), (4), (5), and (6) depend on the operations and the set involved. Therefore, if V is a vector space and we consider a nonempty subset W of V under the same addition and multiplication as V, then the six axioms (2), (3), (7)-(10) hold for W because they only depend on addition and multiplication. Therefore, to verify that W is a subspace of V, we only need to verify that the addition and multiplication satisfy the four axioms (1), (4), (5), and (6). The following result shows that if the axioms (1) and (6) hold, then the axioms (4) and (5) hold. Hence, we only need to verify that the axioms (1) and (6) hold to show that W is a subspace of V. Let V be a vector space and W be a nonempty subset of V. Then W is a subspace of V if and only if the following conditions hold. Proof N ot a. if u, υ ∈ W, then u + υ ∈ W . b. if u ∈ W, k ∈ R, then ku ∈ W . It is sufficient to show that under (a) and (b), the axioms (4), (5) to W hold. Indeed, let k = 0 and u ∈ W . Because V is a vector space ku = 0u = 0. By (b), ku = 0 ∈ W . Therefore, 0 ∈ W . Because V is a vector space, 0 + u = u + 0 = u for each u ∈ W . Hence, the axiom (4) holds. By (b), −u = (−1)u ∈ W for each u ∈ W . Because V is a vector space, u + (−u) = (−u) + u = 0. Hence, the axiom (5) holds. In Theorem 9.2.1 , if V = Rn , then Theorem 9.2.1 is the same as Definition 9.1.1 . We refer to n Section 9.1 for the discussion about subspaces in R . In the following, we give examples of subspaces that https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/11 8/30/2020 9.2 Vector Spaces are not in Rn . Example 9.2.7. Let fo rD is tri bu tio n W = {A : A is an n × n symmetric matrix}. Show that W is a vector subspace of Mn×n , where Mn×n is the vector space of all n × n matrices. Solution By (2.3.3) , A is symmetric if and only if AT Theorem 2.1.1 , a. (A + B) b. (kA) T T T = A T = kA T +B = kA, = A + B, 1. Let N ot Exercises Let A, B ∈ W. Then A = AT and B = BT . By thus A + B ∈ V ; thus kA ∈ V . It follows that V is a subspace of Mn×n . = A. C(R) = {f : f : R → R is continuous on R}. Show that C(R) is a vector space under the following operations. 1. + : (f + g)(x) = f(x) + g(x) for x ∈ R . 2. (kf)(x) = k(f(x)) for x ∈ R . 2. Let https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/11 8/30/2020 9.2 Vector Spaces C[a, b] = {f : f : [a, b] → R is continuous}. Show that C[a, b] is a vector space under the standard addition and scalar multiplication: 1. For f, g ∈ C[a, b], (f + g)(x) = f(x) + g(x) for x ∈ [a, b] . 2. For k ∈ R and f ∈ C[a, b], (k ⋅ f)(x) = kf(x) for x ∈ [a, b] fo rD is tri bu tio n 3. Let n be a nonnegative integer and let n P n = {p : p(x) = a0 + a1 x + ⋯ + an x Show that Pn is a subspace of C(R). 4. Let V = {A : A is a polynomial of degree n}. is a 3 × 3 matrix}. We define the following addition and scalar multiplication. 1. Addition: For A, B ∈ V , A + B = AB, where AB is the standard product of A and B in Section 2.5 . 2. Scalar multiplication: For A ∈ V and k ∈ R, k ⋅ A = kA, where kA is the standard scalar multiplication given in Definition 2.1.4 . N ot Then V is not a vector space under the above addition and scalar multiplication. 5. Let C 1 [a, b] = {f : f ′ ∈ C[a, b]}, where f ′ (x) denotes the derivative of f at x. Show that C 1 [a, b] is a vector subspace of C[a, b]. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/11 8/30/2020 Chapter 10 Complex numbers 10.1 Complex numbers We want to solve the following quadratic equation (10.1.1) 2 aλ fo rD is tri bu tio n Chapter 10 Complex numbers + bλ + c = 0, where a, b, c are real numbers and a ≠ 0. Multiplying (10.1.1) 2 2 4a λ by 4a implies + 4abλ + 4ac = 0. N ot Applying completing the square to the above equation implies (10.1.2) (2aλ + b) If Δ := b2 − 4ac ≥ 0, 2 = b 2 − 4ac. then − − − − − − − 2 2aλ + b = √ b − 4ac − − − − − − − 2 or 2aλ + b = −√ b − 4ac . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/11 8/30/2020 Chapter 10 Complex numbers It follows that (10.1.3) 2 −b+√b −4ac and λ2 := 2a are all the real roots of the quadratic equation (10.1.1) 2 −b−√b −4ac 2a . fo rD is tri bu tio n λ1 := If Δ < 0, then (10.1.1) has no real roots. Hence, when Δ < 0, we need to extend the set of real numbers R to a larger set denoted by C such that (10.1.1) still has roots in the larger set. What is the set C ? If Δ < 0, then −(b2 − 4ac) > 0 − − − − − − − − − − − 4ac) and √−(b2 (10.1.4) is a real number. We introduce a symbol − − − i = √−1. Then By (10.1.3) N ot − −− −− −− −− −− −− − − − − − − − − − − − − − − − − − − − − − − − − − − 2 2 2 √Δ = √ (−1)[−(b − 4ac)] = √−1√ −(b − 4ac) = i√ −(b − 4ac). , we obtain two roots of the quadratic equation (10.1.1) (10.1.5) 2 2 −b+i√−(b −4ac) λ1 = Hence, the quadratic equation (10.1.1) 2a : −b−i√−(b −4ac) and λ2 = 2a . always has two roots: λ1 and λ2 given in (10.1.5) . Example 10.1.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/11 8/30/2020 Chapter 10 Complex numbers Find the roots of 2 λ + 4λ + 13 = 0. Solution b = 4, c = 13, fo rD is tri bu tio n Because a = 1, − − − − − − − − − − −− − − − − − − − − − − − − − − − 2 2 √Δ = √ b − 4ac = √ 4 − 4 × 13 = √−36 = √−1√36 = 6i. By (10.1.5) By (10.1.5) , λ1 = −4+6i 2 = −2 + 3i and λ2 = −4−6i 2 = −2 − 3i . , we see that λ1 and λ2 can be written into the following form x + yi, where x and y are real numbers. If y ≠ 0, then x + yi is not a real number. Hence, we introduce Definition 10.1.1. The number of the form N ot z = x + yi is called a complex number, where x and y are real numbers and i is given in (10.1.4) . x is called the real part of z and is denoted by Re z = x and y is called the imaginary part of z and is denoted by Im z = y. − − − 2 i = √−1 is called the imaginary number and i = −1. We denote by C the set of all complex numbers. Example 10.1.2. Find Re(−1 + 2i) and Im(−1 − 2i) . Solution https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/11 8/30/2020 Chapter 10 Complex numbers By Definition 10.1.1 , we have Re(−1 + 2i = 1) and Im(−1 − 2i) = Im(−2 + i(−2)) = −2. Definition 10.1.2. Let z1 = x1 + y1 i and z2 = x2 + y2 i. Then z1 Example 10.1.3. Use the following information to find x and y. a. 3x + 4i = 6 + 2yi b. (x + y) + (x − y)i = 5 − i Solution Let z1 = x1 + y1 i and z2 if and only if x1 = x2 and y1 = y2 . , 3x = 6 and 4 = 2y. This implies that x = y = 2 . , we have x + y = 5 and x − y = −1. Solving the system, we obtain Operations on complex numbers Definition 10.1.3. = z2 N ot a. By Definition 10.1.2 b. By Definition 10.1.2 x = 2 and y = 3 . fo rD is tri bu tio n The following definition shows that two complex numbers are equal if and only if their corresponding real parts and imaginary parts are the same. = x2 + y2 i . i. (Addition) z1 + z2 = (x1 + x2 ) + (y1 + y2 )i . ii. (Subtraction) z1 − z2 = (x1 − x2 ) + (y1 − y2 )i . iii. (Multiplication) z1 ⋅ z2 = (x1 x2 − y2 y2 ) + (x1 y2 + y1 x2 )i . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/11 8/30/2020 Chapter 10 Complex numbers iv. (Division) If z2 ≠ 0, then z1 = ( x1 x2 + y1 y2 2 z2 x +y 2 2 )+( y2 x2 − x1 y2 2 x 2 2 +y 2 ) i. 2 Remark 10.1.1. fo rD is tri bu tio n The multiplication (iii) can be obtained by the following way. z 1 z 2 = (x1 + y1 i)(x2 + y2 i) = x1 (x2 + y2 i) + (y1 i)(x2 + y2 i) = x1 x2 + x1 y2 i + x2 y1 i + y1 y2 i 2 = x1 x2 + x1 y2 i + x2 y1 i − y1 y2 = (x1 x2 − y1 y2 ) + (x1 y2 + x2 y1 )i. Hence, we have the following three formulas. 1. (x + yi)(x − yi) = x2 2. (x + yi) 2 3. (x − yi) 2 2 = x 2 = x − (yi) 2 2 = x + 2x(yi) + (yi) − 2x(yi) + (yi) 2 2 2 +y . 2 = x 2 = x + 2xyi − y − 2xyi − y Remark 10.1.2. 2 2 2 = (x 2 = (x 2 − y ) + 2xyi. 2 − y ) − 2xyi. z1 z2 = N ot By using the above formula (1), the division (iv) can be obtained by the following way. x1 +y1 i x2 +y i = 2 = (x1 +y1 i)(x2 −y2 i) (x2 +y i)(x2 −y i) 2 2 1 2 2 2 2 x +y 2 x1 x2 −x1 y2 i+x2 y1 i−y1 y2 i 2 = 2 2 (x2 ) −(y i) 2 (x1 x2 +y y )+(x2 y −x1 y )i 1 = 2 x1 x2 +y y 1 2 2 2 2 x +y 2 + y x2 −x1 y 1 2 2 2 2 x +y 2 i. Example 10.1.4. Let z1 = 4 − 5i and z2 = −1 + 6i. Express the following complex numbers in the form x + yi . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/11 8/30/2020 Chapter 10 Complex numbers 1. z1 + z2 2. z1 − z2 3. 3z1 4. −z1 5. z1 z2 z 6. z 1 Solution fo rD is tri bu tio n 2 1. z1 + z2 = (4 − 5i) + (−1 + 6i) = (4 − 1) + (−5 + 6)i = 3 + i. 2. z1 − z2 = (4 − 5i) − (−1 + 6i) = (4 − (−1)) + (−5 − 6)i = 5 − 11i. 3. 3z1 = 3(4 − 5i) = 3 × 4 − 3 × 5i = 12 − 15i. 4. −z1 = −(4 − 5i) = −4 + 5i. z 1 z 2 = (4 − 5i)(−1 + 6i) = 4(−1 + 6i) − (5i)(−1 + 6i) = −4 + 24i + 5i − 30i z1 6. z2 = = 4−5i −1+6i = −34−19i 1+36 2 (4−5i)(−1−6i) (−1+6i)(−1−6i) = − Definition 10.1.4. Let z = x + yi. Then, 34 37 − 19 37 i. = −4 + 29i + 30 = 26 + 29i. = −4−24i+5i−30 2 2 (−1) −(6i) N ot 5. − − − − − − ∣ ∣ 2 2 z = √x + y ∣ ∣ is called the modulus of z or the absolute value of z. Example 10.1.5. Let z = 3 − 4i. Find |z|. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/11 8/30/2020 Chapter 10 Complex numbers Solution − − − − − − − − − − − ∣ ∣ − − − − − 2 2 √ ∣z∣ = 3 + (−4) = √9 + 16 = √25 = 5. ∣ ∣ The following result gives some properties of the modulus of complex numbers. Let z, z 1 , z 2 ∈ C. fo rD is tri bu tio n Proposition 10.1.1. Then the following assertions hold. 1. |z1 z2 |z1 ||z2 |. 2. |z n | = |z|n for n ∈ N. 3. If z2 ≠ 0, z1 then ∣∣ z 2 ∣ = ∣ |z1 | |z2 | . 4. |αz| = |α||z| for α ∈ R. 5. |z1 + z2 | ≤ |z1 | + |z2 |. Example 10.1.6. Find the modulus of (1 + i)(1 − 2i) . N ot i(1 − i)(2 + 3i) Solution ∣ ∣ (1+i)(1−2i) i(1−i)(2+3i) ∣ = ∣ |1+i||1−2i| |i||1−i||2+3i| = √2√5 (1)√2√13 = √65 13 . Definition 10.1.5. The complex number z = x − yi is called the conjugate of z = x + yi. z̄ reads “z bar.” Example 10.1.7. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/11 8/30/2020 Chapter 10 Complex numbers For each of the following complex numbers, find its conjugate. z 1 = 3 + 2i; z 2 = −4 − 2i; z 3 = i; z 4 = 4. Solution The conjugate numbers have the following properties. Proposition 10.1.2. Let z, z1 , and z2 be complex numbers. Then ¯¯¯ ¯¯¯¯ z1 ) = vii. ( z 2 viii. ¯¯ ¯ z 1 z2 . 2 ∣ ∣z ∣ ∣ = zz̄ . Proof We only prove (ii)and (viii). ii. Assume that z̄ = z. N ot i. z̄¯ = z. ii. z̄ = z if and only if z is a real number. iii. z̄ = −z if and only if z is a pure imaginary. ¯¯¯¯¯¯¯¯ iv. ¯ z¯ z¯1 − ¯¯ z¯2 . 1 + z 2 = ¯¯ ¯¯¯¯¯¯¯¯ v. ¯ z¯ z¯1 − ¯¯ z¯2 . 1 − z 2 = ¯¯ vi. ¯ z¯ ¯¯ ⋅¯¯ z¯ ¯ = ¯¯ z¯1 ⋅ ¯¯ z¯2 and ¯ αz ¯¯ ¯ = αz̄ for α ∈ R. 1 2 fo rD is tri bu tio n ¯¯ z¯1 = 3 − 2i, ¯¯ z¯2 = −4 + 2i, ¯¯ z¯3 = −i and ¯¯ z¯4 = 4. Let z = x + yi. Then z̄ = x − yi and x + yi = x − yi. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/11 8/30/2020 Chapter 10 Complex numbers This implies that 2yi = 0 and y = 0. Hence, z = x + 0i = x is a real number. Conversely, if z is a real number, that is, z = x + 0i, where x is a real number, then z̄ = x − 0i = x = z . viii. Let z = x + yi. Then z̄ = x − yi and 2 Example 10.1.8. Verify the following identities. ¯¯¯¯ ¯¯ = z − 7i 1. ¯ z̄¯¯ + 7i ¯¯¯ i+3z ¯¯¯¯ ¯ ¯¯ ¯ 2. ( i−3z¯ ) = i−3z i+3z Solution ¯¯¯¯ ¯¯ = z̄ ¯ = z − 7i. ¯ + ¯¯ 1. ¯ z̄¯¯ + 7i 7i ¯¯¯¯¯¯¯ ¯ ¯¯ ¯ 2. ( i+3z ) = ¯ i−3z ¯ ¯ ¯¯ ¯¯ ¯ ¯ ¯ i+3z ¯ ¯ ¯¯ ¯¯ ¯ ¯ i−3z ¯ ¯ ¯ ¯ ¯ ¯ ¯ i +3z = ¯ ¯ ¯ ī −¯ 3z ¯ = −i+3z −i−3z = Example 10.1.9. 3 + 2λ 2 +λ + 5λ + 6. Solution We prove i+3z 2 2 = x +y 2 = ∣ ∣z 2 ∣ ∣. . Show that if z is a root of p(λ), then z̄ is a root of p(λ),. N ot Let p(λ) = λ4 i−3z 2 − xyi + xyi − y i fo rD is tri bu tio n zz̄ = (x + yi)(x − yi) = x (10.1.6) ¯¯¯¯¯ ¯ p(z) = p(z̄ ) Indeed, by Proposition 10.1.2 for z ∈ C. , we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/11 8/30/2020 Chapter 10 Complex numbers ¯¯¯ ¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ ¯¯4 ¯ ¯¯¯¯¯ ¯ 4¯ 3 2 ¯¯¯¯ ¯ 3 ¯¯2 ¯ ¯¯ + 6̄ p(z) = z + 2z + z + 5z + 6 = z + 2z + z + ¯ 5z = (z̄ ) 4 + 2(z̄ ) If z is a root of p(λ), then p(z) = 0. By (10.1.6) 3 + (z̄ ) 2 + 5z̄ + 6 = p(z̄ ). , we have and z̄ is a root of p(λ). fo rD is tri bu tio n ¯¯¯¯¯ ¯ p(z̄ ) = p(z) = 0̄ = 0 Similar to Example 10.1.9 , we can show that if z ∈ C is a root of a polynomial of degree n, then its conjugate z̄ is also a root of this polynomial. This implies that the quadratic equation (10.1.1) has either of two real roots, which could possibly be the same or two different complex roots. Exercises 1. Solve each of the following equations. a. z 2 + z + 1 = 0 b. z 2 − 2z + 5 = 0 c. z 2 − 8z + 25 = 0 N ot 2. Use the following information to find x and y. a. 2x − 4i = 4 + 2yi b. (x − y) + (x + y)i = 2 − 4i 3. Let z1 = 2 + i and z2 = 1 − i. Express each of the following complex numbers in the form x + yi, calculate its modulus, and find its conjugate. 1. z1 + 2z2 2. 2z1 − 3z2 3. z1 z2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/11 8/30/2020 Chapter 10 Complex numbers 4. z1 z2 4. Express each of the following complex numbers in the form x + yi . b. c. d. ¯¯¯¯ ¯¯¯ ¯ i ( ) 1−i 2−i (1−i)(2+i) 1+i i(1+i)(2−i) 1+i 1−i 5. Find i2 , − 3 2−i 1+i 4 5 i , i , i , i 1−i 8 6. Let z = ( 1+i ) . 6 and compute (−i − i4 Calculate z 66 + 2z 7. Find z ∈ C a. iz = 4 + 3i b. (1 + i)z̄ = (1 − i)z 8. Find real numbers x, y such that 33 − 2. fo rD is tri bu tio n a. +i 5 6 −i ) 1000 . x + 1 + i(y − 3) = 1 + i. 5 + 3i (3+4i)(1+i) 5 i (2+4i) 2 12 . Find |z|. N ot 9. Let z = https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 11/11 8/30/2020 10.2 Polar and exponential forms 10.2 Polar and exponential forms 10.1 (I)). y in the xy-plane (see Figure N ot fo r D is tri b Figure 10.1: (I) () ut io n A complex number z = x + yi can be interpreted as a point (x, y) or a vector x When complex numbers are viewed in an xy-coordinate system, the x-axis is called the real axis, the y-axis is called the imaginary axis with unit i, and the plane is called the complex plane. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 1/10 8/30/2020 10.2 Polar and exponential forms Definition 10.2.1. Let z = x + yi and z ≠ 0. The angle θ from the positive x-axis to the vector () x y is called an argument of z and is denoted by (See Figure 10.1 ut io n θ = argz (II)). If θ satisfies D is tri b − π < θ ≤ π, then θ is called the principal argument of z and is denoted by θ = Arg z. The principal argument Arg z of z is unique and fo r (10.2.1) N ot argz = Argz + 2kπ for k ∈ ℤ, where ℤ = {0, ± 1, ± 2, ⋯} is the set of integers. Definition 10.2.1 does not give the definitions of an argument and a principal argument for z = 0. Hence, if z = 0, then z does not have an argument nor a principal argument. Theorem 10.2.1. Let x, y ∈ ℝ be real numbers. Then the following assertions hold. 1. If z = x and x > 0, then Argz = 0 and argz = 2kπ; 2. If z = x and x < 0, then Argz = π and argz = π + 2kπ; https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 2/10 8/30/2020 10.2 Polar and exponential forms π π 3. If z = yi and y > 0, then Argz = 2 and argz = 2 + 2kπ; and π π 4. If z = yi and y < 0, then Argz = − 2 and argz = − 2 + 2kπ, where k ∈ ℤ. Example 10.2.1. ut io n For each of the following numbers, find its principal argument and argument. D is tri b a. 1 b. 2i c. − 1 d. − 2i Solution By Theorem 10.2.1 , we have a. Arg1 = 0 and arg1 = 2kπ for k ∈ ℤ. π π π π fo r b. Arg(2i) = 2 and arg(2i) = 2 + 2kπ for k ∈ ℤ. c. Arg( − 1) = π and arg( − 1) = π + 2kπ for k ∈ ℤ. N ot d. Arg( − 2i) = − 2 and arg( − 2i) = − 2 + 2kπ for k ∈ ℤ. The following result gives the formulas of the principal arguments of complex numbers (see Figure 10.2 (I), (II), (III), (IV)). Figure 10.2: (I), (II), (III), (IV) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 3/10 10.2 Polar and exponential forms ut io n 8/30/2020 ( ) π Assume that z = x + yi, x ≠ 0, y ≠ 0, and ψ ∈ 0, 2 satisfy (10.2.2) || y . x fo r tanψ = D is tri b Theorem 10.2.2. N ot Then the following assertions hold. 1. If z is in the first quadrant, then Argz = ψ and argz = ψ + 2kπ. 2. If z is in the second quadrant, then Argz = π − ψ and argz = π − ψ + 2kπ. 3. If z is in the third quadrant, then Argz = − π − ψ and argz = − π − ψ + 2kπ. 4. If z is in the fourth quadrant, then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 4/10 8/30/2020 10.2 Polar and exponential forms Argz = − ψ and argz = − ψ + 2kπ, where k ∈ ℤ. ( ) π (10.2.3) || y if and only if ψ = arctan x . D is tri b || y tanψ = x Example 10.2.2. Find Arg z and arg z for each of the following numbers. a. z = 3 + √3i fo r b. z = − 2 − 2i. ( ) N ot Solution a. Let z = x + yi = 3 + √3i. Then x = 3 and y = π √3. Let ψ ∈ 0, 2 tanψ = π ut io n Note that if ψ ∈ 0, 2 , then || y x = √3 3 . Then ψ = 6 . Because x > 0 and y > 0, z is in the first quadrant and by Theorem 10.2.2 (1), https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 5/10 8/30/2020 10.2 Polar and exponential forms π π Argz = ψ = 6 and argz = 6 + 2kπ for k ∈ ℤ. ( ) π b. Let z = x + yi = − 2 − 2i. Then x = − 2 and y = − 2. Let ψ ∈ 0, 2 | | −2 = 1. −2 ut io n tanψ = satisfy π π 3 (3), D is tri b Then ψ = 4 . Because x < 0 and y < 0, z is in the third quadrant and by Theorem 10.2.2 3 Argz = − π + ψ = − π + 4 = − 4 π and argz = − 4 π + 2kπ for k ∈ ℤ. The symbol e iθ, or exp(iθ), is defined by means of Euler’s formula as fo r (10.2.4) where θ is measured in radians. Proposition 10.2.1. N ot cosθ + isinθ = e iθ, 1. e iθ = 1 if and only if θ = 2kπ for k ∈ ℤ. 2. e iθ = e iψ if and only if θ = ψ + 2kπ for k ∈ ℤ. Proof https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 6/10 8/30/2020 10.2 Polar and exponential forms 1. Assume that e iθ = 1. Then cosθ + isinθ = 1. This implies that cosθ = 1 and sinθ = 0. Hence, k ∈ ℤ. Conversely, if θ = 2kπ for k ∈ ℤ, then e iθ = e i ( 2kπ ) = cos2kπ + isin2kπ = 1. Theorem 10.2.3. Let z = x + yi and let θ be an argument of z. Then | z | (cosθ + isinθ) = | z | e iθ. D is tri b z= ut io n 2. It is easy to see that e iθ = e iψ if and only if e i ( θ − ψ ) = 1 if and only if θ − ψ = 2kπ for k ∈ ℤ. Proof Because x = | x | cosθ and y = | z | sinθ, fo r z = x + yi = | z | cosθ + (rsinθ)i = | z | (cosθ + isinθ). The second equality follows from Euler’s formula (10.2.4) . (10.2.5) N ot For each nonzero complex number z = x + yi, there exist a unique number r = | z | > 0 and arguments θ = Argz + 2kπ for k ∈ ℤ. We call (r, θ) the polar coordinates of the complex number z = x + yi or the point (x, y). Conversely, for a polar coordinate (r, θ), we define a complex number z by z = (r cosθ) + (r sinθ)i. Then |z | = r and θ is an argument of z. Indeed, https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 7/10 8/30/2020 10.2 Polar and exponential forms || z = √r 2cos 2θ + r 2sin 2θ = √r 2(cos 2θ + sin 2θ) = √r 2 = r, where we use the identity: cos 2θ + sin 2θ = 1. Let z be a complex number. The following form (10.2.6) is called the polar form of z and the following form (10.2.7) is called the exponential form of z. N ot It is obvious that if r = 1, then fo r z = re iθ D is tri b z = r(cosθ + isinθ) ut io n Definition 10.2.2. z = cosθ + isinθ = e iθ. Such a complex number z is called a unit complex number. By Theorem 10.2.3 , we see that the polar form and the exponential form of a complex number z can be converted into one another easily. Moreover, by (10.2.5) , we see that a polar form of a complex number can be easily transferred into the standard form. However, it is not trivial to transfer the standard form of a complex number z = x + yi into a polar form of z. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 8/10 8/30/2020 10.2 Polar and exponential forms The following are the steps for finding a polar form of a complex number z = x + yi. 1. Step 1. Find |z|. 2. Step 2. Find Arg z. 3. Step 3. Write z as z = r(cosθ + isinθ), where θ = Argz. ut io n In Steps 2 and 3, the principal argument Arg z can be replaced by the argument arg z of z. We always use Arg z instead of arg z if there are no indications. Example 10.2.3. D is tri b Express z = 1 − √3i in polar and exponential forms. Solution || z = √12 + ( − √3) ( ) π 2. Step 2. Let ψ ∈ 0, 2 2 = √4 = 2. such that tanψ = | | √−3 = π √3. Then ψ = 3 . Because x = 1 > 0 and fo r 1. Step 1. r = 1 N ot y = − √3 < 0, z is in the second quadrant and by Theorem 10.2.2 Argz = π − ψ = π − π 3 = 2π 3 (2), . ( () ( ) 2π 3. Step 3. z = 2 cos 3 2π + isin 3 . Hence, the polar form of 1 − √3i is ( 1 − √3i = 2 cos 2π 2π + isin 3 3 ) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f… 9/10 8/30/2020 10.2 Polar and exponential forms 2π and its exponential form is 1 − √3i = 2e i 3 . Exercises 1 ut io n 1. For each of the following numbers, find its principal argument and argument. a. − 4 b. − i √3 c. 2 − 2 i d. 1 + √3i D is tri b e. − 1 + √3i f. − √3 − i f. − 1 − i N ot fo r 2. Express each of the following complex numbers in polar and exponent forms. a. 1 b. i c. − 2 d. − 3i e. 1 + √3i https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml… 10/10 8/30/2020 10.3 Products in polar and exponential forms 10.3 Products in polar and exponential forms Let z1 and z2 = r1 (cos θ 1 + i sin θ 1 ) = r2 (cos θ 2 + i sin θ 2 ). Then z 1 z 2 = r1 r2 [cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 )] = r1 r2 e ≠ 0, then z1 = z2 r1 r2 i(θ 1 +θ 2 ) D is tri b and if z2 ut io n Theorem 10.3.1. [cos(θ 1 − θ 2 ) + i sin(θ 1 − θ 2 )] = Proof r1 e i(θ 1 −θ 2 ) . r2 fo r z 1 z 2 = [r1 (cos θ 1 + i sin θ 1 )][r2 (cos θ 2 + i sin θ 2 )] = r1 r2 (cos θ 1 + i sin θ 1 )(cos θ 2 + i sin θ 2 ) r1 r2 [(cos θ 1 cos θ 2 − sin θ 1 sin θ 2 ) + i(sin θ 1 cos θ 2 + cos θ 1 sin θ 2 )] N ot = = r1 r2 [cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 )] = r1 r2 e z1 z2 = i(θ 1 +θ 2 ) r1 e r2 e . iθ 1 iθ 2 = r1 e iθ 1 r2 e iθ 2 = r1 e r2 iθ 1 e iθ 2 e iθ2 e iθ 2 = r1 e r2 i(θ 1 −θ 2 ) e i0 = r1 e i(θ 1 −θ 2 ) . r2 Example 10.3.1. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/8 8/30/2020 10.3 Products in polar and exponential forms Let z1 = 4 (cos π 4 + i sin π 4 and z2 ) = 2 (cos π 6 π + i sin 6 ). Express z1 z2 and z1 in polar, z2 exponential, and standard forms. Solution i + 4 π 6 ) + i sin ( π 8 [(cos π 4 π cos √2 √3 2 2 z2 z1 z2 2 [cos ( = 2e i − sin 6 − √2 1 2 2 π 4 sin ) + i( π 6 ) + i (sin √2 √3 2 2 + 2 z2 + i sin 5π 12 ). 6 ) + i sin ( π 4 − π 6 π 4 π 4 − π 6 cos √2 √3 2 2 + ) + i sin ( π + sin 6 √2 1 2 2 π 4 π − 4 sin π 6 π 6 π 4 √2 1 2 2 cos 6 + cos )] )] = 2 [cos )] π π 4 sin π 6 ) + i (sin π 4 π 12 cos π 6 + i sin − cos π 4 π 12 )] ]. sin π 6 )] ) + i( √2 √3 2 2 − √2 1 2 2 )] – – [(√3 + 1) + i(√3 − 1)]. , we obtain the following result on the arguments of products and quotients. Theorem 10.3.2. Let z1 , 12 N ot By Theorem 10.3.1 √2 π − 4 . 2 [(cos = 2 [( = π π 12 = 2 [cos ( = 5π fo r z1 = )] = 8 (cos . – – – = 2√2[(√3 − 1) + i(√3 + 1)]. z2 6 5π 12 = 8 [( z1 π + 4 ut io n z 1 z 2 = 8e z1 z2 = π (4)(2) [cos ( D is tri b z1 z2 = be two nonzero complex numbers. Then the following assertions hold. 1. arg(z1 z2 ) = arg z1 2. arg ( z 2 z 1 + arg z 2 = Arg z 1 + Arg z 2 + 2kπ ) = arg z 1 − arg z 2 = Arg z 1 − Arg z 2 + 2kπ for ∈ Z. for k ∈ Z. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/8 8/30/2020 10.3 Products in polar and exponential forms Proof We only prove (1). Let θ1 = arg z1 for n ∈ Z. By Theorem 10.3.1 , arg(z 1 z 2 ) = = Arg z 1 + 2mπ for m ∈ Z and θ2 θ 1 + θ 2 = arg z 1 + arg z 2 = (Arg z 1 + 2mπ) + (Arg z 2 + 2mπ) (Arg z 1 + Arg z 2 ) + 2(m + n)π = (Arg z 1 + Arg z 2 ) + 2kπ, where k = m + n. Because m, n can be any values in Z, k can be any integer in Z. Hence, = 1+i and z2 Find arg(z1 z2 ), = −1 − i. arg ( Solution π Arg z 1 = Arg(1 + i) = 4 arg ( z1 z2 z2 z1 ), arg ( z2 z1 ), and arg 1 z1 π . 3π = − 4 , 4 that arg(z 1 z 2 ) = ( arg ( z2 and Arg z 2 = Arg(−1 − i) = −π + N ot it follows from Theorem 10.3.2 z1 fo r Because for k ∈ Z. D is tri b arg(z 1 z 2 ) = (Arg z 1 + Arg z 2 ) + 2kπ Example 10.3.2. ut io n = Let z1 = arg z 2 = Arg z 2 + 2nπ ) = π 4 π 4 ) = − − 3π 4 − (− 3π 4 − ) + 2kπ = − 3π 4 π 4 π 2 + 2kπ for k ∈ Z, ) + 2kπ = π + 2kπ for k ∈ Z, + 2kπ = −π + 2kπ for k ∈ Z, https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/8 8/30/2020 10.3 Products in polar and exponential forms and arg In Theorem 10.3.2 1 z1 = − π 4 + 2kπ for k ∈ Z. , the arguments corresponding to k = 0 may not be the principal arguments if they are not in the interval (−π, π]. z1 Therefore, to find Arg(z1 z2 ) or Arg ( z ), we need to find a suitable integer k0 ∈ Z such that the corresponding argument belongs to (−π, π], ut io n 2 which is the principal argument. Corollary 10.3.1. z2 be two nonzero complex numbers. Then the following assertions hold. 1. Arg(z1 z2 ) = Arg z 1 + Arg z 2 + 2k0 π, D is tri b Let z1 , where k0 such that ∈ Z −π < Arg z 1 + Arg z 2 + 2k0 π ≤ π. z1 2. Arg ( z 2 ) = Arg z 1 − Arg z 2 + 2k0 π, where k0 ∈ Z such that fo r −π < Arg z 1 − Arg z 2 + 2k0 π ≤ π. Let z1 , z2 N ot Example 10.3.3. be the same as in Example 10.3.2 Solution By Example 10.3.2 z1 . Find Arg ( z ), Arg ( 2 z2 z1 ), , we obtain arg ( z1 z2 ) = π + 2kπ for k ∈ Z. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/8 8/30/2020 10.3 Products in polar and exponential forms Only when k = 0, the above angle belongs to (−π, Arg ( z1 π]. Hence, ) = π. z2 arg ( z2 z1 When k = 1, the above angle is in (−π, ) = −π + 2kπ π]. for k ∈ Z, z2 Hence, Arg ( z ) = π . D is tri b 1 Example 10.3.4. Let z1 = −1 and z2 = i. Use Corollary 10.3.1 to find Arg(z1 z2 ) Solution , we have arg(z 1 z 2 ) = π + 3π 2 ∉ (−π, π], so 3π 2 2 + 2kπ = 3π 2 + 2kπ for k ∈ Z. is not the principal argument of z1 z2 .. When k = −1, we have 3π 2 N ot Because π fo r By Theorem 10.3.2 ut io n Similarly, we have + 2kπ = 3π − 2π = − 2 π ∈ (−π, π]. 2 Hence, Arg(z1 z2 ) = − π2 . Theorem 10.3.3 (Demoivre’s formulas) 1. Let z = r(cos θ + i sin θ) and n ∈ N. Then https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 5/8 8/30/2020 10.3 Products in polar and exponential forms z n n = r (cos nθ + i sin nθ). 2. Let z = reiθ and n ∈ N. Then z n n = r e i(nθ) z −n −n = r e i(−nθ) . Proof z 2 3 = z ⋅ z = (r ⋅ r)[cos(2θ + θ) + i sin(2θ + θ)] = r (cos(3θ) + i sin(3θ)). and 3 = z 2 2 = z ⋅ z = (r 3 ⋅ r)[cos(2θ + θ) + i sin(2θ + θ)] = r (cos(3θ) + i sin(3θ)). fo r z Repeating the process implies that = 1 z n = [r(cos θ + i sin θ)] , z −1 1 = z In Theorem 10.3.3 n n = r N ot z Because z −1 , we have D is tri b We only prove (i) because (ii) follows from (i). By Theorem 10.3.1 ut io n and if z ≠ 0, then n 1 = n r e i(nθ) (cos nθ + i sin nθ). −n = r e i(−nθ) . , if r = 1, then we obtain the following formulas. Corollary 10.3.2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 6/8 8/30/2020 10.3 Products in polar and exponential forms 1. (cos θ + i sin θ) 2. (eiθ ) n = e i(nθ) n = cos nθ + i sin nθ for n ∈ N . for n ∈ N . Example 10.3.5. 4 ut io n Express (−2 − 2i) in x + yi form. Solution 1. Step 1. Find the polar form of z = −2 − 2i . Let ψ ∈ (0, π 2 ) D is tri b − −−−−−− −−− − – ∣ ∣ 2 2 – √ r = z = (−2) + (−2) = √8 = 2√2. ∣ ∣ be such that − − − ∣ √−2 ∣ tan ψ = ∣ ∣ = 1. ∣ −2 ∣ 4 . Because x = −2 < 0 and y = −2 < 0, Theorem 10.2.2 fo r π (3), N ot Then ψ = Arg z = −π + ψ = −π + Hence, z π 4 is in the third quadrant and by = − 3π . 4 3π 3π – −2 − 2i = 2√2 [(cos (− ) + i sin (− )] . 4 4 2. Step 2. Use Theorem 10.3.3 to find the polar form. By Theorem 10.3.3 , we have https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 7/8 8/30/2020 10.3 Products in polar and exponential forms (−2 − 2i) 4 – = {2√2 [cos (− 3π 4 ) + i sin (− – 4 = (2√2) [cos 4 (− 3π 4 3π 4 4 )]} ) + i sin 4 (− 3π 4 )] = 64[cos(−3π) + i sin(−3π)] = 64[cos(3π) + i0] = 64 cos π = −64. ut io n Exercises 1. Express the following numbers in x + yi form. 1. 3 – (1 + √3i) 2. (−1 − i) 3. (1 + i) = 3. Let z = 1+i √2 √2 2 −8 D is tri b 2. Let z1 4 . and z2 (1 − i). – = √3 − i. Find z 100 Find the exponential form of z1 z2 . +z 50 + 1. 1 z = (− + 2 600 i) 1 + (− − 2 – √3 60 i) . 2 N ot 2 – √3 fo r 4. Compute the following value https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 8/8 8/30/2020 10.4 Roots of complex numbers 10.4 Roots of complex numbers fo rD is tri bu tio n Given a complex number z, and an integer n ≥ 2, we want to find a complex number w ∈ C such that (10.4.1) n w If w ∈ C satisfies (10.4.1) , we write (10.4.2) = z. 1 w = z 1 z n n n = √z . is called the nth root of z. It is obvious that the nth root of z is a root of equation (10.4.1) Theorem 10.4.1. . N ot Let z = r(cos θ + i sin θ) Then z has n nth roots in exponential form: (10.4.3) (θ+2k)π n n √z = √re n for k = 0, 1, 2, ⋯ , n − 1, or in polar form https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/8 8/30/2020 10.4 Roots of complex numbers (10.4.4) n n √z = √r (cos θ+2kπ + i sin n θ+2kπ n ) for k = 0, 1, 2, ⋯ , n − 1, Proof n ρ e By Proposition 10.2.1 i(nψ) (2), we obtain ρ n = r n = (ρe θ+2kπ iψ n ) = re iθ . for k ∈ Z. n N ot ψ = (10.4.5) it follows from Theorem 10.3.3 and nψ = θ + 2kπ for k ∈ Z. This implies that ρ = √r and Hence, = z, fo rD is tri bu tio n Let w = ρeiψ . Because z = r(cos θ + i sin θ) = reiθ and wn (2) that θ+2kπ n w = √re Note that for each k ∈ Z, there exist j ∈ Z and l ∈ {0, n for k ∈ Z. 1, 2, ⋯ , n − 1} such that k = jn + l. This, together with (10.4.5) and Proposition 10.2.1 (1), implies that https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/8 8/30/2020 10.4 Roots of complex numbers θ+2(jn+l)π θ+2kπ n w = √re n θ+2l+2jn)π n = √re n = √re n (θ+2l)π n n = √re n e i(2j) (θ+2l)π n = √re . n Replacing l by k, we obtain for k ∈ {0, 1, 2, ⋯ , n − 1}, (θ+2k)π Because the right side of (10.4.4) n θ + 2kπ + i sin θ + 2kπ n ). n depends on k, we can write n (√z )k = √r (cos By Theorem 10.4.1 n = √r (cos n fo rD is tri bu tio n n w = √re θ+2kπ n + i sin , we see that equation (10.4.1) n ) n n √z = {(√z )0 , (√z )1 , N ot Corollary 10.4.1. Let z = r(cos θ + i sin θ). Then (√z ) = √r (cos 0 If z is a real number, then by Corollary 10.4.1 √z = √r (cos θ 2 θ 2 for k = 0, 1, 2, ⋯ , n − 1. has n distinct roots given by (10.4.4) (10.4.6) n θ+2kπ + i sin θ 2 . We write n ⋯ , (√z ) }. n−1 ) and (√z ) = −(√z ) . 1 0 , + i sin θ 2 ) or √z = −√r (cos θ 2 + i sin θ 2 ). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/8 8/30/2020 10.4 Roots of complex numbers It means that the square root of a complex number has a positive root (√z)0 and a negative root (√z)1 . Hence, when we consider complex numbers, the square root √z represents two roots, which are (10.4.7) Example 10.4.1. − − − Find √−1. Solution and √z = −(√z ) . 0 fo rD is tri bu tio n √z = (√z )0 Because −1 = cos π + i sin π. It follows from Corollary 10.4.1 − − − (√−1) − − − Therefore, √−1 = {i, − i} 0 = cos π 2 + i sin . π 2 = i − − − that − − − and (√−1) 1 = −i. − − − From Example 10.4.1 , we see that in (10.1.4) , i = √−1 can be interpreted as the positive root of √−1, that is, i is the positive root of the following equation 2 Example 10.4.2. Solve z 2 – − 1 − √3i = 0 for z. = −1. N ot w Solution z 2 – − 1 − √3i = 0 − −−−− − – if and only if z = √1 + √3i . Because https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/8 8/30/2020 10.4 Roots of complex numbers π π – 1 + √3i = 2 (cos + i sin ), 3 3 z0 that − −−−− − – := (√ 1 + √3i ) 0 – – – √3 √6 √2 π π 1 – – = √2 (cos + i sin ) = √2 ( + i) = + i 6 6 2 2 2 2 fo rD is tri bu tio n it follows from Theorem 10.4.1 and − −−−− − – z 1 : = (√ 1 + √3i ) = − √6 2 √2 − 2 1 – = √2 [cos ( i. 6 + π) + i sin ( – − 1 − √3i = 0 Hence, z0 and z1 are the solutions of z 2 Example 10.4.3. − − − Find all √−1. 3 π Solution − − 3 − (√−1) 1 2 = cos N ot − − 3 − (√−1) 0 = cos = cos = cos − − − Therefore, √−1 = { 12 3 +i √3 2 , − 1, π 3 + i sin π+2π 3 π+4π 3 π 3 1 2 3 = + i sin + i sin − i sin , π −i π 3 √3 2 2 +i π+2π 3 π+4π 3 = } 1 1 2 6 – + π)] = √2 (− √3 2 − 1 2 i) . Because −1 = cos π + i sin π. It follows from Theorem 10.4.1 − − 3 − (√−1) π √3 2 that . = cos π + i sin π = −1. = cos (2π − −i √3 2 π 3 ) + i sin (2π − π 3 ) . . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 5/8 8/30/2020 10.4 Roots of complex numbers Theorem 10.4.2. Let a, b, c ∈ C be complex numbers and a ≠ 0. Then the solution of the following quadratic equation (10.4.8) is (10.4.9) z = 2 + bz + c = 0 fo rD is tri bu tio n az − − − − − − − 2 −b + √ b − 4ac . 2a Proof The proof is the same as (10.1.2) and is omitted. − − − − − − − − 4ac Example 10.4.4. Solve the following equation contains two values. Hence, N ot Note that if b2 − 4ac ≠ 0, then by (10.4.7) , we see that √b2 (10.4.9) is consistent with (10.1.3) or (10.1.5) . (1 − i)z 2 − 2z + (1 + i) = 0. Solution By (10.4.9) , the solution of the equation is (10.4.10) https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 6/8 8/30/2020 10.4 Roots of complex numbers − − − − − − − − − − − − − − − 2 + √ 4 − 4(1 − i)(1 + i) z = = − − − 2 + √−4 2(1 − i) − − − 2(1 − i) − − − − − − , √−1 = i or √−1 = −i. If √−1 = i, by (10.4.10) z0 = − − − If √−1 = −i. by (10.4.10) , we have 1+i 1−i (1 + i) = = (1 − i)(1 + i) z1 = Hence, z0 = i and z1 = 1 1−i 2i , we have = i. 2 = 1. 1−i are the solutions of (1 − i)z 2 Exercises 2 . (1 − i) fo rD is tri bu tio n By Example 10.4.1 = − − − 1 + √−1 − 2z + (1 + i) = 0 . 1. Find all roots for each of the following complex numbers. – a. √1 − − − b. √−1 − − − − c. √1 − i − − d. √−i 4 6 3 e. – (−1 + √3i) 1 4 N ot 3 2. Find all solutions of z 3 = −8 . 3. Solve the following equations. a. (1 − i)z 2 + 2z + (1 + i) = 0 b. z 2 + 2z + (1 − i) = 0 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 7/8 10.4 Roots of complex numbers N ot fo rD is tri bu tio n 8/30/2020 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 8/8 8/30/2020 Appendix A Solutions to the Problems N ot fo r D is tri b ut io n Appendix A Solutions to the Problems https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/1 8/30/2020 A.1 Euclidean spaces A.1 Euclidean spaces Section 1.1 D is tri b ut io n 1. Write a three-column zero vector, and a three-column nonzero vector, and a five-column zero vector and a five-column nonzero vector. 2. Suppose that the buyer for a manufacturing plant must order different quantities of oil, paper, steel, and plastics. He will order 40 units of oil, 50 units of paper, 80 units of steel, and 20 units of plastics. Write the quantities in a single vector. 3. Suppose that a student’s course marks for quiz 1, quiz 2, test 1, test 2, and the final exam are 70, 85, 80, 75, and 90, respectively. Write his marks as a column vector. ( ) () x − 2y 2z 5. → a= () 6. → a= ( ) |x| y2 → and b = x−y 4 → → − 2 . Find all x, y, z ∈ ℝ that → a = b. 1 () ( ) and b = 1 4 fo r → and b = → . Find all x, y ∈ ℝ such that → a = b. 2 x+y N ot 2x − y 4. Let → a= 2 → . Find all x, y ∈ ℝ such that → a − b is a nonzero vector. → → 7. Let P(1, x ), Q(3, 4), P 1(4, 5), and Q 1(6, 1). Find all possible values of x ∈ ℝ such that PQ = P 1Q 1. 8. A company with 553 employees lists each employee’s salary as a component of a vector → a in ℝ 553. If a 6% salary increase has been approved, find the vector involving → a that gives all the new salaries. 2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/4 8/30/2020 A.1 Euclidean spaces () 110 9. Let → a= 88 denote the current prices of three items at a store. Suppose that the store announces a sale so that the price of each item is reduced by 20%. a. Find a three-vector that gives the price changes for the three items. b. Find a three-vector that gives the new prices of the three items. () () () 10. Let → a= 2 → 2 , b= 3 3 −2 , → c = 0 0 , α ∈ ℝ. Compute −1 → i. 2→ a − b + 5→ c; → → ii. 4 a + αb − 2→ c ( ) ( ) () 11. Find x, y, and z such that 4y 2z 3x + 8 −6 0 = fo r 9 D is tri b −1 ut io n 40 0 . 0 N ot → 12. Let → u = (1, − 1, 0, 2), → v = ( − 2, − 1, 1, − 4), and w = (3, 2, − 1, 0). → a. Find a vector → a ∈ ℝ 4 such that 2→ u − 3→ v−→ a = w. b. Find a vector → a 1 → → [2 u − 3→ v+→ a] = 2w +→ u − 2→ a. 2 → 13. Determine whether → a ǁ b, → 1. → a = (1, 2, 3) and b = ( − 2, − 4, − 6). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/4 8/30/2020 A.1 Euclidean spaces → 2. → a = ( − 1, 0, 1, 2) and b = ( − 3, 0, 3, 6). → 3. → a = (1, − 1, 1, 2) and b = ( − 2, 2, 2, 3). () () () 16. Let → a= 2 2 → , b= 3 −2 , → c = 0 3 0 , and α ∈ ℝ. −1 D is tri b −1 ut io n 14. Let x, y, a, b ∈ ℝ with x ≠ y. Let P(x, 2x), Q(y, 2y), P 1(a, 2a), and Q 1(b, 2b) be points in ℝ 2. Show → → that PQ ǁ P 1Q 1. → → 2 2 15. Let P(x, 0), Q(3, x ), P 1(2x, 1), and Q 1(6, x ). Find all possible values of x ∈ ℝ such that PQ ǁ P 1Q 1. Compute → a. → a ⋅ b; → → b. a ⋅ c; → c. b ⋅ → c; → fo r d. → a ⋅ (b + → c). N ot → 17. Let → u = (1, − 1, 0, 2), → v = ( − 2, − 1, 1, − 4), and w = (3, 2, − 1, 0). → → → → → → Compute u ⋅ v, u ⋅ w, and w ⋅ v. 18. Assume that the percentages for homework, test 1, test 2, and the final exam for a course are 10%, 25%, 25%, and 40%, respectively. The total marks for homework, test 1, test 2, and the final exam are 10, 50, 50, and 90, respectively. A student’s corresponding marks are 8, 46, 48, and 81, respectively. What is the student’s final mark out of 100? 19. Assume that the percentages for homework, test 1, test 2, and the final exam for a course are 14%, 20%, 20%, and 46%, respectively. The total marks for homework, test 1, test 2, and the final exam are 14, 60, 60, and 80, respectively. A student’s corresponding marks are 12, 45, 51, and 60, respectively. What is the student’s final mark out of 100? https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/4 8/30/2020 A.1 Euclidean spaces 20. A manufacturer produces three different types of products. The demand for the products is denoted by the vector → a = (10, 20, 30). The price per unit for the products is given by the vector → b = ($200, $150, $100). If the demand is met, how much money will the manufacturer receive? 21. A company pays four groups of employees a salary. The numbers of the employees for the four groups are expressed by a vector → a = (5, 20, 40). The payments for the groups are expressed by a → D is tri b ut io n vector b = ($100, 000, $80, 000, $60, 000). Use the dot product to calculate the total amount of money the company paid its employees. 22. There are three students who may buy Calculus or algebra books. Use the dot product to find the total number of students who buy both calculus and algebra books. 23. There are n students who may buy a calculus or algebra books (n ≥ 2). Use the dot product to find the total number of students who buy both calculus and algebra books. 24. Assume that a person A has contracted a contagious disease and has direct contacts with four people: P1, P2, P3, and P4. We denote the contacts by a vector → a : = (a 11, a 12, a 13, a 14), where if the person A has made contact with the person Pj, then a 1j = 1, and if the person A has made no contact with the person Pj, then a 1j = 0. Now we suppose that the four people then have had a variety → fo r of direct contacts with another individual B, which we denote by a vector b : = (b 11, b 21, b 31, b 41), where if the person Pj has made contact with the person B, then b j1 = 1, and if the person Pj has made no contact with the person B, then b j1 = 0. N ot i. Find the total number of indirect contacts between A and B. ii. If → a = (1, 0, 1, 1) and → a = (1, 1, 1, 1), find the total number of indirect contacts between A and B. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/4 8/30/2020 A.2 Matrices A.2 Matrices Section 2.1 ⎛ 5 8 1 4 ) C = ( 10 3 10 3 0 8 ⎜0 ) D = ⎜ ⎜0 4 ⎝ 6 0 10 ⎞ 2 ⎟ ⎟ 1 ⎟ 0 9 10 ⎠ D is tri b A = (2) B = ( 1 ut io n 1. Find the sizes of the following matrices: 2. Use column vectors and row vectors to rewrite each of the following matrices. 3 2 A = ⎜1 8 1 ⎟ ⎛ ⎝ 5 4 10 ⎞ ⎛ 1 B = ⎜3 ⎠ ⎝ 7 2 10 8 7⎟ 3 10 4 4 3. Find the transposes of the following matrices: ⎛ C = (4) B = ( 3 11 2 ) 6 5. Let C = ( ⎛ 6. Let E 3 and B = ( 12 5 25 19 4 6 120 126 9 6 x = ⎜ 123 ⎝ 2 ) 25 ) 12 3 ⎞ 124 125 ⎟ 127 128 ⎠ 7 ⎞ ⎜6 C = ⎜ ⎜3 5 4 ⎠ 2 1 −1 −2 ⎝ 3 ⎛ 9 D = ⎜ −2 ⎠ ). 0 ⎝ 5 −19 ⎞ ⎟ ⎟ ⎟ ⎠ ⎞ 8 −7 ⎟ 3 −9 ⎠ Find all x ∈ R such that A = B. 4 and D = ( 122 8 N ot 9 ⎞ C = ⎜ 8 ⎟ ⎝ 4. Let A = ( 33 9 ⎛ 9 fo r 3 5 x 19 4 6 ⎛ and F 2 12 = ⎜ ⎝ ). 120 Find all x ∈ R such that C 2 x 123 124 26x − 4 127 122 3 x 128 = D. ⎞ ⎟. Find all x ∈ R such that ⎠ E = F. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/3 8/30/2020 A.2 Matrices a b 3x + 2y −x + y 7. Let A = ( ) and B = ( 1 1 3 6 ). Find all a, b, x, y ∈ R such that A = B. 8. Let 2b − a x − 2y 5x + 3y Find all a, b, x, y ∈ R 9. Let A = ( −2 3 4 6 −1 −8 ) and D = ( = D. and B = ( 7 −8 9 0 −1 0 ). Compute A + B; −A; 4A − 2B; 100A + B. 9 5 1 2 2 2 10. Let A = ⎜ 8 0 0 ⎟, B = ⎜0 0 0⎟, 3 2 6 8 ⎝ 0 ⎞ ⎛ ⎠ ⎝ 4 ⎠ 3. (4A + 1 2 T , B − C) T . 11. Find the matrix A if [(3AT ) − ( and C 0 8 14 ). 0 = ⎜0 3 1⎟. 0 2 ⎝ 0 ⎞ ⎠ N ot [3(A + B)] 7 ⎛ Compute 1. 3A − 2B + C, 2. 4 ⎞ fo r ⎛ 6 D is tri b i. ii. iii. iv. such that C ) ut io n a+b C = ( −7 −2 ) −6 9 T T ] = ( −5 −10 33 12 ). 12. Find the matrix B if https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/3 8/30/2020 A.2 Matrices ⎡ 1 ⎢ 6 B + ⎜8 ⎝ 1 3 T ⎞⎤ ⎛ 3 ⎟⎥ − 3⎜ 4 ⎠⎦ ⎝ −5 8 −4 6 T ⎞ −9 ⎟ 2 ⎠ = ( 23 −16 17 −16 26.5 2 ). N ot fo r D is tri b ut io n ⎣ 2 ⎛ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/3 8/30/2020 A.3 Determinants A.3 Determinants 1. Let A = ( a b c d ), where a, b, c, d ∈ R. Show that 2 ∣ ∣λI − A∣ ∣ = λ − tr(A)λ + ∣ ∣A∣ ∣ 2. Let A1 2 4 = ( 1 ), 3 2 A2 = ( −2 ), −1 D is tri b for each λ ∈ R. −1 1 −1 −3 A3 = ( 4 ut io n Section 3.1 For each i = 1, ) 2, 3, find all such that |λI − Ai | = 0. 3. Use (3.1.3) and (3.1.4) to calculate each of the following determinants. λ ∈ R |A| = ∣ ∣ 2 3∣ −2 1 2 3 −1 ∣ ∣ 4∣ ∣2 −2 |B| = ∣ 4∣ ∣ 4 3 1 ∣0 1 2∣ ∣ ∣0 |C| = ∣ ∣ −1 3∣ ∣ 4 1 2 ∣0 0 1∣ ∣ ∣ N ot ∣ 1 fo r ∣ 4. Compute the determinant |A| by using its expansion of cofactors corresponding to each of its rows, where ∣ |A| = ∣ ∣ ∣ 1 2 3∣ −2 1 2 3 −1 ∣ ∣ . 4∣ 5. Compute each of the following determinants. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/2 8/30/2020 A.3 Determinants ∣2 |A| = ∣ ∣ 0 6 0 −4 ∣ 0 5 ∣ ∣ ∣ −2 |B| = ∣ ∣ ∣ ∣ 0 0∣ 1 1 0 0 3 8∣ ∣ ∣ N ot fo r D is tri b ut io n ∣0 3 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/2 8/30/2020 A.4 Systems of linear equations A.4 Systems of linear equations Section 4.1 – − −√2x + 6 2 3 y = 4 − 3z 3x1 + 2x2 + 4x3 + 5x4 = 1 D is tri b c. d. e. ut io n 1. Determine which of the following equations are linear. a. x + 2y + z = 6 b. 2x − 6y + z = 1 2xy + 3yz + 5z = 8 2. For each of the following systems of linear equations, find its coefficient and augmented matrices 1. { 2x1 − x2 = 6 4x1 + x2 = 3 2. fo r ⎧ ⎪ −x1 + 2x2 + 3x3 = 4 ⎨ 3x1 + 2x2 − 3x3 = 5 ⎩ ⎪ 2x1 + 3x2 − x3 = 1 3. N ot ⎧ x1 − 3x3 − 4 = 0 ⎪ ⎨ 2x2 − 5x3 − 8 = 0 ⎩ ⎪ 3x1 + 2x2 − x3 = 4 3. For each of the following augmented matrices, find the corresponding linear system. ⎛ 1 ⎜ −1 ⎝ 0 2 1 −1 3∣ 0 ∣ ∣ 1∣ 1 ⎞ −2 ⎟ 0 ⎠ ⎛ 2 ⎜0 ⎝ 0 1 −1 3 ∣ ∣ ∣ 1 ⎞ −2 ⎟ ∣ −1 ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/3 8/30/2020 A.4 Systems of linear equations → → 4. For each of the following linear systems, write it into the form A X combination of the column vectors of A. = b and express → b as a linear ⎧ ⎪ −x1 + x2 + 2x3 = 3 a. ⎨ 2x1 + 6x2 − 5x3 = 2 ⎩ ⎪ −3x1 + 7x2 − 5x3 = −1 b. c. ut io n ⎧ ⎪ x1 − x2 + 4x3 = 0 ⎨ −2x1 + 4x2 − 3x3 = 0 ⎩ ⎪ 3x1 + 6x2 − 8x3 = 0, { x1 − 2x2 + x3 = 0 5. For each of the following points, P1 (6, − 1 2 D is tri b −x1 + 3x2 − 4x3 = 0 ) , P 2 (8, 1), P 3 (2, 3), is a solution of the system { and P4 (0, 0), verify whether it x − 2y = 7, fo r x + 2y = 5. 6. Consider the system of linear equations N ot ⎧ ⎪ −x1 + x2 + 2x3 = 3 ⎨ 2x1 + 6x2 − 5x3 = 2 ⎩ ⎪ −3x1 + 7x2 − 5x3 = −1 1. Verify that (1, 2. Verify that ( 6 , 5 3, 7 6 − 1) , 4 3 ) is a solution of the system. is a solution of the system. → 3. Determine if the vector b = (3, coefficient matrix of the system. 2, − 1) T is a linear combination of the column vectors of the https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/3 8/30/2020 A.4 Systems of linear equations → T 4. Determine if the vector b = (3, 2, − 1) belongs to the spanning space of the column vectors of the coefficient matrix of the system. 7. Solve each of the following systems. 1. { x1 − x2 = 6 2. { ut io n x1 + x2 = 5 x1 + 2x2 = 1 2x1 + x2 = 2 3. { x1 + 2x2 = 1 N ot fo r D is tri b x1 + 2x2 = 2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/3 8/30/2020 A.5 Linear transformations A.5 Linear transformations Section 5.1 2 A1 = ⎜ −1 ⎝ ⎛ 4 −1 2 3 0 0 1 1 −2 0 3 1 −2 3 2 −1 ⎜ −2 A3 = ⎜ ⎜ 3 ⎝ −2 1 2 −1 ⎞ ⎛ ⎟ A2 = ⎜ ⎠ ⎝ 2 1 1 −3 ⎟ 1 0 −2 −1 1 ⎜ 2 A4 = ⎜ ⎜ −1 0 ⎞ ⎟ ⎟ ⎟ −2 ⎛ ⎠ ⎝ determine the domain and codomain of TA i 1 −1 0 1 ⎞ ⎠ ⎞ D is tri b ⎛ ut io n 1. For each of the following matrices, −2 ⎟ ⎟ 1 ⎟ 0 1 ⎠ , i = 1, 2, 3, 4. Moreover, compute TA (0, 1, 1, 2), TA (1, −2, 1), TA (−1, 0, 1), TA (−1, 1, 1). 2. Let A = ( 3. Let A = ( 3 2 −1 3 −1 2 1 3 0 6 ). Find TA ( ). 1 4 ) 0 → and TA ( → fo r 2 1 0 ) . 1 → Find TA ( e1 ), TA ( e2 ), TA ( e3 ), where N ot 3 → → → e1 , e2 , e3 are the standard vectors in . 4. For each of the following linear transformations, express it in (5.1.1) . a. T (x1 , x2 ) = (x1 − x2 , 2x1 + x2 , 5x1 − 3x2 ). b. T (x1 , x2 , x3 , x4 ) = (6x1 − x2 − x3 + x4 , x2 + x3 − 2x4 , 3x3 + x4 , 5x4 ). c. T (x1 , x2 , x3 ) = (4x1 − x2 + 2x3 , 3x1 − x3 , 2x1 + x2 + 5x3 ). d. T (x1 , x2 , x3 ) = (8x1 − x1 + x3 , x2 − x3 , 2x1 + x2 , x2 + 3x3 ). 3 R 5. Let → → X = (2, −2, −2, 1), Y = (1, −2, 2, 3) and https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/2 8/30/2020 A.5 Linear transformations ⎛ A = ⎜ ⎝ → 1 0 1 0 1 −2 1 1⎟. −1 1 2 0 ⎞ ⎠ → be linear transformations such that T (1, 2, 3) = (1, 1, 0, −1) and S(1, 2, 3) = (2, −1, 2, 2). Compute (3T − 2S)(1, 2, 3) . 8. Compute (T + 2S)(x1 , x2 , x3 , x4 ), where 3 : R 4 → R D is tri b 7. Let S, T ut io n Compute TA ( X − 3 Y ) . 6. Let S, T : R3 → R3 be linear operators such that T (1, 2, 3) = (1, −4) and S(1, 2, 3) = (4, 9). Compute (5T + 2S)(1, 2, 3) . T (x1 , x2 , x3 , x4 ) = (x1 − x2 + x3 , x1 − x2 + 2x3 − x4 ), S(x1 , x2 , x3 , x4 ) = (x2 − x3 + 2x4 , x1 + x3 − x4 ). + 3x2 + x3 ). N ot fo r 9. For each pair of linear transformations TA and TB , find TB TA . a. TA (x1 , x2 ) = (x1 + x2 , x1 − x2 ), TB (x1 , x2 ) = (x1 − x2 , 2x1 − 3x2 , 3x1 + x2 ). b. TA (x1 , x2 , x3 ) = (−x1 − x2 + x3 , x1 + x2 ), TB (x1 , x2 ) = (x1 − x2 , 2x1 + 3x2 ). c. TA (x1 , x2 , x3 ) = (−x1 + x2 − x3 , x1 − 2x2 , x3 ), TB (x1 , x2 , x3 ) = (x1 − x2 + x3 , 2x1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/2 8/30/2020 A.6 Lines and planes in Math Equation A.6 Lines and planes in R3 → → a × b → a , → × → b ) ⊥ a 1. → → a = (1, 2, 3); b = (1, 0, 1), 2. → a = (1, 0, 3. → a = (0, 2, 1); 4. 2. Let → and verify ( a → a = (1, 1, → a = (1, → b , → c → and ( a b = (−1, 0, b = (−5, 0, 1), → − 1); b = (−1, 1, 0), → − 2, 3), b = (−1, − 4, 0), and 2. → c = (2, 3, 1). . → a = (1, 0, 0), → b b = (−1, 0, 1); = (−1, 1, → a → − 1, 0), → a = (−1, b = (−1, 0, 2), b → b . − 2). → → a , b, → c = (0, → − 1, 0), and Find the scalar triple product of fo r → a = (1, 0, 1), → → a = (1, . → 4. Find the volume of the parallelpiped determined by 1. b − 2), N ot 2. → b ) ⊥ → − 3), 3. Find the area of the parallelogram determined by 1. → × D is tri b 1. Find ut io n Section 5.4 = (−1, 1, − 2), and − 1, → c . − 1). → c = (−1, 1, 1). https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/1 8/30/2020 A.7 Bases and dimensions A.7 Bases and dimensions Section 7.1 = (2, 4, 6) T → T , a2 = (0, 0, 0) , → and a3 = (1, 2, 3) T . → Determine whether {a1 , is linearly dependent. → → → a2 , a3 , a4 } ⎛ ⎛ 1 ⎞ → ⎜2⎟ a1 = ⎜ ⎟, ⎜1⎟ → ⎜0⎟ a2 = ⎜ ⎟, ⎜1⎟ ⎝ ⎝ 0 ⎠ ⎝ 1 3 −4 ⎞ ⎟, 0 ⎠ ⎛ ⎞ ⎛ 0 ⎞ → ⎜2⎟ a3 = ⎜ ⎟, ⎜0⎟ → ⎜2⎟ a4 = ⎜ ⎟. ⎜0⎟ ⎝ ⎝ ⎠ 0 1 ⎠ is linearly independent, where 7 ⎛ ⎞ → a2 = ⎜ 9 ⎟ , ⎠ → → → = {a1 , a2 , a3 } 0 fo r → → → → = {a1 , a2 , a3 , a4 } ⎛ → a1 = ⎜ 4. Determine that S ⎞ −1 ⎛ ⎞ → a3 = ⎜ −2 ⎟ , N ot 3. Determine whether S 1 is linearly dependent, where D is tri b → 2. Determine whether {a1 , → → a2 , a3 } ut io n → 1. Let a1 ⎝ 8 ⎠ ⎝ 1 6 ⎛ ⎞ → a4 = ⎜ −2 ⎟ . ⎠ ⎝ 5 ⎠ is linearly independent, where ⎛ → a1 = ⎜ ⎝ −1 0 −1 ⎞ ⎟, ⎠ 3 ⎛ ⎞ → a2 = ⎜ −1 ⎟ , ⎝ −6 ⎠ 1 ⎛ ⎞ → a3 = ⎜ 2 ⎟ . ⎝ 3 ⎠ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/2 A.7 Bases and dimensions N ot fo r D is tri b ut io n 8/30/2020 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/2 8/30/2020 A.8 Eigenvalues and Diagonalizability A.8 Eigenvalues and Diagonalizability Section 8.1 1 0 2 1 0 0 B = ⎜1 1 0 ⎛ ) ⎝ 2 ⎛ ⎞ ⎟ −1 0 0 0 0 2 0 ⎜ C = ⎜ ⎜ −3 −4 ⎠ ⎝ ⎞ 0 0 −1 0⎟ ⎟ 0⎟ 0 0 0 D is tri b A = ( 2 ut io n 1. For each of the following matrices, find its eigenvalues and determine which eigenvalues are repeated eigenvalues. 0 ⎠ 2. For each of the following matrices, find its eigenvalues and eigenspaces. 7 3 1 ) B = ( −1 1 ) C = ( 5 3I − A, N ot 3. Let A be a 3 × 3 matrix. Assume that 2I − A, 1. Find |A| and tr(A). 2. Prove that I + 5A is invertible. 3 fo r A = ( 5 3 0 0 3 1 2 3 ) D = ⎜2 1 3⎟ 1 2 ⎛ ⎝ 1 ⎞ ⎠ and 4I − A are not invertible. 4. Assume that the eigenvalues of a 3 × 3 matrix A are 1, 2, 3. Compute ∣∣A3 2 − 2A + 3A + I ∣ ∣ . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/1 8/30/2020 A.9 Vector spaces A.9 Vector spaces = {(x, y) ∈ R 2. Let W = {(x, y) ∈ R 2 2 + : x − y = 0}. : x + y ≤ 1}. Show that W is a subspace of R2 . Show that W is not a subspace of R2 . 3. Let + i. Show that if → 2 a ∈ R+ and → b = {(x, y) : x ≥ 0 2 ∈ R+ , then and y ≥ 0}. → → 2 a + b ∈ R+ . D is tri b 2 R ii. Show that R2+ is not a subspace of R2 by using Definition 9.1.1 4. Let 2 W = R + 2 ∪ (−R + ). 9.1.1 3 = {(x, y, z) ∈ R and Theorem 9.1.1 . N ot i. Show that if a ∈ W and k is a real number, then k a ∈ W . ii. Show that W is not a subspace of R2 by using definition 9.1.1 5. Let W . → fo r → ut io n 1. Let W : x − 2y − z = 0}. Show that W is a subspace of R3 by Definition , respectively. 6. Show that W is a subspace of R4 , where 4 W = {(x, y, z, w) ∈ R : { x − 2y + 2z − w = 0, }. −x + 3y + 4z + 2w = 0 7. Show that W is a subspace of R3 , where https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/2 8/30/2020 A.9 Vector spaces x + 2y + 3z = 0, ⎧ ⎪ ⎧ ⎪ 3 W = ⎨(x, y, z) ∈ R ⎩ ⎪ 3 = {(x, y, z) ∈ R : x − 2y − z = 3}. : ⎨ −x + y − 4z = 0, ⎬ . ⎩ ⎭ ⎪ ⎪ x − 2y + 5z = 0 Show that W is not a subspace of R3 by Definition 9.1.1 . 9. Let W = [0, 1]. Show that W is not a subspace of R. 10. Show that all the subspaces of R are {0} and R . 11. Let W = {(0, 0, ⋯ , 0)} ∈ Rn . Show that W is a subspace of Rn . → n υ ∈ R be a given nonzero vector. Show that D is tri b 12. Let ut io n 8. Let W ⎫ ⎪ → W = {t υ : t ∈ R} . N ot fo r is a subspace of Rn by Definition 9.1.1 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/2 8/30/2020 A.10 Complex numbers A.10 Complex numbers Section 10.1 ut io n 1. Solve each of the following equations. a. z 2 + z + 1 = 0 b. z 2 − 2z + 5 = 0 c. z 2 − 8z + 25 = 0 D is tri b 2. Use the following information to find x and y. a. 2x − 4i = 4 + 2yi b. (x − y) + (x + y)i = 2 − 4i fo r 3. Let z1 = 2 + i and z2 = 1 − i. Express each of the following complex numbers in the form x + yi, calculate its modulus, and find its conjugate. 1. z1 + 2z2 2. 2z1 − 3z2 3. z1 z2 z 4. z N ot 1 2 4. Express each of the following complex numbers in the form x + yi . a. b. c. d. ¯¯¯¯ ¯¯¯ ¯ i ( ) 1−i 2−i (1−i)(2+i) 1+i i(1+i)(2−i) 1+i 1−i − 2−i 1+i https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/2 8/30/2020 A.10 Complex numbers 5. Find i2 , 3 4 5 i , i , i , i 1−i 8 6. Let z = ( 1+i ) . 6 and compute (−i − i4 Calculate z 66 + 2z 33 +i 5 6 −i ) 1000 . − 2. 8. Find real numbers x, y such that x + 1 + i(y − 3) = 1 + i. 5 + 3i 5 2 . Find |z|. N ot fo r i (2+4i) 12 D is tri b 9. Let z = (3+4i)(1+i) ut io n 7. Find z ∈ C a. iz = 4 + 3i b. (1 + i)z̄ = (1 − i)z https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/2 8/30/2020 Index Index matrix, 14 3 × 3 determinant, 95 n R -space, 1 n-column vector, 2 n-dimensional space, 221 n-dimensional subspace, 221 n × n determinant, 102 2×2 determinant of a 2 × 2 matrix, 14 adjoint, 115 algebraic multiplicity, 248 angle, 19 area of the parallelogram, 26 argument, 293 augmented matrix, 126 axioms, 281 back-substitution, 132 basic variable, 132 basis of a solution space, 221 basis of a spanning space, 221 basis vectors of the solution space, 144 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 1/7 8/30/2020 Index Cauchy-Schwarz inequality, 15 change of bases, 239 change of coordinates, 239 characteristic equation, 247 characteristic polynomial, 247 coefficient matrix, 126 cofactor, 97, 101 column matrix, 39 column spaces, 230 complex number, 287 conjugate, 291 consistency, 156, 166 coordinates, 237 Cramer’s rule, 150 Demoivre’s formulas, 301 determinants by row operations, 105 diagonal matrix, 57 distance between a point and a plane, 212 distance between two points, 17 dot product, 8 eigenvalue, 246 eigenvector, 246 elementary matrices, 78 equal matrices, 40 equal vector, 4 equations of lines, 200 equations of planes, 200 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 2/7 8/30/2020 Index Euclidean space, 1 Euler’s formula, 296 expansion of cofactors, 98 expansion of minors, 97 exponential form, 293 first row operation, 63 free variable, 132 Gauss-Jordan elimination, 141 Gaussian elimination, 137 geometric multiplicity, 248 Gram determinant, 14 Gram-Schmidt process, 243 homogeneous system, 127 identity matrix, 57 image, 172 imaginary axis, 293 imaginary number, 288 imaginary part, 288 inverse image, 172 inverse matrix method, 150 inverse of a square matrix, 85 kernel, 180 Lagrange’s identity, 27 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 3/7 8/30/2020 Index leading 1, 61 leading entry, 61 linear combination, 31, 166 linear operator, 172 linear system, 126 linear transformation, 172 linearly dependent, 217 linearly independent, 217 lower triangular matrix, 57 matrix, 38 matrix of cofactors, 115 midpoint formula, 18 minor, 97, 101 modulus, 290 multiplication, 172 nonhomogeneous system, 127 nontrivial solution, 129 nonzero row, 61 norm, 12 normal vector, 204 null space, 129, 248 nullity, 74 one to one, 180 operations of vectors, 4 orthogonal vectors, 19 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 4/7 8/30/2020 Index parametric equation, 200 pivot columns, 75 point-normal, 204 polar form, 293 powers of a square matrix, 58 principal argument, 294 product of a matrix and a vector, 45 product of two matrices, 45, 49 projection, 24 projection operator, 184 properties of products of matrices, 52 Pythagorean Theorem, 20 range, 178 rank, 74, 156 real axis, 293 real part, 288 reflection operator, 184 root of a complex number, 302 rotation operator, 184 row echelon matrix, 62 row equivalent, 76 row matrix, 39 row operations of matrices, 63 row spaces, 230 scalar triple product, 196 second row operation, 63 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 5/7 8/30/2020 Index sets, 279 similar matrices, 272, 448 size of a matrix, 38 sizes of two matrices, 51 solution space, 129, 274 spanning space, 35 spanning spaces, 31, 277 square matrix, 56 standard basis, 221 standard matrix, 172 standard vectors, 2 subspace, 273 subspaces of Rn, 273 subspaces of vector spaces, 285 symmetric matrix, 56 system of n variables, 125 third row operation, 63 trace, 253 transition matrix, 239 transpose, 40 triangular matrix, 57 trivial solution, 129 upper triangular matrix , 57 vector spaces, 279, 281 zero matrix, 39 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 6/7 8/30/2020 Index zero row, 61 zero vector, 2 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/fil… 7/7