Linear transformations in sec.3.3 Theorem 10’. a. Let T : ℝ2 → ℝ2 : the linear transformation determined by a 2×2 matrix A . If S : a region in ℝ2 with finite area, then the area of 𝑇 𝑆 = det 𝐴 ∙ the area of 𝑆 . b. If T is determined by a 3×3 matrix A , and if S : a solid in ℝ3 with finite volume, then the volume of 𝑇 𝑆 = det 𝐴 ∙ the volume of 𝑆 . Pf. (a) : 1 Quiz. Q. Find the eigenvalues of the given matrix A and a basis for the corresponding eigenspace. Sol. 2 𝐴 = −4 3 4 3 −6 −3 3 1 2 Section 5.3 Diagonalization(대각화) If A is similar to a diagonal matrix D , that is, if 𝑨 = 𝑷𝑫𝑷−𝟏 for some invertible matrix P and some diagonal matrix D , def! ⟺ A square matrix A is diagonalizable(대각화 가능하다) : - Diagonalization enables us to compute 𝐴𝑘 quickly for large values of 𝑘 . 𝐴𝑘 = 𝑃𝐷𝑃−1 𝑃𝐷𝑃−1 × ⋯ × 𝑃𝐷𝑃−1 = 𝑃𝐷𝑘 𝑃−1 3 The diagonalization theorem Theorem 5( The Diagonalization Theorem: 대각화 정리 ). An 𝑛 × 𝑛 matrix A is diagonalizable. ⟺ A has 𝒏 linearly independent eigenvectors. ⟺ Span{𝐯1 , … , 𝐯𝑛 } = ℝ𝑛 . → {𝐯1 , … , 𝐯𝑛 } : eigenvector basis of ℝ𝑛 . In fact, 𝐴 = 𝑃𝐷𝑃−1 with a diagonal matrix D ⟺ 𝑃 = 𝐯1 … 𝐯𝑛 and each 𝐯𝑖 ∶ an eigenvector corresponding to 𝜆𝑖 . 4 The diagonalization theorem Pf. (⟹) If P is any 𝑛 × 𝑛 matrix with columns 𝐯1 , … , 𝐯𝑛 and D is any diagonal matrix with diagonal entries 𝜆1 , … , 𝜆𝑛 , then Suppose A is diagonalizable. i.e., 𝐴 = 𝑃𝐷𝑃−1 . ⟶ 𝐴𝑃 = 𝑃𝐷 ⟶ (1)=(2) ⟶ Since P : invertible, its columns 𝐯1 , … , 𝐯𝑛 : linearly independent. 5 The diagonalization theorem Since these columns are nonzero, by (4), 𝜆1 , … , 𝜆𝑛 : eigenvalues and 𝐯1 , … , 𝐯𝑛 : corresponding eigenvectors. (⟸) Suppose 𝐯1 , … , 𝐯𝑛 : linearly independent eigenvectors corresponding to eigenvalues 𝜆1 , … , 𝜆𝑛 . ⟶ By (1)&(2), 𝐴𝑃 = 𝑃𝐷 . Since the eigenvectors 𝐯1 , … , 𝐯𝑛 : linearly independent, P : invertible by Thm.8 in sec.2.3. ⟶ 𝐴 = 𝑃𝐷𝑃−1 . ∴ 𝐴 : diagonalizable. 6 Diagonalizing matrices(행렬의 대각화) Example 3’. (a) Diagonalize the following matrix, if possible. That is, find an invertible matrix P and a diagonal matrix D such that 𝐴 = 𝑃𝐷𝑃−1 . Sol. Step 1. : Find the eigenvalues of A . Step 2. : Find three linearly independent eigenvectors of A . 1 Since a eigenvector corresponding to 𝜆 = 1 is 𝐯1 = −1 1 and 7 Diagonalizing matrices −1 −1 Eigenvectors corresponding to 𝜆 = −2 are 𝐯2 = 1 , 𝐯3 = 0 . 0 1 Since 𝐴 = 𝐯1 𝐯2 𝐯3 with det A ≠ 0 , by Thm.8. in sec.2.3, {𝐯1 , 𝐯2 , 𝐯3 } : linearly independent. Step 3. : Construct P from the vectors in step 2. Step 4. : Construct D from the corresponding eigenvalues. 8 Diagonalizing matrices (b) Use (a) to compute 𝐴4 . 1 1 Sol. Since 𝑃−1 = 1 2 −1 −1 1 −1 𝐴 = 𝑃𝐷 𝑃−1 = −1 1 1 0 1 Then, 𝐴4 = 𝑃𝐷4 𝑃−1 = −1 1 1 −15 −15 = 15 31 15 . −15 −15 1 1 1 , 0 −1 1 0 0 1 1 0 0 −2 0 1 2 1 0 0 −2 −1 −1 −1 −1 1 0 0 1 1 0 0 16 0 1 0 1 0 0 16 −1 1 1 . 0 1 1 2 1 −1 0 9 Diagonalizing matrices Theorem 6. An 𝑛 × 𝑛 matrix with 𝒏 distinct eigenvalues is diagonalizable. Pf. Let 𝐯1 , … , 𝐯𝑛 : eigenvectors corresponding to the 𝑛 distinct eigenvalues of a matrix A. Then, by Thm.2 in sec.5.1, { 𝐯1 , … , 𝐯𝑛 } : linearly independent. So, by Thm.5, A : diagonalizable. Example 5. Determine whether the matrix A is diagonalizable. 5 −8 1 𝐴= 0 0 7 . 0 0 −2 Sol. Since A : triangular matrix, eigenvalues are 5, 0, and − 2. So, by Thm.6, A is diagonalizable. 10 Section 5.3 : Exercises 1, 7,9, 11,12,17, 27, 31,32 11 Section 4.1 Vector spaces(벡터 공간) Definition. Let V(≠ ∅) be a set of elements, called vectors, on which vector addition and scalar multiplication are defined. def! Then V ⟺ a vector space if the followings are satisfied : For all vectors u, v, and w in V and for all scalars c and d , 1. 𝐮 + 𝐯 ∈ 𝑉, 2. 𝐮 + 𝐯 = 𝐯 + 𝐮 , 3. 𝐮 + 𝐯 + 𝐰 = 𝐮 + (𝐯 + 𝐰) , 4. ∃ zero vector 𝟎 ∈ 𝑉 such that 𝐮 + 𝟎 = 𝐮 , 5. For each 𝐮 ∈ 𝑉, ∃ − 𝐮 ∈ 𝑉 such that 𝐮 + (−𝐮) = 𝟎 , 6. 𝑐𝐮 ∈ 𝑉, 7. 𝑐 𝐮 + 𝐯 = 𝑐𝐮 + 𝑐𝐯 , 9. 𝑐(𝑑𝐮) = (𝑐𝑑)𝐮 , 8. 𝑐 + 𝑑 𝐮 = 𝑐𝐮 + 𝑑𝐮 , 10. 1𝐮 = 𝐮 . 12 Vector spaces Example. (a) ℝ𝑛 , 𝑛 ≥ 1 ∶ vector space, (b) ℙ𝑛 = 𝐩 𝑡 = 𝑎0 + 𝑎1 𝑡 + ⋯ + 𝑎𝑛 𝑡 𝑛 , 𝑎𝑖 ∈ ℝ, 𝑖 = 0,1, … , 𝑛 : the set of all polynomials of degree at most 𝑛(≥ 0) (차수가 𝑛 보다 같거나 작은 모든 다항식들의 집합) ⟶ vector space. (∵) See Ex.4 in sec.4.1. (c-d) If the following each set is a vector space, write “V ” in the blank. Otherwise, write “S ” . 𝑥 𝑥 (c) 𝐴 = 𝑦 𝑥 ≥ 0 , 𝑦 ≥ 0 : (d) 𝐵 = 𝑦 𝑥 + 𝑦 = 0 : Definition. For a vector space V , if a set B is linearly independent and Span{B } = V : def! ⟺ B : a basis(기저) for V and the number of vectors in a basis B for V def! ⟺ the dimension(차원) of V ; dim V . 13 Coordinate system(좌표 시스템) in sec.4.4 Definition. Suppose B ={ b1 , ... , 𝐛𝑛 } is a basis for V . For each x ∈ V , the weights 𝒄𝟏, … , 𝒄𝒏 such that 𝐱 = 𝑐1𝐛1 + ⋯ + 𝑐𝑛𝐛𝑛 . def! ⟺ the coordinate of x relative to the basis B . (기저 B 에 대한 x 의 좌표 : the B–coordinate of x ) x의 B 좌표 − If 𝑐1, … , 𝑐𝑛 : the B–coordinate of x , then the vector in ℝ𝑛 𝑐1 . def! 𝐱 B = . ⟺ the coordinate vector of x relative to B . . (기저 B 에 대한 x 의 좌표 벡터 or 𝑐𝑛 the B–coordinate vector of x : x의 B 좌표 벡터) def! − The mapping 𝐱 ⟼ 𝐱 B ⟺ the coordinate mapping(좌표 사상) . 14 The coordinate mapping(좌표 사상) in sec.4.4 Theorem 8. Let B = 𝐛1 , … , 𝐛𝑛 ∶ a basis for a vector space V . Then the coordinate mapping 𝐱 ⟼ 𝐱 B is a one-to-one linear transformation from V onto ℝ𝑛 . Pf. Take u = 𝑐1 b1 + . . . + 𝑐𝑛 𝐛𝑛 and w = 𝑑1 b1 + . . . + 𝑑𝑛 𝐛𝑛 in V . Then, u + w = (𝑐1 + 𝑑1 )b1 + . . . + 𝑐𝑛 + 𝑑𝑛 𝐛𝑛 . 𝑐1 𝑐1 + 𝑑1 𝑑1 ⋮ ⟶ 𝐮+𝐰 B = = ⋮ + ⋮ = 𝐮B + 𝐰B . 𝑐𝑛 𝑐𝑛 + 𝑑𝑛 𝑑𝑛 15 The coordinate mapping So the coordinate mapping preserves addition. If 𝑟 is any scalar, then 𝑟u = 𝑟(𝑐1 b1 + . . . + 𝑐𝑛 𝐛𝑛 ) = 𝑟𝑐1 b1 + . . . + 𝑟𝑐𝑛 𝐛𝑛 . 𝑟𝑐1 ⟶ 𝑟𝐮 B = ⋮ = 𝑟 𝐮 B . 𝑟𝑐𝑛 So the coordinate mapping preserves scalar multiplication and hence is a linear transformation. 𝑐1 Suppose that 𝐮 B = 𝐰 B = ⋮ . 𝑐𝑛 Then, u = v = 𝑐1 b1 + . . . + 𝑐𝑛 𝐛𝑛 in V . ∴ one-to-one . For ∀ 𝐲 = 𝑦1 , … , 𝑦𝑛 ∈ ℝ𝑛 , let u = 𝑦1 b1 + . . . + 𝑦𝑛 𝐛𝑛 in V . ⟶ 𝐮 B = 𝐲 . ∴ The coordinate mapping is onto ℝ𝑛 . 16 Section 5.4 Eigenvectors and linear transformations Let 𝑉 ∶ 𝑛-dimensional vector space and 𝑊 ∶ 𝑚-dimensional vector space. Consider a linear transformation 𝑇 ∶ 𝑉 → 𝑊 . Choose basis B for V and basis C for W . For ∀ 𝐱 ∈ 𝑉, the coordinate vector 𝐱 B is in ℝ𝑛 and the coordinate vector of its image, [T(x)]C , is in ℝ𝑚 : 17 The matrix of a linear transformations Let { b1 , ... , 𝐛𝑛 } : the basis B for V . 𝑟1 If x = 𝑟1 b1 + . . . + 𝑟𝑛 𝐛𝑛 , then 𝐱 B = ⋮ 𝑟𝑛 . (1) Then, 𝑇 𝐱 = 𝑇 𝑟1 b1 + . . . + 𝑟𝑛 𝐛𝑛 = 𝑟1𝑇 𝐛1 + ⋯ + 𝑟𝑛 𝑇 𝐛𝑛 . ⟶ 𝑇 𝐱 ∴ (2) C = 𝑟1 𝑇 𝐛1 C + ⋯ + 𝑟𝑛 [𝑇(𝐛𝑛 )]C 𝑟1 𝑟2 = [ 𝑇 𝐛1 C 𝑇 𝐛2 C ... 𝑇 𝐛𝑛 C ] ⋮ = 𝑀 𝐱 B . (4) def! =: 𝑴 ⟺ the 𝑚 × 𝑛 matrix for 𝑇 𝑟𝑛 relative to the bases B and C . 𝑇 𝐱 C =𝑀 𝐱B . 18 The matrix of a linear transformations 19 The matrix of a linear transformations Example 1. Suppose B = {b1, b2} : a basis for V and C = {c1, c2, c3} : a basis for W . Let T : V →W : a linear transformation with 𝑇 𝐛1 = 3𝐜1 − 2𝐜2 + 5𝐜3 and 𝑇 𝐛2 = 4𝐜1 + 7𝐜2 − 𝐜3 . Find the matrix M for T relative to B and C . Sol. Since 𝑇 𝐛1 𝐶 3 = −2 and 5 𝑇 𝐛2 𝐶 4 = 7 , −1 3 4 𝑀 = −2 7 . 5 −1 20 The matrix of a linear transformations If W = V and the basis C is the same as B , def! then the matrix M in (4) ⟺ the matrix for T relative to B , or the B–matrix for T ; 𝑇 B . i.e., B–matrix for T : 𝑉 → 𝑉 such that 𝑇(𝐱) B = 𝑇 B 𝐱 B for ∀ 𝐱 ∈ 𝑉 . 21 The matrix of a linear transformations Example 2. The mapping T : ℙ2 → ℙ2 defined by 𝑇 𝑎0 + 𝑎1𝑡 + 𝑎2𝑡2 = 𝑎1 + 2𝑎2𝑡 is a linear transformation. a. Find the B–matrix for T , when B is the basis {1, t , 𝑡2 } . b. Verify that T(p) B = T B p B . Sol. (a) : Since T(1) = 0, T(t ) = 1, and T(𝑡2) = 2t , 0 T(1) B = 0 , 0 ∴ 1 T(t ) B = 0 , and 0 0 1 T B= 0 0 0 0 0 T(𝑡2) B = 2 . 0 0 2 . 0 22 The matrix of a linear transformations (b) : Let p(t ) = 𝑎0 + 𝑎1𝑡 + 𝑎2𝑡2 . 𝑎1 Since T(p(t )) = 𝑎1 + 2𝑎2t , T(p) B = 𝑎1 + 2𝑎2t B = 2𝑎2 . 0 𝑎1 0 1 0 𝑎0 Since T B p B = 0 0 2 𝑎1 = 2𝑎2 , T(p) B = T B p B . 0 0 0 0 𝑎2 23 Linear transformations on ℝ𝑛 Theorem 8. Suppose A = PD𝑃−1 , D : a diagonal 𝑛 × 𝑛 matrix. If B is the basis for ℝ𝑛 formed from the columns of P , then D is the B–matrix for the transformation x ⟼ 𝐴𝐱. Pf. Let P = 𝐛1 ⋯ 𝐛𝑛 . Then, B = { 𝐛1 , ⋯ , 𝐛𝑛 } . If 𝐱 = 𝑐1 𝐛1 + 𝑐2 𝐛2 + ⋯ + 𝑐𝑛 𝐛𝑛 , 𝑐1 (∗) ⋮ ⟺ 𝐱 = 𝐛1 ⋯ 𝐛𝑛 = 𝑃 𝐱 B . ⟶ 𝐱 B = 𝑃−1 𝐱 . 𝑐𝑛 If T(x) = Ax for x in ℝ𝑛 , then (4) 𝑇 B = [ [T(b1)]𝛽 … [T(𝐛𝑛 )]𝛽 ] = [ [Ab1]𝛽 … [A𝐛𝑛 ]𝛽 ] (∗) = [𝑃−1 Ab1 … 𝑃−1 A𝐛𝑛 ] = 𝑃−1 A [b1 … 𝐛𝑛 ] = 𝑃−1 AP = D . 24 Linear transformations on ℝ𝑛 25 Linear transformations on ℝ𝑛 7 2 . −4 1 Find a basis B for ℝ2 s.t. B–matrix for T is a diagonal matrix. Sol. In Ex.2 in sec.5.3, 1 1 5 0 A = PD𝑃−1 , 𝑃 = and 𝐷 = . −1 −2 0 3 b1 b2 By Thm.8, a diagonal matrix D is the B–matrix for T when B = {b1, b2} . 1 1 ∴ B ={ , }. −1 −2 Note. If A is a similar to a matrix C with A = PC𝑃−1 , then C = 𝑃−1 AP : B–matrix for the transformation x ⟼ 𝐴𝐱 when the basis B is formed from the columns of P . Example 3. Define T : ℝ2→ ℝ2 by T(x)=Ax, 𝐴 = 26 Section 5.4 : Exercises 1,3,4,5,7,8,9,11,13,14 27