1
MODULE 1
MATRICES OVER A FIELD 𝑭
Introduction
In this lesson, we will discuss the different operations on matrices and its properties,
the transpose of a matrix, the different types of matrices and the special types of square
matrices.
Objectives
1.
2.
3.
4.
5.
After going through this chapter, you are expected to be able to do the following:
Define matrix.
Give the different types of matrices.
Perform fundamental operations on matrices.
Find the transpose of a given matrix.
Identify special types of square matrices.
1.1 Definition of a Field
Definition 1.1.1: By a field 𝐹 we mean a nonempty set of elements with two laws of
combination, which we call addition and multiplication, satisfying the following conditions:
𝐹1 Closure Properties
To every pair of elements 𝑎, 𝑏 ∈ 𝐹, 𝑎 + 𝑏 ∈ 𝐹
To every pair of elements 𝑎, 𝑏 ∈ 𝐹, 𝑎𝑏 ∈ 𝐹
𝐹2 Commutative Laws
𝑎+𝑏 =𝑏+𝑎
𝑎𝑏 = 𝑏𝑎 ∀𝑎, 𝑏 ∈ 𝐹
𝐹3 Associative Laws
(𝑎 + 𝑏) + 𝑐 = 𝑎 + (𝑏 + 𝑐)
(𝑎𝑏)𝑐 = 𝑎(𝑏𝑐) ∀𝑎, 𝑏, 𝑐 ∈ 𝐹
𝐹4 Distributive Laws
(𝑎 + 𝑏)𝑐 = 𝑎𝑐 + 𝑏𝑐
𝑐(𝑎 + 𝑏) = 𝑐𝑎 + 𝑐𝑏 ∀𝑎, 𝑏, 𝑐 ∈ 𝐹
𝐹5 Identity Elements
∃0 ∈ 𝐹 such that 𝑎 + 0 = 𝑎 ∀𝑎 ∈ 𝐹
∃1 ∈ 𝐹, 1 ≠ 0 such that 𝑎 ∙ 1 = 𝑎 ∀𝑎 ∈ 𝐹
2
𝐹6 Inverse Elements
∀𝑎 ∈ 𝐹, ∃ − 𝑎 ∈ 𝐹 such that 𝑎 + (−𝑎) = 0
∀𝑎 ∈ 𝐹, 𝑎 ≠ 0, ∃𝑎−1 ∈ 𝐹 such that 𝑎𝑎 −1 = 1
The elements of 𝐹 are called scalars. The set of real numbers ℝ and complex numbers ℂ are
examples of fields under the usual addition and multiplication in these sets
1.2 Definition of a Matrix
Definition 1.2.1: A matrix over a field 𝐹 is a rectangular array of elements from 𝐹 arranged
in 𝑚 horizontal rows and 𝑛 vertical columns:
𝑎11 𝑎12 … 𝑎1𝑛
𝑎21 𝑎22 … 𝑎2𝑛
𝐴=[ ⋮
⋮
⋮ ]
𝑎𝑚1 𝑎𝑚2 … 𝑎𝑚𝑛
(1)
The ith row of A is
[𝑎𝑖1 𝑎𝑖2 … 𝑎1𝑛 ] (1 ≤ 𝑖 ≤ 𝑚);
the jth column of A is
𝑎1𝑗
𝑎2𝑗
[ ⋮ ] (1 ≤ 𝑗 ≤ 𝑛).
𝑎𝑚𝑗
The symbol 𝑀𝑚,𝑛 (𝐹) denotes the collection of all 𝑚 x 𝑛 matrices over 𝐹. We usually denote
matrices by capital letters. If 𝑚 is equal to 𝑛, then the matrix is called a square matrix of
order 𝒏.
Example 1: Below are examples of matrices of different sizes.
2 −5
i) 𝐴 = [
] is a 2 x 2 matrix
3
4
3
ii) 𝐵 = [−1
0
(A is a square matrix of order 2)
4 −8 −5
5
3
1 ] is a 3 x 4 matrix
7
2 −4
−8
The second row is [−1 5 3 1] and the third column is [ 3 ]
2
3
iii) 𝐶 = [
1
2
1/2
0
] is a 2 x 3 matrix
3
4/3
We will use the notation 𝐴 = [𝑎𝑖𝑗 ] to denote a matrix A with entries 𝑎𝑖𝑗 . The subscripts 𝑖
and 𝑗 will be used to denote the position of an entry 𝑎𝑖𝑗 in a matrix. Thus, the 𝑖𝑗th element
or entry of a matrix A is the number appearing in the 𝑖th row and 𝑗th column of A.
1 0
1 0 4
Example 2: Let 𝐴 = [
] and 𝐵 = [5 1
−2 2 3
3 2
−1
4 ].
−2
Then A is a 2 x 3 matrix with 𝑎12 = 0 (the element appearing in the first row and the second
column), 𝑎13 = 4, and 𝑎23 = 3; B is a 3 x 3 matrix with 𝑏12 = 0, 𝑏13 = −1 , 𝑏21 = 5 , 𝑏23 =
4, 𝑏31 = 3 , and 𝑏32 = 2 .
If A is a square matrix of order 𝑛, then the numbers 𝑎11 , 𝑎22 , … , 𝑎𝑛𝑛 form the main
diagonal. Thus in matrix B above, the elements 𝑏11 = 1, 𝑏22 = 1, and 𝑏33 = −2 form the
main diagonal.
Definition 1.2.2: (Equality of Matrices) Two 𝑚 x 𝑛 matrix 𝐴 = [𝑎𝑖𝑗 ] and 𝐵 = [𝑏𝑖𝑗 ] are said
to be equal if 𝑎𝑖𝑗 = 𝑏𝑖𝑗 , 1 ≤ 𝑖 ≤ 𝑚 , 1 ≤ 𝑗 ≤ 𝑛 , that is, if corresponding elements agree.
Example 3:
1 2
𝐴=[
2 −5
−3
1 2
] and 𝐵 = [
4
2 𝑥
𝑤
] are equal if 𝑥 = −5 and 𝑤 = −3.
4
1.3 Matrix Operations and Its Properties
Definition 1.3.1: (Addition of Matrices) If 𝐴 = [𝑎𝑖𝑗 ] and 𝐵 = [𝑏𝑖𝑗 ] are 𝑚 x 𝑛 matrices, then
the sum of A and B is the 𝑚 x 𝑛 matrix 𝐶 = [𝑐𝑖𝑗 ] defined by
[𝑐𝑖𝑗 ] = [𝑎𝑖𝑗 ] + [𝑏𝑖𝑗 ]
(1  𝑖 ≤ 𝑚, 1  𝑗 ≤ 𝑛)
That is, C is obtained by adding the corresponding elements of A and B.
4
1 2 4 
0 2 4 
Example 1: Let A  
and B  


3 1 5 
1 3 1 
Then 𝐴 + 𝐵 = [
1+0
3+1
(−2) + (−2) 4 + 4
1
]= [
(−1) + (−3) 5 + 1
4
−4 8
]
−4 6
Note that only matrices with the same size can be added.
Definition 1.3.2: (Scalar Multiplication) If 𝐴 = [𝑎𝑖𝑗 ] is an 𝑚 x 𝑛 matrix and 𝑟 is a real
number, then the scalar multiple of A by 𝑟, 𝑟A, is the 𝑚 x 𝑛 matrix 𝐵 = [𝑏𝑖𝑗 ] , where
bij  ra ij
(1  𝑖 ≤ 𝑚, 1  𝑗 ≤ 𝑛)
That is, B is obtained by multiplying each element of A by 𝑟.
 2
4 3
Example 2: Let 𝑟 = −3 and A  
 . Then
2  5 0 
4
𝑟𝐴 = −3 [
2
−3(4) −3(3)
3 −2
] =[
−3(2) −3(−5)
−5
0
−3(−2)
−12 −9 6
]= [
]
−3(0)
−6 15 0
Definition 1.3.3: (Matrix Multiplication) If 𝐴 = [𝑎𝑖𝑗 ] is an 𝑚 x 𝑛 matrix and 𝐵 = [𝑏𝑖𝑗 ] is an
𝑛 x 𝑝 matrix, then the product of A and B is the 𝑚 x 𝑝 matrix 𝐶 = [𝑐𝑖𝑗 ], defined by
𝑐𝑖𝑗 = ∑𝑛𝑘=1 𝑎𝑖𝑘 𝑏𝑘𝑗
To illustrate this, let us consider an 𝑚 x 𝑛 matrix 𝐴 and an 𝑛 x 𝑝 matrix 𝐵.
𝑗th column of 𝐵
𝑖th row of 𝐴
𝑎11
𝑎21
⋮
𝑎𝑖1
⋮
[ 𝑎𝑚1
𝑎12
𝑎22
⋮
𝑎𝑖2
⋮
𝑎𝑚2
… 𝑎1𝑛
𝑏11 𝑏12 … 𝑏1𝑗 … 𝑏1𝑝
… 𝑎2𝑛
𝑏21 𝑏22 … 𝑏2𝑗 … 𝑏2𝑝
⋮
⋮
⋮
⋮
⋮
… 𝑎𝑖𝑛
𝑏𝑛1 𝑏𝑛2 … 𝑏𝑛𝑗 … 𝑏𝑛𝑝
⋮
[
]
… 𝑎𝑚𝑛 ]
Note that the 𝑖th row of 𝐴 and the 𝑗th column of 𝐵 must have the same number of
components for product AB to be defined.
5
If we let AB = C then the 𝑖𝑗th element of AB is
𝑐𝑖𝑗 = [𝑎𝑖1 𝑎𝑖2
Example 3: If 𝐴 = [
𝑏1𝑗
𝑏
… 𝑎𝑖𝑛 ] 2𝑗 = 𝑎𝑖1 𝑏𝑖𝑗 + 𝑎𝑖2 𝑏2𝑗 + … + 𝑎𝑖𝑛 𝑏𝑛𝑗
⋮
[𝑏𝑛𝑗 ]
1 3
−2 6
] and 𝐵 = [
] then
−2 5
4 7
−2
𝑐11 = [1 3] [ ] = 1(−2) + 3(4) = 10
4
−2
𝑐21 = [−2 5] [ ] = (−2)(−2) + 5(4) = 24
4
6
𝑐12 = [1 3] [ ] = 1(6) + 3(7) = 27
7
6
𝑐22 = [−2 5] [ ] = (−2)(6) + 5(7) = 23
7
10
Thus 𝐴𝐵 = [
24
Similarly, 𝐵𝐴 = [
27
]
23
(−2)(1) + 6(−2) (−2)(3) + 6(5)
−2 6
1 3
][
]= [
]
4(1) + 7(−2)
4(3) + 7(5)
4 7 −2 5
−14
= [
−10
24
]
47
Example 3 shows that matrix multiplication, in general, is not commutative.
7 −1
2
0 5
Example 4: Let 𝐴 = [
] and 𝐵 = [−2
5
4 −3 1
3
2
Find a) AB and b) BA.
1
6
4 −4].
−3
2
6
Solution:
a. Note that the number of columns of A is equal to the number of rows of B hence the
product AB is defined. By Definition 1.3.3, AB is a 2 x 4 matrix.
2
𝐴𝐵 = [
4
= [
= [
7 −1
0 5
] [−2
5
−3 1
3
2
14 + 0 + 15
28 + 6 + 3
29
8
37 −17
1
6
4 −4]
−3
2
(−2) + 0 + 10
(−4) + (−15) + 2
2 + 0 + (−15)
4 + (−12) + (−3)
12 + 0 + 10
]
24 + 12 + 2
−13 22
]
−11 38
b. The product BA is not defined because the number of columns of B is not equal to
number of rows of A.
the
SAQ 1-1
−2 5
3 −1
] and 𝐵 = [−4 3]. Find
1 −4
3 1
a. 𝐴𝐵 and 𝐵𝐴
1
Let 𝐴 = [
2
Solve SAQ1-1 in your notebook and compare your answer with ASAQ1-1.
ASAQ 1-1
The number of columns of 𝐴 is equal to the number of rows of 𝐵 hence 𝐴𝐵 is defined.
Similarly, the number of columns of 𝐵 is equal to the number of rows of 𝐴 thus 𝐵𝐴 is also
defined. Multiplying we have
1(2)  3(4)  (1)(3) 1(5)  3(3)  (1)(1) 
AB  

 2(2)  1(4)  (4)(3) 2(5)  1(3)  (4)(1) 
 17 13


 20 9 
7
and
 2 5
8 1 18
1 3 1 


BA   4 3 
  2 9 8 

2 1 4
3 1  
5 10 7 
Remarks: If 𝐴 = [𝑎𝑖𝑗 ] is an 𝑚 x 𝑛 matrix and 𝐵 = [𝑏𝑖𝑗 ] is an 𝑛 x 𝑝 matrix then,
1. BA may not be defined as in Example 4.b; this will take place if 𝑝 ≠ 𝑚.
2. If BA is defined, which means 𝑝 = 𝑚, then BA is 𝑛 x 𝑛 while AB is 𝑚 x 𝑝; thus if 𝑛 ≠
𝑝, AB and BA are of different size (see ASAQ1-1).
3. If AB and BA are both of the same size, they may be equal.
4. If AB and BA are both of the same size, they may be unequal (see Example 3).
Definition 1.3.4: (Additive Inverse of a Matrix) Let 𝐴 = [𝑎𝑖𝑗 ]. Then – 𝐴 is the matrix
obtained by replacing the elements of 𝐴 by their additive inverses; that is
−𝐴 = −[𝑎𝑖𝑗 ] = [−𝑎𝑖𝑗 ]
Example 5. Let 𝐴 = [
−2
5
1/2
] then
3 2/3 −17
–𝐴 = [
2
−5
−1/2
]
−3 −2/3
17
Definition 1.3.5: (The Zero Matrix) An 𝑚 x 𝑛 matrix 𝐴 = [𝑎𝑖𝑗 ] is called a zero matrix and is
denoted by 0 if 𝑎𝑖𝑗 = 0 , 1  𝑖 ≤ 𝑚, 1  𝑗 ≤ 𝑛; that is all entries are equal to zero.
Properties of Matrix Operations
Theorem 1.3.1: (Properties of Matrix Addition) Let A, B, and C be 𝑚 x 𝑛 matrices then:
(a) A + B = B + A
(commutative law for matrix addition)
(b) A + ( B + C ) = ( A + B ) + C
(associative law for matrix addition)
(c) A + O = A
(The matrix O is the 𝒎 x 𝒏 zero matrix)
(d) there is a unique 𝑚 x 𝑛 matrix (−𝐴) such that
𝐴 + (−𝐴) = 𝑂
We will prove part (a) and leave the proof of the remaining parts as an exercise.
8
Proof of part (a):
Let 𝐴 = [𝑎𝑖𝑗 ] and 𝐵 = [𝑏𝑖𝑗 ] be 𝑚 x 𝑛 matrices. By Definition 1.3.1
(1  𝑖 ≤ 𝑚, 1  𝑗 ≤ 𝑛)
𝐴 + 𝐵 = 𝑎𝑖𝑗 + 𝑏𝑖𝑗
= 𝑏𝑖𝑗 + 𝑎𝑖𝑗
since a, b ∈ 𝑅 then a + b = b + a for any real
numbers a and b.
= 𝐵 + 𝐴
Therefore, matrix addition is commutative.
Example 6: To illustrate part (a) of Theorem 1.3.1, let
 4 1
 4 6
A
and B  


3 2 
10 2 
Then,
 4 1  4 6   8 5 
A B  



 3 2  10 2 13 4
and
 4 6   4 1  8 5 
B A 



10 2   3 2  13 4
Example 7: To illustrate part (c) of Theorem 1.3.1, let
1 2 1
0 0 0 


A   3 2 2 and O  0 0 0 . Note that O is the 3 x 3 zero matrix.
 4 5 3 
0 0 0
Then,
1
𝐴 + 𝑂 = [3
4
2 −1
0 0
2 −2] + [0 0
5
3
0 0
0
1
0] = [3
0
4
Example 8: To illustrate part (d) of Theorem 1.3.1, let
2 2 3 
 2 2 3
A
and  A  

2 
3 1 2 
 3 1
2 −1
2 −2] = 𝐴
5
3
9
Then,
 2 2 3   2 2 3 0 0 0 
A  ( A)  


2 0 0 0 
3 1 2  3 1
Theorem 1.3.2: (Properties of Matrix Multiplication) Let 𝐴 = [𝑎𝑖𝑗 ] be an 𝑚 x 𝑛 matrix, 𝐵 =
𝑏𝑖𝑗 be an 𝑛 x 𝑝 matrix, and 𝐶 = 𝑐𝑖𝑗 be a 𝑝 x 𝑞 matrix. Then
(a) (𝐴𝐵)𝐶 = 𝐴(𝐵𝐶)
(associative law for matrix multiplication)
(b) 𝐴(𝐵 + 𝐶) = 𝐴𝐵 + 𝐴𝐶
(left distributive law for matrix multiplication)
(c) (𝐴 + 𝐵)𝐶 = 𝐴𝐶 + 𝐵𝐶
(right distributive law for matrix multiplication)
Proof of part (a):
Note that AB is an 𝑚 x 𝑝 matrix hence (AB)C is an 𝑚 x 𝑞 matrix. Similarly, BC is an 𝑛 x 𝑞
matrix hence A(BC) is an 𝑚 x 𝑞 matrix. We see that (AB)C and A(BC) have the same size.
Now we must show that the corresponding parts of (AB)C and A(BC) are equal. Let AB = D =
[𝑑𝑖𝑗 ]. Then
𝑑𝑖𝑗 = ∑𝑛𝑘=1 𝑎𝑖𝑘 𝑏𝑘𝑗
The 𝑖𝑗th component of (AB)C = DC is
∑𝑝𝑠=1 𝑑𝑖𝑠 𝑐𝑠𝑗 = ∑𝑝𝑠=1(∑𝑛𝑘=1 𝑎𝑖𝑘 𝑏𝑘𝑠 )𝑐𝑠𝑗
𝑝
= ∑𝑠=1 ∑𝑛𝑘=1 𝑎𝑖𝑘 𝑏𝑘𝑠 𝑐𝑠𝑗
Now we let BC = E = [𝑒𝑖𝑗 ] then
𝑝
𝑒𝑘𝑗 = ∑𝑠=1 𝑏𝑘𝑠 𝑐𝑠𝑗
Thus the 𝑖𝑗th component of A(BC) = AE is
∑𝑛𝑘=1 𝑎𝑖𝑘 𝑒𝑘𝑗 = ∑𝑛𝑘=1 ∑𝑝𝑠=1 𝑎𝑖𝑘 𝑏𝑘𝑠 𝑐𝑠𝑗 = 𝑖𝑗th component of (AB)C.
This shows that (𝐴𝐵)𝐶 = 𝐴(𝐵𝐶).
10
Example 9: To illustrate part (a) of Theorem 1.3.2 we let
1
 2 3 1 0 
2
3 2 4 



A
 , B  0 2 2 2 , and C  0

1

2
3


1 0 1 3 

 2
0 2 
3 0 
1
3 

0 0 
Then,
 8 10 1
18 11
3 2 4 
  4
A( BC )  
0

8
6





 1 2 3   5 1 1   23 23 8 


and
1
3 8   2
10 5
( AB)C  

1 7 2 5 0

 2
0 2 
3 0   4
18 11

1
3   23 23 8 

0 0 
This shows that matrix multiplication is associative.
Proof of part (b):
Let 𝐴 = [𝑎𝑖𝑗 ] be an 𝑚 x 𝑛 matrix and let 𝐵 = [𝑏𝑖𝑗 ] and 𝐶 = [𝑐𝑖𝑗 ] be 𝑛 x 𝑝 matrices. Then
the 𝑘𝑗th element of B + C is
𝑏𝑘𝑗 + 𝑐𝑘𝑗
and the 𝑖𝑗th element of A(B + C) is
∑𝑛𝑘=1 𝑎𝑖𝑘 (𝑏𝑘𝑗 + 𝑐𝑘𝑗 ) = ∑𝑛𝑘=1 𝑎𝑖𝑘 𝑏𝑘𝑗 + ∑𝑛𝑘=1 𝑎𝑖𝑘 𝑐𝑘𝑗
= 𝑖𝑗th component of AB + 𝑖𝑗th component of AC
Thus, A(B + C) = AB + AC. The proof of part (c) is similar to the proof of part (b) and
is left as an exercise.
0
1
0 2 
2 2 4


Example 10: Let A  
, B   2 3  , and C   1 1 

 4 1 3 
5 1 
 2 0 
11
Then
2
1
2 2 4 
 24 16 
A( B  C )  
3 4  



 4 1 3  7 1   20 9


and
18 10 6 6   24 16 
AB  AC  



13 0  7 9  20 9
Theorem 1.3.3: (Properties of Scalar Multiplication) If r and s are real numbers and A and B
are matrices then
(a) r(sA) = (rs)A
(b) (r + s)A = rA + sA
(c) r(A + B) = rA + rB
(d) A(rB) = r(AB) = (rA)B
We prove part (c) of the theorem and leave the proof of the other parts as an exercise.
Proof:
Let 𝐴 = [𝑎𝑖𝑗 ] and 𝐵 = [𝑏𝑖𝑗 ] be 𝑚 x 𝑛 matrices. Then the
𝑖𝑗th entry of 𝑟(𝐴 + 𝐵) = 𝑟(𝑎𝑖𝑗 + 𝑏𝑖𝑗 )
definition of scalar multiplication
= 𝑟𝑎𝑖𝑗 + 𝑟𝑏𝑖𝑗
distributive property of real numbers
= 𝑖𝑗th entry of 𝑟𝐴 + 𝑖𝑗th entry of 𝑟𝐵
definition of scalar
multiplication
Hence 𝑟(𝐴 + 𝐵) = 𝑟𝐴 + 𝑟𝐵.
Example 11: To illustrate part (d) of Theorem 1.3.3 we let
Then,
3
1 3 2
𝑟 = −3, 𝐴 = [
], and 𝐵 = [2
2 −1 3
0
𝐴(𝑟𝐵) = [
3 −4
1
3 2
] ((−3) [2
1 ])
2 −1 3
0
5
−9
12
1
3 2
= [
] [−6 −3 ]
2 −1 3
0 −15
−4
1]
5
12
−27 −27
= [
]
−12 −18
and
3 −4
1
3 2
𝑟(𝐴𝐵) = −3 ([
] [2
1 ])
2 −1 3
0
5
= −3 [
9 9
]
4 6
−27 −27
= [
]
−12 −18
Hence 𝐴(𝑟𝐵) = 𝑟(𝐴𝐵).
1.4 Transpose of a Matrix
Definition 1.4.1: (The Transpose of a Matrix) If 𝐴 = [𝑎𝑖𝑗 ] is an 𝑚 x 𝑛 matrix, then the 𝑛 x 𝑚
matrix AT  [aij ] , where
T
aij  a ji
T
(1  𝑖 ≤ 𝑚, 1  𝑗 ≤ 𝑛)
is called the transpose of A. Thus the transpose of A is obtained by interchanging the rows
and columns of A.
4 −2 −3
Example 1: Let 𝐴 = [
], 𝐵 = [3
0 −5 −2
4
0
𝐴𝑇 = [−2 −5]
−3 −2
2
],
and
𝐶
=
[
−1]. Then
5 −1
4
Note that the first row of A became the first column of 𝐴𝑇 and the second row of A became
the second column of 𝐴𝑇 . Similarly
3
𝐵 𝑇 = [ 5 ] and 𝐶 𝑇 = [2
−1
−1 4]
13
Theorem 1.4.1: (Properties of Transpose) If r is a scalar and A and B are matrices, then
(a) (AT)T = A
(b) (A + B)T = AT + BT
(c) (AB)T = BTAT
(d) (rA)T = rAT
Proof of part (a):
Let 𝐴 = [𝑎𝑖𝑗 ]. Then the 𝑖𝑗th entry of A is
𝑎𝑖𝑗 = 𝑎𝑗𝑖𝑇
definition of transpose of a matrix
= (𝑎𝑇 )𝑇𝑖𝑗
definition of a transpose of a matrix
𝑇 𝑇
Hence (𝐴 ) = 𝐴.
Proof of part (b):
Let [(𝐴 + 𝐵)𝑇 ]𝑖𝑗 be the 𝑖𝑗th component of (𝐴 + 𝐵)𝑇 and (𝐴 + 𝐵)𝑗𝑖 denote the 𝑗𝑖th
component of 𝐴 + 𝐵. Then
[(𝐴 + 𝐵)𝑇 ]𝑖𝑗 = (𝐴 + 𝐵)𝑗𝑖
definition of a transpose
= 𝑎𝑗𝑖 + 𝑏𝑗𝑖
definition of addition of matrices
= (𝐴𝑇 )𝑖𝑗 + (𝐵 𝑇 )𝑖𝑗
definition of a transpose
= 𝑖𝑗th entry of 𝐴𝑇 + 𝑖𝑗th entry of 𝐵 𝑇
Thus (𝐴 + 𝐵)𝑇 = 𝐴𝑇 + 𝐵 𝑇 .
The proof of parts (c) and (d) is left as an exercise.
Example 2: To illustrate part (c) of Theorem 1.4.1 we let
0 1
B   2 2
3 1
2 
1
0 2 3  
0 7 
  0 7 
T T
T

3

1
Then ( AB)  
and B A  




 
 3 3
 1 2 1  2 3  3 3


1 3 2
A
 and
 2 1 3 
14
1.5 Special Types of Square Matrices
Definition 1.5.1: A square matrix 𝐴 = [𝑎𝑖𝑗 ] for which every term off the main diagonal is
zero, that is, 𝑎𝑖𝑗 = 0 for 𝑖 ≠ 𝑗, is called a diagonal matrix.
Example 1: The following are diagonal matrices.
 5 0 0
3 0 
and H   0 8 0 
G

0 1
 0 0 3
Definition 1.5.2: A diagonal matrix 𝐴 = [𝑎𝑖𝑗 ], for which all terms on the main diagonal are
equal, that is, 𝑎𝑖𝑗 = 𝑐 for 𝑖 = 𝑗 and 𝑎𝑖𝑗 = 0 for 𝑖 ≠ 𝑗, is called a scalar matrix.
Example 2: The following are examples of scalar matrices.
3
𝐵 = [0
0
0 0
−7
0
]
3 0] and 𝐶 = [
0 −7
0 3
Definition 1.5.3: A matrix 𝐴 = [𝑎𝑖𝑗 ] is called upper triangular if a ij  0 for i > j. It is called
lower triangular if a ij  0 for i < j.
Definition 1.5.4: A matrix 𝐴 = [𝑎𝑖𝑗 ] is called strictly upper triangular if a ij  0 for i
is called strictly lower triangular if a ij  0 for i
Example 3:
0
𝐷 = [0
0
8
0
0
j. It
j.
 2 0 0
1 3 5 
0 0 0




Let A = 0 2  1 , B =  5  1 0 , 𝐶 = [2 0 0] and
3 7 0
 6  3 4
0 0  4
4
−9]
0
Matrix A is upper triangular since all entries below the main diagonal are zero. Matrix B is
lower triangular since all entries above the main diagonal are zero while C and D are strictly
lower triangular and strictly upper triangular, respectively. Strictly upper/lower triangular
matrices are upper/ lower triangular matrices with all entries on the main diagonal equal to
zero.
15
There are two important types of square matrices that can be defined in terms of
the transpose operation. These are symmetric and skew symmetric matrices which we
define as follows:
Definition 1.5.5: A matrix 𝐴 = [𝑎𝑖𝑗 ] is called symmetric if AT = A.
 1 2 3
1 −2 −3


𝑇
Example 4: Let A   2 5 0  then 𝐴 = [−2
5
0 ]. Since 𝐴𝑇 = 𝐴 then A is
−3
0
4
 3 0 4 
symmetric.
SAQ 1-2
Let 𝐴 and 𝐵 be symmetric matrices. Show that 𝐴 + 𝐵 is symmetric.
ASAQ 1-2
(𝐴 + 𝐵)𝑇 = 𝐴𝑇 + 𝐵 𝑇
=𝐴+𝐵
Part (b) of Theorem 1.4.1
Definition of symmetric matrix
Hence 𝐴 + 𝐵 is symmetric.
Definition 1.5.6: A matrix 𝐴 = [𝑎𝑖𝑗 ] is called skew symmetric if 𝐴𝑇 = −𝐴.
 0 3 2 
Example 5: The matrix B   3 0 4  is skew symmetric because
 2 4 0 
0 3
𝐵 𝑇 = [−3 0
2 4
−2
−4] = −𝐵.
0
16
SAQ 1-3
Let 𝐴 and 𝐵 be skew-symmetric matrices. Show that 𝐴 + 𝐵 is skew-symmetric.
ASAQ 1-3
(𝐴 + 𝐵)𝑇 = 𝐴𝑇 + 𝐵 𝑇
= (−𝐴) + (−𝐵)
= −(𝐴 + 𝐵)
Part (b) of Theorem 1.4.1
Definition of skew-symmetric matrix
Thus 𝐴 + 𝐵 is skew-symmetric.
Definition 1.5.7: (The Identity Matrix) The 𝑛 x 𝑛 matrix 𝐼𝑛 = [𝑎𝑖𝑗 ] defined by 𝑎𝑖𝑗 = 1 if 𝑖 =
𝑗, 𝑎𝑖𝑗 = 0 if 𝑖 ≠ 𝑗, is called the 𝑛 x 𝑛 identity matrix of order 𝒏.
Example 6: Below are examples of identity matrices.
1 0 0 
I 3  0 1 0
0 0 1 
1
0
I4  
0

0
is a 3 x 3 identity matrix of order 3
0 0 0
1 0 0 
is a 4 x 4 identity matrix of order 4
0 1 0

0 0 1
Note that an identity matrix is a scalar matrix where all the elements on the main diagonal
are 1.
17
Remarks:
The identity matrix 𝐼𝑛 functions for 𝑛 x 𝑛 matrices the way the number 1 functions
for real numbers. In other words, the identity matrix 𝐼𝑛 is actually a multiplicative identity
for 𝑛 x 𝑛 matrices. That is
𝐴𝐼𝑛 = 𝐼𝑛 𝐴 = 𝐴
for every 𝑛 x 𝑛 matrix A.
4 0 1
Example 7: Let A   5 3 2  . Then
 2 1 4 
 4 0 1  1 0 0   4 0 1 
AI3  5 3 2 0 1 0  5 3 2   A
 2 1 4 0 0 1   2 1 4
and
1 0 0   4 0 1   4 0 1 
I 3 A  0 1 0 5 3 2  5 3 2  A
0 0 1   2 1 4  2 1 4
Hence, AI3 = I3A = A
Suppose that A is a square matrix. If p is a positive integer, then we define
1. Ap = A.A…A
p factors
If A is n x n, we also define
2. A0 = In
3. ApAq = Ap+q
4. (Ap)q = Apq
NOTE:
1. The rule (AB)P = APBP holds only if AB = BA.
2. The rule “if AB = 0 then either A = 0 or B = 0” is not true for matrices.
18
2
Example 8: Let A  
 8
Then,
2
AB  
 8
3
3 6
and B  


12 
 2 4
3  3 6  0 0

12   2 4 0 0
ACTIVITY
1
1. Let 𝐴 = [
3
2 0
3 5 1
2 4
3 5
], 𝐵 = [3 4 ], 𝐶 = [−2 7 2 ], 𝐷 = [
],
−1 5
2 −4
1 −5
6 4 −3
2 4 5
−5 4
𝐸 = [0 1 4], and 𝐹 = [
].
2 3
3 −2 1
If possible, compute
a. ( BT + A )C
b. AB
c. 2D – 3F
d) ( C + E )T
e) AB + DF
f) ( 3C – 2E )TB
2. Let 𝐴 and 𝐵 be skew-symmetric matrices, show 𝐴𝐵 is symmetric if and only if
𝐴𝐵 = 𝐵𝐴.
3. Let 𝐴 be an 𝑛 x 𝑛 matrix. Show that
a. 𝐴 + 𝐴𝑇 is symmetric.
b. 𝐴 − 𝐴𝑇 is skew-symmetric.
c. 𝐴𝐴𝑇 and 𝐴𝑇 𝐴 are symmetric.
 4 2
4. Let A  
 . Find
1 3 
a) A2 + 3A
b) 2A3 + 3A2 + 4A + 5I2
19
MODULE 2
LINEAR EQUATIONS AND MATRICES
Introduction
In this lesson we will discuss the Gauss-Jordan reduction method and the Gaussian
Elimination method and their application to the solution of linear systems, inverse of a
matrix and the practical method for finding the inverse of a matrix, determinants and its
properties, cofactor expansion and Cramer’s rule.
Objectives
After going through the lessons in this chapter, you are expected to be able to do the
following:
1. Explain solution of system of linear equations by substitution and by elimination through
simultaneous equations.
2. Transform a given matrix into a row-echelon and reduced row-echelon form.
3. Solve systems of linear equations using the Gauss-Jordan reduction method and the
Gaussian Elimination method.
4. Define homogenous systems.
5. Differentiate between singular and non-singular matrices.
6. Enumerate the properties of the Inverse.
7. Find the inverse of a given matrix.
8. Solve systems of linear equations using the inverse of a matrix.
9. Define permutation.
10. Evaluate the determinant of a matrix using permutation.
11. Reduce the problem of evaluating determinants by co-factor expansion.
12. Find the inverse of a matrix using determinants and co-factor expansion.
13. Solve systems of linear equations by using the Cramer’s Rule.
2.1 Solutions of Systems of Linear Equations
In this section we introduce a systematic technique of eliminating unknowns in
solving systems of linear equations. To illustrate this technique, let us consider the following
system:
Example 1:
3𝑥1 + 2𝑥2 − 𝑥3 = −2
2𝑥1 − 3𝑥2 + 2𝑥3 = 14
𝑥1 + 2𝑥2 + 3𝑥3 = 6
(1)
20
Interchanging the first and third equations gives us
𝑥1 + 2𝑥2 + 3𝑥3 = 6
2𝑥1 − 3𝑥2 + 2𝑥3 = 14
3𝑥1 + 2𝑥2 − 𝑥3 = −2
(2)
If we multiply the first equation in (2) by – 2 and add the result to the second equation we
get
−2𝑥1 − 4𝑥2 − 6𝑥3 = −12
2𝑥1 − 3𝑥2 + 2𝑥3 = 14
−7𝑥2 − 4𝑥3 = 2
The new equation obtained may replace either the first or second equation in system (2),
(the two equations used to obtain it). Next let us multiply the first equation in (2) by – 3
and add the result to the third equation. This gives us
−3𝑥1 − 6𝑥2 − 9𝑥3 = −18
3𝑥1 + 2𝑥2 − 𝑥3 = −2
or
−5𝑥2 − 10𝑥3 = −20
𝑥2 + 2𝑥3 = 4
(dividing both sides of the
equation by –5 )
The new equation obtained may replace either the first or the third equation in system (2).
Thus if we replace the second equation in (2) by −7𝑥2 − 4𝑥3 = 2 and the third equation
by 𝑥2 + 2𝑥3 = 4, we obtain the new system:
𝑥1 + 2𝑥2 + 3𝑥3 = 6
−7𝑥2 − 4𝑥3 = 2
𝑥2 + 2𝑥3 = 4
(3)
Note that the variable 𝑥1 has been eliminated from the second and third equations
of system (3). This new system is equivalent to the original system (1).
Next we multiply the third equation of (3) by 7 and add the result to the second
equation. This gives us
7 𝑥2 + 14𝑥3 = 28
−7𝑥2 − 4𝑥3 = 2
or
10𝑥3 = 30
𝑥3 = 3
(dividing both sides of the
equation by 10)
21
The new equation obtained may replace either the second or third equation in (3).
Replacing the third equation we get an equivalent system
𝑥1 + 2𝑥2 + 3𝑥3 = 6
−7𝑥2 − 4𝑥3 = 2
𝑥3 = 3
(4)
To solve for the variable 𝑥2 we substitute 𝑥3 = 3 to the second equation in (4):
−7𝑥2 – 4(3) = 2
𝑥2 = −2
To solve for the variable 𝑥1 , we substitute 𝑥2 = −2 and 𝑥3 = 3 to the first equation in
(4):
𝑥1 + 2(−2) + 3(3) = 6
𝑥1 = 1
Thus the solution to the system (1) is the ordered triple (1, 2, 3).
The method used here is called the Gaussian elimination. The objective of Gaussian
elimination is to reduce a given system to triangular or echelon (staircase pattern) form
and then use back substitution to find the solution of the system. The Gaussian elimination
was modified by Carl Friedrich Gauss (1777 – 1855) and Wilhelm Jordan (1842 – 1899). They
called it the Gauss-Jordan elimination. To illustrate it, let us start from system (3) of
Example 1.
Example 2:
𝑥1 + 2𝑥2 + 3𝑥3 = 6
−7𝑥2 − 4𝑥3 = 2
𝑥2 + 2𝑥3 = 4
(3)
Interchange the second and third equations in (3):
𝑥1 + 2𝑥2 + 3𝑥3 = 6
𝑥2 + 2𝑥3 = 4
−7𝑥2 − 4𝑥3 = 2
(4)
Now let us multiply the second equation in (4) by – 2 and add the result to the first
equation. This gives us
𝑥1 + 2𝑥2 + 3𝑥3 = 6
−2 𝑥2 − 4𝑥3 = −8
𝑥1
− 𝑥3 = −2
22
Next we multiply the second equation in (4) by 7 and add the result to the third equation.
This gives us
7 𝑥2 + 14𝑥3
−7𝑥2 − 4𝑥3
10𝑥3
or
𝑥3
=
=
=
=
28
2
30
3
Replacing the first equation in (4) by 𝑥1 − 𝑥3 = −2 and the third equation by 𝑥3 = 3, we
obtain the new system
𝑥1
− 𝑥3 = −2
𝑥2 + 2𝑥3 = 4
𝑥3 = 3
(5)
Note that the variable 𝑥2 has been eliminated from the first and third equations of (5). This
time let us add the first equation in (5) to the third equation. This gives us
𝑥1
𝑥1
− 𝑥3 = −2
𝑥3 = 3
= 1
Finally we multiply the third equation in (5) by – 2 and add the result to the second
equation.
𝑥2 + 2𝑥3 = 4
−2𝑥3 = 3
𝑥2
= −2
System (5) now becomes
𝑥1
𝑥2
= 1
= −2
𝑥3 = 3
Thus the solution to the system is the ordered triple (1, -2, 3). The Gauss-Jordan elimination
is an algorithm that reduces a given system to reduced row echelon form or row canonical
form. This method is less efficient than the Gaussian elimination in solving systems of
equations however, it is well suited for calculating the inverse of a matrix which we will
discuss in a later section.
23
2.2 Elementary Row Operations
A system of linear equations can be placed in matrix form. This notation is very
efficient especially when we are dealing with large systems. To begin, let us consider an 𝑚 x
𝑛 system of linear equations
𝑎11 𝑥1 + 𝑎12 𝑥2 + … + 𝑎1𝑛 𝑥𝑛 = 𝑏1
𝑎21 𝑥1 + 𝑎22 𝑥2 + … + 𝑎2𝑛 𝑥𝑛 = 𝑏2
⋮
𝑎𝑚1 𝑥1 + 𝑎𝑚2 𝑥2 + … + 𝑎𝑚𝑛 𝑥𝑛 = 𝑏𝑚
(1)
The coefficients of this linear system can be written in a rectangular array having 𝑚 rows
and 𝑛 columns, and we designate this array as 𝐴:
𝑎11 𝑎12 … 𝑎1𝑛
𝑎21 𝑎22 … 𝑎2𝑛
⋮
⋮
⋮
𝐴=
𝑎𝑚1 𝑎𝑚2 … 𝑎𝑚𝑛
[
]
This 𝑚 x 𝑛 matrix 𝐴 is called the coefficient matrix for the given system (1) above. If we add
another column in 𝐴 to include the constants b1, b2, …, bm , we will have a matrix that
expresses compactly all the relevant information contained in (1). Such a matrix is called the
augmented matrix for (1), and is usually denoted as [A│B]. If we let C = [A│B] be the
augmented matrix for the system (1), then C is the 𝑚 x ( 𝑛+1 ) matrix given by
𝑎11 𝑎12 … 𝑎1𝑛 𝑏1
𝑎21 𝑎22 … 𝑎2𝑛 𝑏2
C=
⋮
⋮
⋮
⋮
𝑎𝑚1 𝑎𝑚2 … 𝑎𝑚𝑛 𝑏𝑚
[
]
We also write 𝐴𝑋 = 𝐵, where
𝑎11 𝑎12 … 𝑎1𝑛
𝑎21 𝑎22 … 𝑎2𝑛
⋮
⋮
⋮
𝐴=
and
𝑎𝑚1 𝑎𝑚2 … 𝑎𝑚𝑛
[
]
represent the coefficients and
𝑥1
𝑥2
𝑋=[ ⋮ ]
𝑥𝑛
𝑏1
𝑏2
𝐵=[ ]
⋮
𝑏𝑚
24
represents the variables. For example, the array
1 −3 0 2
[0
4 5 −1
2 −1 3 0
2
−3]
4
represents the system of 3 linear equations
𝑥1 − 3𝑥2
+ 2𝑥4 = 2
4𝑥2 + 5𝑥3 − 𝑥4 = −3
2𝑥1 − 𝑥2 + 3𝑥3
= 4
The task of this section is to manipulate the augmented matrix representing a given
linear system into reduced row echelon form. But before we continue, let us introduce
some terminology. We have seen from Examples 1 and 2 that three important operations
were applied to solve the given system. These are: multiplying or dividing both sides of an
equation by a nonzero number, adding a multiple of one equation to another equation, and
interchanging two equations. These three operations when applied to the rows of an
augmented matrix are called elementary row operations and will result in the matrix of an
equivalent system.
Definition 2.2.1: An elementary row operation on an 𝑚 x 𝑛 matrix 𝐴 = [𝑎𝑖𝑗 ] is any one of
the following operations:
1. Interchange rows 𝑖 and 𝑗 of A.
2. Multiply row 𝑖 of 𝐴 by any nonzero real number.
3. Add 𝑐 times row 𝑖 of 𝐴 to row 𝑗 of 𝐴, 𝑖 ≠ 𝑗.
The process of applying elementary row operations to simplify an augmented matrix
is called row reduction. The following notations will be used to denote the elementary row
operation used:
1. 𝑅𝑖
𝑅𝑗 means interchanging the 𝑖th row and the 𝑗th row
2. 𝑐𝑅𝑖
𝑅𝑖 means that the 𝑖th row is replace with 𝑐 times the 𝑖th row.
3. 𝑐𝑅𝑖 + 𝑅𝑗
𝑅𝑗 means that the 𝑗th row is replaced with the sum of the 𝑗th row
and the 𝑖th row multiplied by 𝑐.
Before we illustrate the three elementary operations, let us define first a matrix in
reduced row echelon form.
25
Definition 2.2.2: An 𝑚 x 𝑛 matrix is said to be in reduced row echelon form when it satisfies
the following properties:
a) All rows consisting entirely of zeros, if any, are at the bottom of the matrix.
b) The first nonzero entry in each row that does not consist entirely of zeros is a 1,
called the leading entry(pivot) of its row.
c) If rows 𝑖 and 𝑖 + 1 are two successive rows that do not consist entirely of zeros, then
the leading entry of row 𝑖 + 1 is to the right of the leading entry of row 𝑖.
d) If a column contains a leading entry of some row, then all other entries in that
column are zero.
Note that a matrix in reduced row echelon form might not have any rows that consist
entirely of zeros.
Example 1: Below are matrices in reduced row echelon form:
1 0 0 4 
A  0 1 0 3 
0 0 1 2 
 The first nonzero entry in each row is 1 called the
pivot.
 The leading entry (pivot) for row 2 is to the right of
the leading entry for row 1 and the leading entry for
row 3 is to the right of the leading entry for row 2.
 All other entries for the columns containing a leading
entry are zero.
 There are entries other than zero in column 4 since
there is no pivot in that column.
1 2 0 0
B  0 0 1 0
0 0 0 1 
 The leading entry for row 2 appears in column 3 and
the leading entry for row 3 appears in column 4.
 Column 2 does not contain a pivot.
1
0
C
0

0
 All rows consisting of zeros are at the bottom of the
matrix.
 Columns 1 and 3 contain a pivot hence all other
entries in those columns are zero.
0 0 1/ 3
0 1 0 
0 0 0 

0 0 0 
26
Example 2: Below are matrices not in reduced row echelon form:
1 2 0 4
D  0 0 0 0
0 0 1 3 
 Fails property (a), the row consisting entirely of zeros
must be at the bottom of the matrix.
 Fails property (b), the first nonzero entry in row 2 is
not equal to 1.
 Fails property (d), column 3 contains a leading entry
but the other entries in that column are not equal to
zero.
4
1 0 3
E  0 2  2 5 
0 0 1
2
1
0
F
0

0
0 3 4
1  2 5
1 2 2

0 0 0
 Fails property (c), the leading entry of the third row
must be to the right of the leading entry of row 2.
Example 3: The three elementary operations are illustrated below:
0 1 0 2 
Let A  2 3 0  4
2 1 6 9 
(a) Interchanging rows 1 and 3 of A, (𝑅1
𝑅3 ) , we obtain:
2 1 6 9 
B  2 3 0  4
0 1 0 2 
(b) Replacing row 1 of B by ½ times row 1 of B, ( 12𝑅1
1
9
1 2 3 2
𝐶 = [ 2 3 0 − 4]
0 1 0 2
𝑅1 ) , we obtain:
27
(c) Replacing row 2 by the sum of -2 times row 1 of 𝐶 plus row 2 of 𝐶,
(−2𝑅1 + 𝑅2
𝑅2 ) we obtain:
1
9
1 2
3
2
𝐷 = [ 0 2 − 6 − 13]
0 1
0 2
Definition 2.2.3: An 𝑚 x 𝑛 matrix 𝐴 is said to be row equivalent to an 𝑚 x 𝑛 matrix 𝐵 if 𝐵
can be obtained by applying a finite sequence of elementary row operations to 𝐴.
Hence in Example 3, 𝐴 is row equivalent to 𝐷 because 𝐷 is obtained from 𝐴 after
applying a series of elementary row operations on A.
NOTE:
1. Every matrix is row equivalent to itself.
2. If A is row equivalent to B, then B is row equivalent to A.
3. If A is row equivalent to B and B is row equivalent to C, then A is row equivalent to C.
Theorem 2.2.1: Every nonzero 𝑚 x 𝑛 matrix is row equivalent to a unique matrix in reduced
row echelon form.
2.3 The Gauss-Jordan Method
The Gauss-Jordan method is a systematic technique for applying matrix row
transformations in an attempt to reduce a matrix to diagonal form also called reduced row
echelon form.
The following are steps in using the Gauss-Jordan method to put a matrix into reduced row
echelon form:
STEP 1: Form the augmented matrix [ A│B ].
STEP 2: Obtain 1 as the first element of the first column.
STEP 3: Use the first row to transform the remaining entries in the first column to 0.
STEP 4: Obtain 1 as the second entry in the second column.
STEP 5: Use the second row to transform the remaining entries in the second column to 0.
STEP 6: Continue in this manner as far as possible.
28
0 0 1 2 
Example 1: Transform A  2 3 0  2 to reduced row echelon form.
3 3 6  9 
To transform A to reduced row echelon form, we perform Gauss-Jordan elimination
method. The sequence of elementary row operations, beginning with A, follows.
Interchange rows 1 and 3 of 𝐴 to obtain 𝐴1
0 0 1 2
𝐴 = [ 2 3 0 −2 ]
3 3 6 −9
𝑅1
𝑅3
3 3 6 −9
[2 3 0 −2] = 𝐴1
0 0 1 2
(Multiply row 1 of 𝐴1 by 1/3 to obtain 𝐴2 )
1
3
𝑅1
𝑅1
1
[2
0
1 2 −3
3 0 −2] = 𝐴2
0 1 2
(Multiply row 1 of 𝐴2 by -2 and add to row 2 to obtain 𝐴3 )
−2𝑅1 + 𝑅2
𝑅2
1
[0
0
1 2 −3
1 −4 4 ] = 𝐴3
0 1
2
(Multiply row 2 of 𝐴3 by -1 and add to row 1 to obtain 𝐴4 )
1𝑅2 + 𝑅1
𝑅1
1 0
[0 1
0 0
6 −7
−4 4 ] = 𝐴4
1
2
Multiply row 3 of 𝐴4 by −6 and add to row 1; multiply row 3 of 𝐴4 by 4 and add to row 2 to
obtain 𝐴5 )
−6𝑅3 + 𝑅1
4𝑅3 + 𝑅2
𝑅1
𝑅3
1 0 0 −19
[0 1 0 12 ] = 𝐴5
0 0 1
2
𝐴5 is already in reduced row echelon form.
We now apply these results to the solution of linear system.
29
Theorem 2.3.1. Let AX = B and CX = D be two linear systems each of 𝑚 equations in 𝑛
unknowns. If the augmented matrices [ A│B ] and [ C│D ] of these systems are row
equivalent, then both linear systems have exactly the same solutions.
Example 2. Solve the linear system
x + y + 2z = -1
x – 2y + z = -5
3x + y + z = 3
by Gauss-Jordan reduction.
Solution: Let the augmented matrix be
1
𝐴 = [1
3
1 2
−2 1
1 1
−1
−5]
3
Applying the Gauss-Jordan elimination we have
1
1 2
[1 −2 1
3
1 1
−1 𝑅 + 𝑅
1
2
−5] 3𝑅 + 𝑅
1
3
3
1
− 3 𝑅2
−𝑅2 + 𝑅1
2𝑅2 + 𝑅3
3
− 13 𝑅3
5
− 3 𝑅3 + 𝑅1
1
− 3 𝑅3 + 𝑅2
𝑅2
𝑅3
1
[0
0
1
2
−3 −1
−2 −5
𝑅2
1
[0
0
1
2
1 1/3
−2 −5
1
[0
0
0
5/3
−7/3
1
1/3
4/3 ]
0 −13/3 26/3
𝑅1
𝑅3
𝑅3
𝑅1
𝑅2
1
[0
0
1
[0
0
0
1
0
0
1
0
−1
−4]
6
5/3
1/3
1
0
0
1
−1
4/3]
6
−7/3
4/3 ]
−2
1
2]
−2
The last matrix is in reduced row echelon form and is row equivalent to the augmented
matrix representing the given system. As an augmented matrix, it represents the system
x
y
z
= 1
= 2
= -2
30
Thus the solution set is {(1, 2, - 2)}. The given system is consistent.
Example 3: Solve the linear system
𝑥1 + 𝑥2 − 𝑥3 = 7
4𝑥1 − 𝑥2 + 5𝑥3 = 4
6𝑥1 + 𝑥2 + 3𝑥3 = 0
Solution: Let the augmented matrix be
1
[4
6
1 −1
−1
5
1
3
7
4]
0
Applying the Gauss-Jordan elimination we have
1
1 −1
[4 −1
5
6
1
3
7 −4𝑅 + 𝑅
4] −6𝑅1 + 𝑅2
1
3
0
1
𝑅2 + 𝑅1
−𝑅2 + 𝑅3
5
1
− 5 𝑅2
𝑅2 1
[
𝑅3 0
0
1 −1
−5
9
−5
9
𝑅1
𝑅3
1
0 4/5
[0 −5
9
0
0
0
𝑅2
1
[0
0
0
1
0
7
−24]
−42
11/5
−24 ]
−18
4/5
−9/5
0
11/5
24/5 ]
−18
The last equation reads 0𝑥1 + 0𝑥2 + 0𝑥3 = −18, which is impossible since 0 ≠ −18.
Hence the given system has no solution. In this case the system is said to be inconsistent.
Example 4: Solve the system
2𝑥1 + 6𝑥2 − 4𝑥3 + 2𝑥4 = 4
𝑥1
− 𝑥3 + 𝑥4 = 5
−3𝑥1 + 2𝑥2 − 2𝑥3
= −2
Solution: Applying the Gauss-Jordan elimination to the augmented matrix, we have
2 6 −4
[ 1 0 −1
−3 2 −2
2
1
0
4
5]
−2
𝑅1
𝑅2
1 0 −1
[ 2 6 −4
−3 2 −2
1
2
0
5
4]
−2
31
𝑅2
𝑅3
1 0
[ 0 6
0 2
𝑅2
𝑅2
1
[ 0
0
−2𝑅2 + 𝑅3
𝑅3
1 0
[ 0 1
0 0
𝑅3
1
1 0
−1
0
[ 0 1 −1/3
0 0
1 −9/13
−2𝑅1 + 𝑅2
3𝑅1 + 𝑅3
1
6
3
− 13 𝑅3
𝑅3 + 𝑅1
1
3
𝑅3 + 𝑅2
𝑅1
𝑅2
−1
−2
−5
1
0
3
0
−1
1 −1/3
2
−5
5
−6]
13
1
0
3
5
−1]
13
−1
1
−1/3 0
−13/3 3
5
−1]
15
1 0 0 4/13
[ 0 1 0 3/13
0 0 1 −9/13
5
−1 ]
−45/13
20/13
−28/13]
−45/13
Column 4 has no pivot hence 𝑥4 is a free variable. If we let 𝑥4 = 𝑟, 𝑟 ∈ 𝑅, then we obtain:
𝑥1 =
20
13
−
4
13
𝑟 , 𝑥2 = −
28
13
20
Thus the solution set is in the form of (
13
−
−
3
13
4
13
𝑟 , and 𝑥3 = −
𝑟, −
28
13
−
3
13
45
13
𝑟, −
+
45
13
9
13
+
𝑟.
9
13
𝑟, 𝑟), 𝑟 ∈
𝑅. The given system has infinitely many solution.
Corollary 2.3.1. If A and C are row equivalent 𝑚 x 𝑛 matrices, then the linear systems
AX = 0 and CX = 0 have exactly the same solutions.
SAQ 2-1
Consider the system
2𝑥1 − 𝑥2 + 3𝑥3 = 𝑎
3𝑥1 + 𝑥2 − 5𝑥3 = 𝑏
−5𝑥1 − 5𝑥2 + 21𝑥3 = 𝑐
Show that the system is inconsistent if 𝑐 ≠ 2𝑎 − 3𝑏.
32
ASAQ 2-1
Reducing the augmented matrix of the given system to row echelon form, we have
2 −1 3
[3
1 −5
−5 −5 21
𝑎
𝑏]
𝑐
1 −1/2
𝑅1 [ 3
1
−5
−5
1
𝑅
2 1
−3𝑅1 + 𝑅2
5𝑅1 + 𝑅3
𝑅2
𝑅3
1
[0
0
3/2
−5
21
−1/2
3/2
5/2
−19/2
−15/2 57/2
𝑎/2
𝑏 ]
𝑐
𝑎/2
−3𝑎+2𝑏
2
5𝑎+2𝑐
]
2
3𝑅2 + 𝑅3
𝑅3
1
[0
0
−1/2
3/2
5/2 −19/2
0
0
𝑎/2
−3𝑎+2𝑏
2
]
−2𝑎 + 3𝑏 + 𝑐
The last equation reads 0𝑥1 + 0𝑥2 + 0𝑥3 = −2𝑎 + 3𝑏 + 𝑐. The given system has no
solution if −2𝑎 + 3𝑏 + 𝑐 ≠ 0. Thus the system is inconsistent if 𝑐 ≠ 2𝑎 − 3𝑏.
2.4 Homogeneous Systems
A linear system of the form
a11 x1  a12 x2  ...  a1n xn  0
a21 x1  a22 x2  ...  a2 n xn  0
(2)
am1 x1  am 2 x2  ...  amn xn  0
is called a homogeneous system. We can write it in matrix form as
𝐴𝑋 = 0
The solution to the homogeneous system (2) is called the trivial solution if
x1  x2  ...  xn  0 and a solution x1 , x2 ,..., x n in which not all the xi are zero is called a
nontrivial solution.
33
Example 1. Consider the homogeneous system
𝑥 + 2𝑦 + 𝑧 = 0
2𝑥 + 5𝑦 + 3𝑧 = 0
3𝑥 + 7𝑦 + 5𝑧 = 0
The augmented matrix of this system is
1 2
[2 5
3 7
1
3
5
0
0]
0
which is row equivalent to (you must do the calculation here)
1 0
[0 1
0 0
0
0
1
0
0]
0
Hence the solution to the given homogeneous system is 𝑥 = 𝑦 = 𝑧 = 0 (a trivial solution).
SAQ 2-2
Find the solution of the homogeneous system
𝑥 + 𝑦 + 𝑧 + 𝑤 = 0
2𝑥 + 𝑦
𝑧
= 0
3𝑥 + 𝑦 + 𝑧 + 2𝑤 = 0
ASAQ 2-2
The augmented matrix of this system is
1 1 1 1
[2 1 −1 0
3 1 1 2
0
0]
0
34
Transforming it to reduced row echelon form we have
1 1 1 1
[2 1 −1 0
3 1 1 2
0 −2𝑅 + 𝑅
0] −3𝑅1 + 𝑅2
1
3
0
−𝑅2
1
𝑅2 1 1
[
𝑅3 0 −1 −3
0 −2 −2
𝑅2
1
[0
0
1
1
1
3
−2 −2
1 0
−2 0]
−1 0
1 0
2 0]
−1 0
𝑅1 1
[
𝑅3 0
0
0 −2
1 3
0 4
−1 0
2 0]
3 0
𝑅
4 3
1
𝑅3 [0
0
0 −2
1 3
0 1
−1 0
2
0]
3/4 0
2𝑅3 + 𝑅1
−3𝑅3 + 𝑅2
1
𝑅1
[
𝑅2 0
0
−𝑅2 + 𝑅1
2𝑅2 + 𝑅3
1
0 0
1 0
0 1
1/2 0
−1/4 0]
3/4 0
The fourth column has no pivot hence 𝑤 is a free variable. If we let 𝑤 = 𝑠, 𝑠 ∈ 𝑅, then the
solution of the given homogeneous system is
1
𝑥 = −2𝑠
𝑦 =
1
4
𝑠
3
𝑧 = −4𝑠
𝑤 = 𝑠
1
1
3
where 𝑠 is any real number. Thus the solution set is in the form of {(− 2 𝑠, 4 𝑠, − 4 𝑠, 𝑠)},
𝑠 ∈ 𝑅. This shows that the system has a nontrivial solution. This result is generalized in the
next theorem.
Theorem 2.4.1: A homogeneous system of 𝑚 equations in 𝑛 unknowns always has a
nontrivial solution if 𝑚 < 𝑛, that is , if the number of unknowns exceeds the number of
equations.
35
ACTIVITY
1. Find all solutions to the given linear systems.
a. 𝑥1 − 𝑥2 − 𝑥3 = 0
b. 2𝑥1 + 3𝑥2 − 𝑥3 = 0
2𝑥1 + 𝑥2 + 2𝑥3 = 0
−4𝑥1 + 2𝑥2 + 𝑥3 = 0
𝑥1 − 4𝑥2 − 5𝑥3 = 0
7𝑥1 + 3𝑥2 − 9𝑥3 = 0
c. 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 6
2𝑥1
− 𝑥3 − 𝑥4 = 4
3 𝑥3 + 6𝑥4 = 3
𝑥1
− 𝑥4 = 5
2. Find all values of 𝑎 for which the resulting system has (a) no solution, (b) a unique
solution, and (c) infinitely many solutions.
𝑥1 + 𝑥2 − 𝑥3 = 3
𝑥1 − 𝑥2 + 3𝑥3 = 4
𝑥1 + 𝑥2 + (𝑎2 − 10)𝑥3 = 𝑎
2.5 The Inverse of a Matrix
When working with real numbers, we know that a number 𝑎 times its inverse is one
provided that 𝑎 ≠ 0. Thus the equation 𝑎𝑥 = 𝑏 could be solved for 𝑥 by dividing both sides
(or multiplying both sides by 1/𝑎) of the equation by 𝑎 to get 𝑥 = 𝑏/𝑎 provided that 𝑎 ≠ 0.
However the matrix equation 𝐴𝑋 = 𝐵 cannot be divided by the matrix 𝐴 on both sides
because there is no matrix division.
This section is concerned on finding a matrix whose function is similar to the inverse
of a real number 𝑎. We like to find a matrix inverse such that 𝐴 times its inverse is equal to
“one”. The matrix equivalent of “one” is called the “identity matrix”.
Definition 2.5.1: An 𝑛 x 𝑛 matrix 𝐴 is called nonsingular (or invertible) if there exists an 𝑛 x
𝑛 matrix 𝐵 such that
𝐴𝐵 = 𝐵𝐴 = 𝐼𝑛
The matrix 𝐵 is called an inverse of 𝐴. If there exists no such matrix 𝐵, then 𝐴 is called
singular (or non-invertible).
36
−2
Example 1: Let 𝐴 = [
3
−4/11
1
] and 𝐵 = [
3/11
4
−2 1 −4/11
𝐴𝐵 = [
][
3 4 3/11
−4/11
𝐵𝐴 = [
3/11
1/11
] then,
2/11
1/11
1 0
] = [
] = 𝐼2
2/11
0 1
1/11 −2
][
2/11
3
1
1
]= [
4
0
and
0
] = 𝐼2
1
Since 𝐴𝐵 = 𝐵𝐴 = 𝐼2 , we conclude that 𝐵 is an inverse of 𝐴 and that 𝐴 is nonsingular.
4
Example 2: Let 𝐴 = [
−3
Solution: Let 𝐴−1 = [
𝑎
𝑐
2
], find 𝐴−1 .
1
𝑏
] then
𝑑
𝐴𝐴−1 = 𝐼2
[
[
4 2 𝑎
][
−3 1 𝑐
4𝑎 + 2𝑐
−3𝑎 + 𝑐
1 0
𝑏
]= [
]
0 1
𝑑
4𝑏 + 2𝑑
1
]= [
0
−3𝑏 + 𝑑
0
]
1
Two matrices are equal if and only if their corresponding parts are equal. Hence if we
equate the corresponding parts, we get
4𝑎 + 2𝑐 = 1
−3𝑎 + 𝑐 = 0
and
4𝑏 + 2𝑑 = 0
−3𝑏 + 𝑑 = 1
Solving the system we have a = 1/10, b = -1/5, c = 3/10, and d = 2/5. Thus
𝐴−1 = [
1/10
3/10
−1/5
]
2/5
You can always check your answer by taking the product 𝐴𝐴−1 and making sure that the
answer is the identity matrix 𝐼2 .
There is a simple procedure for finding the inverse of a 2 x 2 matrix. It can be done
easily as follows:
Let 𝐴 = [
𝑎
𝑐
𝑤
𝑏
]. We are looking for a matrix [ 𝑦
𝑑
𝑥
𝑧 ] such that
37
[
[
𝑎
𝑐
𝑎𝑤 + 𝑏𝑦
𝑐𝑤 + 𝑑𝑦
𝑏 𝑤
][
𝑑 𝑦
𝑥
1 0
𝑧 ] = [0 1]
𝑎𝑥 + 𝑏𝑧
1 0
] = [
]
𝑐𝑥 + 𝑑𝑧
0 1
Equating the corresponding parts we have
𝑎𝑤 + 𝑏𝑦 = 1
(1)
𝑐𝑤 + 𝑑𝑦 = 0
(2)
𝑎𝑥 + 𝑏𝑧 = 0
(3)
𝑐𝑥 + 𝑑𝑧 = 1
(4)
Multiplying equation (1) by 𝑐 and equation (2) by 𝑎 we get:
𝑎𝑐𝑤 + 𝑏𝑐𝑦 = 𝑐
𝑎𝑐𝑤 + 𝑎𝑑𝑦 = 0
(1a)
(2a)
Subtracting equation (1a) from equation (2a) and solving for 𝑦 gives us:
𝑦=
−𝑐
𝑎𝑑−𝑏𝑐
Multiplying equation (3) by 𝑐 and equation (4) by 𝑎 we obtain:
𝑎𝑐𝑥 + 𝑏𝑐𝑧 = 0
𝑎𝑐𝑥 + 𝑎𝑑𝑧 = 𝑎
(3a)
(4a)
Subtracting equation (3a) from equation (4a) and solving for 𝑧 gives us:
𝑧=
𝑎
𝑎𝑑−𝑏𝑐
In a similar manner we can solve for 𝑤 by multiplying equation (1) by 𝑑 and equation (2) by
𝑏:
𝑑
𝑤=
𝑎𝑑−𝑏𝑐
Finally we multiply equation (3) by 𝑑 and equation (4) by 𝑏 to solve for 𝑥:
𝑥=
−𝑏
𝑎𝑑−𝑏𝑐
38
Thus, the inverse of matrix 𝐴 = [
𝑑
−𝑏
[𝑎𝑑−𝑏𝑐
−𝑐
𝑎𝑑−𝑏𝑐
𝑎 ]
𝑎𝑑−𝑏𝑐
𝑎𝑑−𝑏𝑐
=
𝑎
𝑐
𝑏
] is the matrix
𝑑
1
𝑑
[
𝑎𝑑−𝑏𝑐 −𝑐
−𝑏
] provided that 𝑎𝑑 − 𝑏𝑐 ≠ 0.
𝑎
4 2
Example 3: Solve the inverse of matrix 𝐴 = [
] using the formula.
−3 1
Solution: Let 𝐴 = [
4 2
𝑎
]=[
−3 1
𝑐
𝐴−1 =
1
1
[
10 3
𝑏
] then 𝑎𝑑 − 𝑏𝑐 = 4(1) − 2(−3) = 10. Thus
𝑑
1/10
−2
] = [
3/10
4
−1/5
]
2/5
Note that this procedure only works for 2 x 2 matrices. In general, the procedure in finding
the inverse of any 𝑛 x 𝑛 matrix is as follows:
1. Form the 𝑛 x 2𝑛 partitioned matrix [ 𝐴│𝐼𝑛 ] obtained by adjoining the identity matrix
𝐼𝑛 to the given matrix 𝐴.
2. Use elementary row operations to transform the matrix obtained in Step 1 to
reduced row echelon form. Remember that any elementary row operation we do to
a row of 𝐴 we also do to the corresponding row of 𝐼𝑛 .
3. The series of elementary row operations which reduces 𝐴 to 𝐼𝑛 will reduce 𝐼𝑛 to
𝐴−1 . If 𝐴 cannot be reduced to 𝐼𝑛 then 𝐴 is singular and 𝐴−1 does not exist.
1 2
Example 4. Find the inverse of the matrix 𝐴 = [2 5
1 0
3
3].
8
Solution:
We form the 3 x 6 partitioned matrix [A │I3 ] by adjoining the 3 x 3 identity matrix to 𝐴 and
transform it to reduced row echelon form by applying the Gauss-Jordan elimination.
𝐴
1
[2
1
2
5
0
𝐼3
3
3
8
1
0
0
0
1
0
0 −2𝑅 + 𝑅
0] −𝑅 1+ 𝑅 2
1
3
1
𝑅2
𝑅3
39
1
2
[0
1
0 −2
3
−3
5
9
−3
−1
1 0 0
−2 1 0]
−1 0 1
−2𝑅2 + 𝑅1
2𝑅2 + 𝑅3
1
[0
0
0
1
0
1
[0
0
0
9
1 −3
0
1
5 −2
0
−2
1
0]
5 −2 −1
1
[0
0
0
1
0
−40
13
5
0
0
1
5 −2
−2 1
−5 2
0
0] −𝑅3
1
𝑅1
𝑅3
𝑅3
−9𝑅3 + 𝑅1
3𝑅3 + 𝑅2
𝑅1
𝑅2
16
9
−5 −3].
−2 −1
−40
Since 𝐴 has been transformed to 𝐼3 then 𝐴−1 = [ 13
5
16
−5
−2
9
−3].
−1
You can check your answer by taking the product 𝐴𝐴−1 and making sure that the answer is
the identity matrix 𝐼3 .
2
1
Example 5. Find the inverse of the matrix 𝐴 = [ 1 −2
−3 −1
−1
−3].
2
Solution: We form the 3 x 6 matrix [A │I3 ] and transform it to reduced row echelon form
(you must do the calculation here):
2
1 −1 1
[ 1 −2 −3 0
−3 −1
2 0
0 0
1 0
]
is
row
equivalent
to
[
1 0
0 1
0 1
0 0
2/5
1/5
0
−1
1/5
−2/5
0 ]
1
0 −1/5 −1/35 −1/7
Since 𝐴 is row equivalent to a matrix which has a row consisting of zeros, then 𝐴 cannot be
reduced to 𝐼3 . Thus 𝐴 has no inverse and we say that 𝐴 is a singular matrix.
Example 4 shows that an inverse exists if the partitioned matrix [𝐴 𝐼𝑛 ] can be
reduced to [𝐼𝑛 𝐵] where 𝐵 = 𝐴−1 . That is, A is row equivalent to 𝐼𝑛 . This is stated in the
following theorem.
Theorem 2.5.1: An 𝑛 x 𝑛 matrix is nonsingular if and only if it is row equivalent to 𝐼𝑛 .
Theorem 2.5.2: If a matrix has an inverse, then the inverse is unique.
40
Proof:
Suppose 𝐴 has two inverses, say 𝐵 and C. Then by Definition 2.5.1, 𝐴𝐵 = 𝐵𝐴 = 𝐼𝑛 and
𝐴𝐶 = 𝐶𝐴 = 𝐼𝑛 . We now have
𝐵(𝐴𝐶) = (𝐵𝐴)𝐶
Matrix multiplication is associative
Then
𝐵 = 𝐵𝐼𝑛 = 𝐵(𝐴𝐶) = (𝐵𝐴)𝐶 = 𝐼𝑛 𝐶 = 𝐶
By transitivity, we conclude that 𝐵 = 𝐶 and the theorem is proved.
Theorem 2.5.3: ( Properties of the Inverse )
a) If 𝐴 is a nonsingular matrix, then 𝐴−1 is nonsingular and
(𝐴−1 )−1 = 𝐴
b) If 𝐴 and 𝐵 are nonsingular matrices, then 𝐴𝐵 is nonsingular and
(𝐴𝐵)−1 = 𝐵 −1 𝐴−1
c) If A is a nonsingular matrix, then
(𝐴𝑇 )−1 = (𝐴−1 )𝑇
We prove part (b) of the theorem and leave the proof of parts (a) and (c) as an exercise.
Proof of part (b): We have to show that
(𝐴𝐵)(𝐵−1 𝐴−1 ) = (𝐵 −1 𝐴−1 )(𝐴𝐵) = 𝐼𝑛 , that is the inverse of 𝐴𝐵 = 𝐵 −1 𝐴−1 .
(𝐴𝐵)(𝐵−1 𝐴−1 ) = 𝐴(𝐵𝐵 −1 )𝐴−1
Matrix multiplication is associative
= 𝐴(𝐼𝑛 )𝐴−1
Definition of an inverse of a matrix
= 𝐴𝐴−1
𝐼𝑛 is a multiplicative identity for an 𝑛 x 𝑛 matrix
= 𝐼𝑛
Definition of an inverse of a matrix
Similarly
(𝐵 −1 𝐴−1 )(𝐴𝐵) = 𝐵 −1 (𝐴−1 𝐴)𝐵
Matrix multiplication is associative
= 𝐵 −1 (𝐼𝑛 )𝐵
Definition of an inverse of a matrix
= 𝐵 −1 𝐵
𝐼𝑛 is a multiplicative identity for an 𝑛 x 𝑛 matrix
= 𝐼𝑛
Definition of an inverse of a matrix
41
Hence (𝐴𝐵)−1 = 𝐵 −1 𝐴−1.
Corollary 2.5.1: If 𝐴1 , 𝐴2 , ⋯ , 𝐴𝑟 are 𝑛 x 𝑛 nonsingular matrices, then 𝐴1 𝐴2 ⋯ 𝐴𝑟
is nonsingular and
(𝐴1 𝐴2 ⋯ 𝐴𝑟 )−1 = 𝐴𝑟 −1 𝐴𝑟−1 −1 ⋯ 𝐴1 −1
Theorem 2.5.4: Suppose that 𝐴 and 𝐵 are 𝑛 x 𝑛 matrices
a) If 𝐴𝐵 = 𝐼𝑛 then 𝐵𝐴 = 𝐼𝑛 .
b) If 𝐵𝐴 = 𝐼𝑛 then 𝐴𝐵 = 𝐼𝑛 .
2.6 Linear Systems and Inverses
Suppose that matrix 𝐴 is invertible. Consider the system 𝐴𝑋 = 𝐵 where
𝑥1
𝑏1
𝑥2
𝑏
𝑋 = [ ⋮ ] and 𝐵 = [ 2 ] are 𝑛 x 1 matrices where 𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 are variables and
⋮
𝑥𝑛
𝑏𝑛
𝑏1 , 𝑏2 , ⋯ , 𝑏𝑛 are real numbers. Since 𝐴 is invertible then 𝐴−1 exists and
𝐴𝑋 = 𝐵
𝐴−1 𝐴𝑋 = 𝐴−1 𝐵
Multiplying both sides by 𝐴−1
𝐼𝑛 𝑋 = 𝐴−1 𝐵
Definition of an inverse of a matrix
𝑋 = 𝐴−1 𝐵
𝐼𝑛 is the multiplicative identity for
an 𝑛 x 𝑛 matrix
Hence 𝑋 = 𝐴−1 𝐵 is the unique solution of the system.
Example 1. Let 𝐴 = [
1 1
5
], 𝐵 = [ ]. Solve 𝐴𝑋 = 𝐵 by using the inverse of A.
1 2
7
Solution:
𝐴−1 = [
2 −1
] (verify)
−1 1
42
Hence
𝑋 = 𝐴−1 𝐵
2 −1 5
3
𝑋=[
][ ]= [ ]
−1 1 7
2
Hence the unique solution is 𝑋 = (3, 2).
Theorem 2.6.1: If 𝐴 is an 𝑛 x 𝑛 matrix, the homogeneous system 𝐴𝑋 = 0 has a nontrivial
solution if and only if A is singular.
Example 2: Consider the homogeneous system
2𝑥1 + 𝑥2 − 𝑥3 = 0
𝑥1 − 2𝑥2 − 3𝑥3 = 0
−3𝑥1 − 𝑥2 + 2𝑥3 = 0
2
1 −1
where the coefficient matrix 𝐴 = [ 1 −2 −3] is the singular matrix of Example 5 of
−3 −1
2
the previous section. The augmented matrix
2
1 −1
[ 1 −2 −3
−3 −1
2
0
1 0
0] is row equivalent to [0 1
0
0 0
−1
1
0
0
0]
0
which implies that 𝑥1 = 𝑟, 𝑥2 = −𝑟, and 𝑥3 = 𝑟, where 𝑟 ∈ 𝑅. Thus the system has a
nontrivial solution.
Theorem 2.6.2: If 𝐴 is an 𝑛 x 𝑛 matrix, then 𝐴 is nonsingular if and only if the linear system
𝐴𝑋 = 𝐵 has a unique solution for every 𝑛 x 1 matrix 𝐵.
43
SAQ 2-3
1
Let 𝐴 = [1
2
𝐵
1 −1
−1 2 ]. Find the inverse of 𝐴 then solve 𝐴𝑋 = 𝐵 where
1 −1
1
is the 3 x 1 matrix (a) [−2],
3
4
(b) [−3], and
5
5
(c) [ 7 ]
−4
ASAQ 2-3
To solve for 𝐴−1 let us form the 3 x 6 matrix [𝐴 𝐼3 ] and transform it to reduced row echelon form.
1 1 −1
[1 −1 2
2 1 −1
1 0 0 −𝑅 + 𝑅
1
2
0 1 0] −2𝑅 + 𝑅
1
3
0 0 1
𝑅2
−𝑅2
𝑅2 1
[
𝑅3 0
0
1 −1
−2 3
−1 1
1 0
−1 1
−2 0
0
0]
1
1 1 −1
𝑅3 [0 −1 1
0 −2 3
1 0 0
−2 0 1]
−1 1 0
1 1 −1
𝑅2 [0 1 −1
0 −2 3
1
2
−1
0 0
0 −1]
1 0
−1 0
2 0
3 1
1
−1]
−2
−𝑅2 + 𝑅1
2𝑅2 + 𝑅3
𝑅1
𝑅3
1 0
[0 1
0 0
0
−1
1
𝑅3 + 𝑅2
𝑅2
1 0 0
[0 1 0
0 0 1
−1 0 1
5 1 −3]
3 1 −2
−1 0 1
Thus 𝐴−1 = [ 5 1 −3]. Using this to solve the linear system 𝐴𝑋 = 𝐵 for 𝑋 where
3 1 −2
44
1
(a) 𝐵 = [−2] we have
3
𝑋 = 𝐴−1 𝐵
−1 0
= [5 1
3 1
1
1
−3] [−2]
−2 3
2
= [−6]
−5
4
(b) 𝐵 = [−3] we have
5
𝑋 = 𝐴−1 𝐵
4
−1 0 1
= [ 5 1 −3] [−3]
3 1 −2 5
1
= [2]
−1
5
(c) 𝐵 = [ 7 ] we have
−4
𝑋 = 𝐴−1 𝐵
−1 0 1
5
= [ 5 1 −3] [ 7 ]
3 1 −2 −4
−9
= [ 44 ]
30
This only shows that when a matrix is nonsingular, then the system 𝐴𝑋 = 𝐵 has a unique
solution for every 𝑛 x 1 matrix 𝐵.
45
Based on the preceding theorems and examples, it can be noted that the following
statements are equivalent:
1. The matrix 𝐴 is invertible (nonsingular).
2. The system 𝐴𝑋 = 0 has only the trivial solution.
3. The matrices 𝐴 and 𝐼𝑛 are row equivalent.
4. The system 𝐴𝑋 = 𝐵 has a unique solution for every 𝑛 x 1 matrix 𝐵.
ACTIVITY
1. Find the inverse of the following matrices:
 1 2 3 4 
1 1 1
 4 2 1
3 
a. 0 2 3
b. 
 3 0 0  3
5 5 1


 2 0 2 3 
1 4
1 0 5
2. Let 𝐵 = [2 1 4] and 𝐷 = [0 2
3 5
3 4 7
2
1]. Calculate 𝐵𝐷 and 𝐷−1 𝐵 −1 . Verify
3
that (𝐵𝐷)(𝐷−1 𝐵−1 ) = 𝐼3 .
3. Find all values of a
1
A  1
1
-1
exists. What is A ?
for which the inverse of
1 0
0 0 
2 a 
4. Show that if 𝐴, 𝐵, and 𝐶 are invertible matrices, then 𝐴𝐵𝐶 is invertible and
(𝐴𝐵𝐶)−1 = 𝐶 −1 𝐵 −1 𝐴−1 .
5. Prove parts (a) and (c) of Theorem 2.5.4.
46
2.7 Determinants
Definition 2.7.1: Let S = {1, 2, … , n } be the set of integers from 1 to n, arranged in ascending
order. A rearrangement j1 j 2 ... j n of the elements of S is called a permutation of S.
The set of all permutations of S is denoted by 𝑆𝑛 . For example, {1, 2} has two
permutations namely 𝑆2 = { (12), (21) } and {1, 2, 3} has six permutations namely
𝑆3 = { (123), (132), (213), (231), (312), (321) }.
The number of permutations of the set {1, 2, …, 𝑛} can be determined without
writing a list as in the example above. Notice that there are 𝑛 possible positions to be filled.
There are 𝑛 choices for the first position, 𝑛 − 1 for the second, 𝑛 − 2 for the third, and only
one element for the 𝑛th position. Thus the total number of permutations of 𝑛 elements is
𝑛(𝑛 − 1)(𝑛 − 2) … 2.1 = 𝑛!
A permutation j1 j 2 ... j n of S = { 1, 2, … , n } is said to have an inversion if a larger
integer jr precedes a smaller one j s . For example, consider the permutation (23541). The
following pairs of numbers form an inversion: 21, 31, 54, 51.
A permutation is called even if the total number of inversions in it is even and odd if
the total number of inversions in it is odd.
Example 1: The even permutations in S3 are:
123 ( There is no inversion )
231 ( There are two inversions; 21 and 31 )
312 ( There are two inversions; 31 and 32 )
The odd permutations in S3 are:
132 ( There is one inversion; 32 )
213 ( There is one inversion; 21 )
321 ( There are three inversions; 32, 31, and 21 )
Note that the number of odd permutations is equal to the number of even permutations.
Definition 2.7.2: Let A  [ a ij ] be an 𝑛 x 𝑛 matrix. We define the determinant of 𝐴 (written
│𝐴│) by
A    a1 j1a2 j2 ...an jn ,
where the summation ranges over all permutations j1 j 2 ... j n of the set S = {1, 2, …, n}.
47
Note that each term ±𝑎1𝑗1 𝑎2𝑗2 … 𝑎𝑛𝑗𝑛 of |𝐴| is a product of 𝑛 elements of 𝐴
such that one and only one element comes from each row and one and only one element
comes from each column. Thus if the factors come from successive rows then the first
number in the subscripts are in the natural order 1, 2, …, 𝑛. Since all rows are different then
the subscripts 𝑗1 , 𝑗2 , … , 𝑗𝑛 are distinct. This means that {𝑗1 , 𝑗2 , … , 𝑗𝑛 } is a permutation of {1,
2, … , 𝑛 }.
The sign of the term 𝑎1𝑗1 𝑎2𝑗2 … 𝑎𝑛𝑗𝑛 is if the permutation 𝑗1 , 𝑗2 , … , 𝑗𝑛 is even
otherwise, the sign is .
𝑎11
Example 2: Let 𝐴 = [𝑎
21
𝑎12
𝑎22 ].
To illustrate Definition 2.7.2, let us consider exactly one element from each row and each
column. There are two possibilities:
𝑎11
[ ∗
∗
𝑎22 ]
(1)
∗
[𝑎
𝑎12
∗ ]
(2)
21
In (1), the second subscripts form the even permutation 12 hence the sign of the
product 𝑎11 𝑎22 is positive. In (2), the second subscripts form the odd permutation 21 hence
the sign of the product 𝑎12 𝑎21 is negative. Thus
│A│= a11a22  a12 a21 .
The determinant in Example 2 can be solved by writing the terms
𝑎1___ 𝑎2___
and
𝑎1___ 𝑎2___
(there are only two terms because
𝑆2 has only two permutations)
Next we fill in the blanks with the two permutations of 𝑆2 .
𝑎11 𝑎22
and
𝑎12 𝑎21
We assign a sign for the even permutation 12 and a
Taking the sum of the two terms we have
│A│= a11a22  a12 a21 .
sign for the odd permutation 21.
48
Example 3: Let 𝐴 = [
2 −4
] then
3 5
|𝐴| = (2)(5) − (−4)(3) = 22
 a11
Example 4: Let A  a 21
a31
terms
a12
a 22
a32
a13 
a 23  . Since 𝑆3 has 6 permutations then we write the 6
a33 
𝑎1___ 𝑎2___ 𝑎3___ ,
𝑎1___ 𝑎2___ 𝑎3___,
𝑎1___ 𝑎2___ 𝑎3___,
𝑎1___ 𝑎2___ 𝑎3___,
𝑎1___ 𝑎2___ 𝑎3___,
and
𝑎1___ 𝑎2___ 𝑎3___
Next we fill in the blanks with the six permutations of 𝑆3 . The even permutations are 123,
231, and 312 and the odd permutations are 213, 132, and 321. Thus
|𝐴| = 𝑎11 𝑎22 𝑎33 + 𝑎12 𝑎23 𝑎31 + 𝑎13 𝑎21 𝑎32 − 𝑎12 𝑎21 𝑎33 − 𝑎11 𝑎23 𝑎32 − 𝑎13 𝑎22 𝑎31
We can also obtain the determinant of 𝐴 by augmenting the first two columns as
shown below:
𝑎11 𝑎12 𝑎13 𝑎11 𝑎12
𝑎21 𝑎22 𝑎23 𝑎21 𝑎22
𝑎31 𝑎32 𝑎33 𝑎31 𝑎32
− −
− +
+ +
We form the product of each of the three entries joined by the line from left to right and
precede each product by a plus sign. Next we form the product of each of the three entries
joined by the line from right to left and precede each product by a minus sign.
1
2
Example 5: Let 𝐴 = [−2 1
3 −1
1
2 3
|𝐴| = |−2 1 4
3 −1 2
1
−2
3
3
4]. Then
2
2
1|
−1
= (1)(1)(2) + (2)(4)(3) + (3)(-2)(-1) – (3)(1)(3) – (1)(4)(-1) – (2)(-2)(2)
= 35
49
2.8 Properties of Determinants
Theorem 2.8.1: The determinant of a matrix 𝐴 and its transpose are equal.
Example 1. Let A be the matrix of Example 5. Then
1
𝐴 = [2
3
−2 3
1 −1]
4
2
𝑇
The determinant of AT is
1 −2 3
𝐴 = |2 1 −1
3 4
2
𝑇
1 −2
2 1|
3 4
= (1)(1)(2) + (-2)(-1)(3) + (3)(2)(4) – (3)(1)(3) – (1)(-1)(4) – (-2)(2)(2)
= 35
Because of this property we can now replace “row” by “column” in the succeeding
theorems about the determinants of a matrix 𝐴.
Theorem 2.8.2: Let 𝐵 be the matrix obtained from a matrix 𝐴 by
(i)
multiplying a row (column) of 𝐴 by a scalar 𝑐; then |𝐵| = 𝑐|𝐴|.
(ii) interchanging two rows(columns) of |𝐴|; then |𝐵| = −|𝐴|.
(iii) adding a multiple of a row(column) of 𝐴 to another; then |𝐵| = |𝐴|.
2
Example 2. To illustrate part (i) of Theorem 2.8.2, let 𝐵 = [
4
factor of the entries in column 1, then
5
]. Since 2 is a common
−7
2 5
1 5
2
 2(7  10)  34
4 7
2 7
 2 1
5 2
Example 3: To illustrate part (ii) of Theorem 2.8.2, let A  
and B  

 . Then
5 2
 2 1
A  4 + 5 = 1 and B = 5 + 4 = 1. This shows that |𝐵| = −|𝐴|.
50
6
Example 4: To illustrate part (iii) of theorem 2.8.2, let 𝐴 = [−1
3
A
 6 9 12
 1 0 2  3𝑅 + 𝑅
2
3


 3 0 8 
𝑅3
9 −12
0
2 ].
0 −8
B
 6 9 12
 1 0 2 


 0 0 2 
|𝐴| = 54 − 72 = −18 and |𝐵| = 0 − 18 = −18. Thus |𝐴| = |𝐵|.
Theorem 2.8.3: Let 𝐴 be a square matrix.
(i) If two rows (columns) of 𝐴 are equal, then │𝐴│ = 0.
(ii) If a row (column) of 𝐴 consists entirely of zeros, then │𝐴│ = 0.
Proof of part (i):
Suppose the rows 𝑚 and 𝑛 of 𝐴 are equal. Interchange rows 𝑚 and 𝑛 of 𝐴 to obtain a matrix
𝐵. By theorem 2.8.2 (ii), |𝐵| = −|𝐴|. Since rows 𝑚 and 𝑛 are equal then 𝐵 = 𝐴, so |𝐵| =
|𝐴|. Thus
|𝐴| = −|𝐴|
(By substitution)
Hence |𝐴| = 0.
 3 2 1
Example 5: Let A   1 0 4  (row 1 is equal to row 3).The determinant of A is
 3 2 1 
3 2 1
A  1 0 4 = 3(0)(4) + 2(4)(3) +(-1)(2)(1)] – 3(2)(4) - 2(-1)(1) - 1(0)(3)
3 2 1
= 0 + 24 – 2 – 24 + 2 – 0
= 0
51
Proof of part (ii):
Let the 𝑛th row(column) of 𝐴 consist entirely of zero. Since each term in the |𝐴| contains a
factor from each row(column) then each term in the |𝐴| is equal to zero. Hence |𝐴| = 0.
Theorem 2.8.4: If a matrix A  [ a ij ] is upper (lower) triangular, then
A  a11a22 ...ann
that is, the determinant of a triangular matrix is the product of the elements on the main
diagonal.
Example 6: Evaluate the determinant of each matrix by applying theorem 2.8.4.
2
(a) 𝐴 = [0
1
4
2
(b) 𝐵 = [ 3 −2
−2
0
8 −2
3 −4
−4 2 ]
−1 5
3 −4
1
5]
1 −3
6
4
Solution:
(a) We transform matrix 𝐴 into triangular form by applying elementary operations and
taking note of the corresponding changes in the determinant.
2
|0
1
3 −4
−4 2 |
−1 5
𝑅1
𝑅3
−2𝑅1 + 𝑅3
1
𝑅
4 2
5𝑅2 + 𝑅3
1
= − |0
2
(By theorem 2.8.2.ii)
𝑅3
1 −1
5
= − |0 −4
2 |
0 5 −14
(By theorem 2.8.2.iii)
𝑅2
1
= (−4) |0
0
(By theorem 2.8.2.i)
𝑅3
1 −1
5
1/2 |
= ( −4) |0 −1
0 0 −23/2
The last matrix is in triangular form thus
23
−1 5
−4 2 |
3 −4
|𝐴| = (−4)(1)(−1) (− ) = −46
2
−1
5
−1 1/2 |
5 −14
(By theorem 2.8.2.iii)
52
4
2
3
−2
(b) |
−2
0
8 −2
3 −4
1
5| 𝑅
1
1 −3
6
4
𝑅3
2𝑅1 + 𝑅3
4𝑅1 + 𝑅4
1
− 𝑅1
2
1
0
= −(−2) | 3 −2
0
2
0 −2
𝑅1
−𝑅3 + 𝑅4
𝑅2
𝑅3
𝑅4
𝑅4
1 −3
1
5 | (By theorem 2.8.2.ii)
3 −4
6
4
−2
0 1 −3
3
−2
1
5 | (By theorem 2.8.2.iii)
= −|
0
2
5 −10
0 −2 10 −8
𝑅3
𝑅4
−3𝑅1 + 𝑅2
𝑅2 + 𝑅3
−𝑅2 + 𝑅4
−2
0
3
−2
= −|
4
2
8 −2
1
0
= 2 | 0 −2
0
2
0 −2
−1/2 3/2
1
5 | (By theorem 2.8.2.i)
5 −10
10
−8
−1/2
5/2
5
10
1 0 −1/2
0 −2 5/2
= 2|
15/2
0 0
15/2
0 0
3/2
1/2
|(By theorem 2.8.2.iii)
−10
−8
3/2
1/2
|(By theorem 2.8.2.iii)
−19/2
−17/2
3/2
1 0 −1/2
5/2
1/2
= 2 | 0 −2
|(By theorem 2.8.2.iii)
0 0 15/2 −19/2
0 0
0
1
15
Thus |𝐵| = 2(1)(−2) ( 2 ) (1) = −30.
Theorem 2.8.5: The determinant of a product of two matrices is the product of their
determinants; that is
AB  A B
Corollary 2.8.1: If A is nonsingular, then A  0 and
A 1 
1
.
A
53
 2 4
 5 / 22 2 /11
Example 7. Let A  
then A1  
.

 3 / 22 1/11 
3 5 
The determinant of A is │A│= 10 – (-12) = 22 and the determinant of A-1 is
1
A-1│= (5/22)(1/11) – (2/11)(-3/22) = 1/22. Hence |𝐴−1 | =
.
|𝐴|
ACTIVITY
1. Evaluate the determinants of the following matrices.
 4 3 2
a. A  3  2 5
2 4 6
 3  1 2
b. B  4 5 6
7 1 2
4 
 1 2

c.  4 8 16 
 3 0
5 
2. Find all values of  for which
a.
 2
3
2
0
 3
1 0 1
b.  I3  A  0 where A  2 0 1
0 0 1
3. The matrix 𝐴 is called idempotent if 𝐴2 = 𝐴. What are the possible values for the |𝐴| if 𝐴
is idempotent?
4. Prove Corollary 2.8.1.
2.9 Minors and Cofactor Expansion
For large matrices such as 𝑛 ≥ 4, evaluating the determinants using the permutation
formula could be very tedious. We see from the preceding section that transforming a
matrix into a triangular matrix could make the computation easier. Another method that is
also efficient in the computation of determinants of large matrices is by means of cofactor
expansion.
54
Definition 2.9.1: Let A  [ a ij ] be an 𝑛 x 𝑛 matrix. Let M ij be the (𝑛 -1) x (𝑛 – 1) submatrix
of 𝐴 obtained by deleting the 𝑖th row and the 𝑗th column of 𝐴. The determinant M ij is
called the minor of 𝑎𝑖𝑗 . The cofactor Aij of 𝑎𝑖𝑗 is defined as
Aij  (1) i  j M ij
 5 1 6 
Example 1: Let A   3 4 2  . If we delete the first row and second column of 𝐴 we
7 2 1 
3 2
obtain the 2 x 2 submatrix 𝑀12 = [
]. The determinant of 𝑀12 is the minor of 𝑎12 =
7 1
−1. Thus
M 12 
3 2
 3  14  11 .
7 1
Likewise, if we delete the second row and the third column of 𝐴 we obtain the 2 x 2
5 −1
submatrix 𝑀23 = [
]. The determinant |𝑀23 | is called the minor of 𝑎23 = 2. Thus
7 2
M 23 
5 1
 10  (7)  17 .
7 2
By Definition 2.9.1, the cofactor of 𝑎12 is
𝐴12 = (−1)1+2 |𝑀12 |
= −(−11)
= 11
Similarly the cofactor of 𝑎23 is
𝐴23 = (−1)2+3 |𝑀23 |
= (−1)(17)
= −17
55
Theorem 2.9.1: Let A  [ a ij ] be an 𝑛 x 𝑛 matrix. Then for each 1 ≤ 𝑖 ≤ 𝑛,
A  ai1 Ai1  ai 2 Ai 2  ...  ain Ain
(expansion of │𝐴│ about the 𝑖th row)
and for each 1 ≤ 𝑗 ≤ 𝑛,
A  a1 j A1 j  a2 j A2 j  ...  anj Anj
(expansion of │𝐴│ about the 𝑗th column)
2 2 −3 1
0 1
2 −1
Example 2: Evaluate the determinant of 𝐴 = [
] by cofactor expansion.
3 −1
4 1
2 3
0 0
Solution: You can expand about any row or column of your choice, however it is best to
expand about the fourth row because it has the most number of zeros. Thus if we expand
about the fourth row we obtain
|𝐴| = 𝑎41 𝐴41 + 𝑎42 𝐴42 + 𝑎43 𝐴43 + 𝑎44 𝐴44
= 2𝐴41 + 3𝐴42 + 0𝐴43 + 0𝐴44
Notice that the cofactor of a zero entry need not be calculated, so if we can get another
zero on the fourth row then the computation would be a lot easier. Let us apply
elementary operation on the determinant of 𝐴 and take note of the changes in the
determinant.
2 2 −3 1
0 1
2 −1
|
|
3 −1
4 1
2 3
0 0
1
𝐶
2 1
−3𝐶1 + 𝐶2
𝐶1
1
2 −3 1
0
1
2 −1
= 2|
|
3/2 − 1
4 1
1
3
0 0
By theorem 2.8.2.i
1
−1 −3
1
0
1
2 −1
𝐶2 = 2 |
| By theorem 2.8.2.iii
3/2 − 11/2 4
1
1
0
0
0
1
−1 −3
1
0
1
2 −1
Now let us evaluate the determinant |
| . Expanding about the fourth
3/2 − 11/2 4
1
1
0
0
0
row we have
56
1
−1 −3
1
0
1
2 −1
|
| = (1)𝐴41 + 0𝐴42 + 0𝐴43 + 0𝐴44
3/2 − 11/2 4
1
1
0
0
0
=
(1)(−1)4+1
−1 −3 1
| 111 2 −1|
−2
4
1
−1 −3 1
= − | 111 2 −1|
−2
4
1
−1
Next we evaluate | 111
−
2
−3
2
4
1
−1| by expanding about the third row. (Note, you can
1
expand about any row or column of your choice.)
−1
| 111
−
2
−3
2
4
1
−1| = (− 11) (−1)3+1 |−3
2
2
1
1
−1 1
| + (4)(−1)3+2 |
|
−1
1 −1
−1 −3
+ (1)(−1)3+3 |
|
1
2
11
= (− 2 ) (1) + (−4)(0) + (1)(1)
= −9/2
Substituting, we have
1
−1 −3
1
9
0
1
2 −1
|
| = − (− 2) = 9/2
3/2 − 11/2 4
1
1
0
0
0
57
Hence
2 2 −3 1
9
0 1
2 −1
|
| = 2( ) = 9
3 −1
4 1
2
2 3
0 0
SAQ 2-4
4  4 2 1 
1 2
0 3 

Let A 
. Evaluate the determinant by
2 0
3 4


0  3 2 1 
(a) reducing 𝐴 into a triangular matrix, and
(b) expanding about the second row.
ASAQ 2-4
(a) Reducing 𝐴 into a triangular matrix using elementary row operation we have
4 −4
1 2
|
2 0
0 −3
2
0
3
2
1
3
| 𝑅1
4
1
−4𝑅1 + 𝑅2
−2𝑅1 + 𝑅3
3𝐶4 + 𝐶2
−2𝐶4 + 𝐶3
1 2
4 −4
𝑅2 = − |
2 0
0 −3
0
2
3
2
3
1
|
4
1
By theorem 2.8.2.ii
0
3
2 − 11
|
3 −2
2 1
By theorem 2.8.2.iii
1 11 − 6 3
𝐶2
0 − 45 24 − 11
= −|
|
𝐶3
0 − 10 7 − 2
0 0 0
1
By theorem 2.8.2.iii
1
2
𝑅2
0 − 12
= −|
𝑅3
0 −4
0 −3
58
1
9
𝑅2
−2𝑅2 + 𝑅3
−6
1 11
24/9
𝑅2 = −9 |0 −5
0 −10
7
0
0
0
1 11
𝑅3 = −9 | 0 −5
0 −10
0
0
3
−11/9
|
−2
1
By theorem 2.8.2.i
−6
3
24/9 −11/9
| By theorem 2.8.2.iii
5/3
4/9
0
1
5
Hence |𝐴| = −9(1)(−5) (3) (1) = 75
(b) Before we expand about the second row, let us introduce more zeros on the second row
by applying elementary column operations on the determinant.
4 −4
1 2
|
2 0
0 −3
2
0
3
2
1
3 −2𝐶1 + 𝐶2
|
4 −3𝐶1 + 𝐶4
1
4 − 12
𝐶2
1
0
= |
𝐶4
2 −4
0 −3
2 − 11
0
0
|
3 −2
2
1
By theorem 2.8.2.iii
Expanding about the 2nd row we have
4 − 12
1
0
|
2 −4
0 −3
2 − 11
0
0
| = 1𝐴21
3 −2
2
1
−12 2 −11
= 1(−1)2+1 | −4 3 −2 |
−3 2
1
Next we introduce zeros on the third row by using column operations.
−12 2 −11 3𝐶 + 𝐶
3
1
− | −4 3 −2 |
−2𝐶3 + 𝐶2
−3 2
1
−45 24
𝐶1
= − |−10 7
𝐶2
0
0
−11
−2 |
1
−45 24
= −(−1)3+3 |
| expanding about the
−10 7
3rd row
= −(−315 + 240)
= 75
59
SAQ 2-5
Prove: If the 𝑖th row of A is multiplied by a scalar 𝑐, then the
determinant of 𝐴 is multiplied by 𝑐.
ASAQ 2-5
𝑎11
𝑎21
| ⋮
Let |𝐵| = 𝑐𝑎
| 𝑖1
⋮
𝑎𝑛1
𝑎12
𝑎22
⋮
𝑐𝑎𝑖2
⋮
𝑎𝑛2
⋯ 𝑎1𝑛
⋯ 𝑎2𝑛
⋮ |
⋯ 𝑐𝑎𝑖𝑛 |
⋮
⋯ 𝑎𝑛𝑛
If we expand about the 𝑖th row, we have
|𝐵| = 𝑐𝑎𝑖1 𝐴𝑖1 + 𝑐𝑎𝑖2 𝐴𝑖2 + ⋯ + 𝑐𝑎𝑖𝑛 𝐴𝑖𝑛
= 𝑐(𝑎𝑖1 𝐴𝑖1 + 𝑎𝑖2 𝐴𝑖2 + ⋯ + 𝑎𝑖𝑛 𝐴𝑖𝑛 )
= 𝑐|𝐴|
Definition 2.9.2: If A  [ a ij ] is an 𝑛 x 𝑛 matrix, the adjoint of 𝐴 denoted by adj𝐴, is the
transpose of the matrix of cofactors. Thus,
 A11
A
adjA   12


 A1n
A21 ... An1 
A22 ... An 2 


A2 n ... Ann 
60
 2 1 3
Example 3: Let A   1 2 0  . Compute adj 𝐴.
 3 2 1 
Solution:
2 0
1 3
A11  (1)11
2
A21  (1) 21
 7
2 1
2 1
A31  (1)31
1 3
 6
2 0
A12  (1)1 2
1 0
1
3 1
A22  (1) 2 2
2 3
 7
3 1
A32  (1)3 2
2 3
 3
1 0
A13  (1)13
1 2
 4
3 2
A23  (1) 23
2 1
7
3 2
A33  (1)33
2 1
5
1 2
Hence,
 2 7 6 
adj 𝐴 =  1 7 3
 4 7 5 
Theorem 2.9.2: If A  [ a ij ] is an 𝑛 x 𝑛 matrix, then
𝐴(𝑎𝑑𝑗𝐴) = (𝑎𝑑𝑗𝐴)𝐴 = |𝐴|𝐼𝑛
 2 1 3
Example 4: Let A   1 2 0  be the matrix of Example 3. Compute |𝐴| using Theorem
 3 2 1 
2.9.2.
Solution:
 2 1 3  2 7 6
 7 0 0 
1 0 0 






𝐴(adj 𝐴) =  1 2 0  1 7 3 =  0 7 0   7 0 1 0 
 3 2 1   4 7 5 
 0 0 7 
0 0 1 
Thus, A  7
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Corollary 2.9.1: If A is 𝑛 x 𝑛 matrix and A  0 , then
A 1 
1
(adjA)
A
 2 1 3
Example 5: Let A   1 2 0  be the matrix of Example 3, find 𝐴−1 .
 3 2 1 
Solution:
From previous examples we have
 2 7 6 
|𝐴| = −7 and adj 𝐴 =  1 7 3


 4 7 5 
Hence,
−1
𝐴
 2 7 6 
= −7  1 7 3
 4 7 5 
1
−2/7
= [−1/7
4/7
1
6/7
1
3/7 ]
−1 −5/7
You can check your answer by taking the product 𝐴𝐴−1 and making sure that the answer is
the identity matrix 𝐼3 .
Theorem 2.9.3: A matrix A is nonsingular if and only if A  0 .
Corollary 2.9.2: If A is an 𝑛 x 𝑛 matrix, then the homogeneous system AX = 0 has a nontrivial
solution if and only if A  0 .
62
Theorem 2.9.4: (Cramer’s Rule)
Let
a11 x1  a12 x2  ...  a1n xn  b1
a21 x1  a22 x2  ...  a2n xn  b2
an1 x1  an 2 x2  ...  ann xn  bn
be a linear system of 𝑛 equations in 𝑛 unknowns and let A  [ a ij ] be the coefficient matrix
so that we can write the given system as AX = B, where
b1 
b 
B   2
 
 
bn 
If A  0 , then the system has the unique solution
x1 
A1
A
, x2 
A2
A
, … , xn 
An
A
,
where 𝐴𝑖 is the matrix obtained by replacing the 𝑖th column of 𝐴 by 𝐵.
Example 6: Using Cramer’s Rule, determine the solution of the linear system
3x + y – z = 4
- x + y + 3z = 0
x + 2y + z = 1
Solution:
The determinant of the coefficient matrix is (you must do the calculation here)
3 1 1
A   1 1 3 = −8 and
1 2 1
4 1 1
4
A1  0 1 3 = −16 (𝐴1 is obtained by replacing the first column of 𝐴 by 𝐵 = [0])
1
1 2 1
63
3
4 1
A2   1 0
1 1
3
3 =8,
1
1 4
A3   1 1 0 = −8
1 2 1
4
(𝐴2 is obtained by replacing the second column of 𝐴 by 𝐵 = [0])
1
4
(𝐴3 is obtained by replacing the third column of 𝐴 by 𝐵 = [0])
1
By Cramer’s Rule,
x
A1
A

A2
A3  8
8
 16

 1, and z 
 2, y 

1
A
8
8
A
8
Thus the solution set is {(2, -1, 1)}.
NOTE: Cramer’s rule is only applicable to the case where we have 𝑛 equations in 𝑛
unknowns and where the coefficient matrix 𝐴 is nonsingular. Cramer’s rule becomes
computationally inefficient for 𝑛 > 4, and it is better to use the Gauss – Jordan method.
ACTIVITY
I. Solve the following system using Cramer’s Rule.
1. 2𝑥1 + 5𝑥2 − 𝑥3 = −1
4𝑥1 + 𝑥2 + 3𝑥3 = 3
−2𝑥1 + 2𝑥2
= 0
2. 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 6
2𝑥1
− 𝑥3 − 𝑥4 = 4
3 𝑥3 + 6𝑥4 = 3
𝑥1
− 𝑥4 = 5
1 a a2
3. Show that 1 b b 2 = (b-a)(c-a)(c-b). Also, find the inverse of 𝐴 by using the formula
1 c c2
1
𝐴−1 = |𝐴| adj 𝐴.
2
2
4. Find the determinant of 𝐴 = 4
1
[2
1
3
5
0
1
0
1
2
1
0
−1 3
2
5
3 −3 .
2
4
1
0]
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MODULE 3
VECTOR SPACES OVER A FIELD
Introduction
In this chapter we will discuss vectors in ℝ𝑛 and its basic properties, properties and
structure of a vector space and subspace, linear combination and spanning sets.
Objectives
At the end of this chapter, you are expected to be able to do the following:
1. Define vector, vector spaces and subspaces.
2. Enumerate and explain the properties of vector spaces and subspaces.
3. Determine whether a set with given operations is a vector space.
4. Determine whether the given subsets of ℝ𝑛 are subspaces.
5. Define linear combination and spanning.
6. Check whether a given set of vectors spans a vector space V.
3.1 Vectors in the Plane
Vectors are measurable quantities which has magnitude and direction. Example of
vectors are velocity, force, and acceleration.
Vectors in ℝ𝟐
A vector on the plane ℝ𝟐 can be described as an ordered pair 𝐗 = (𝑥1 , 𝑥2 ) where
𝑥1 , 𝑥2 ∈ ℝ. It can also be denoted by a 2 x 1 matrix
𝑥1
𝐗 = [𝑥 ]
2
With X we associate the directed line segment with initial point at the origin O and the
⃗⃗⃗⃗⃗ .
terminal point at 𝑃(𝑥1 , 𝑥2 ), denoted by 𝑂𝑃
The direction of a directed line segment is the angle made with the positive X-axis
and the magnitude of a directed line segment is its length.
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Vector Operations
Definition 3.1.1: Let 𝐗 = (𝑥1 , 𝑥2 ) and 𝒀 = (𝑦1 , 𝑦2 ) be two vectors in the plane. The sum of
the vectors X and Y, denoted by 𝑿 + 𝒀, is the vector
(𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 ).
Example 1: Let X = (2, 3) and Y = (-4, 1). Then
X + Y = (-2, 4)
Definition 3.1.2: If X = (𝑥, 𝑦) and 𝑐 is a scalar (a real number), then the scalar multiple 𝑐𝑿 of
𝑿 by 𝑐 is the vector (𝑐𝑥, 𝑐𝑦).
If 𝑐 > 0, then 𝑐𝑥 is in the same direction as 𝑿, whereas if 𝑑 < 0, then 𝑑𝑥 is in the
opposite direction.
Example 2: Let c = 3, d = -2, and X = (3, -2). Then
cX = (9, -6) and dX = (-6, 4)
NOTE:
1. 𝑂 = (0, 0) is called the zero vector.
2. X + 𝑂 = X
3. X + (-1)X = 𝑂
4. (-1)X = X; the negative of X
5. X + (-1)Y = X – Y; the difference between X and Y.
3.2. 𝒏-Vectors
Definition 3.2.1: An 𝑛-vector is an 𝑛 x 1 matrix
𝑥1
𝑥2
𝑿=[⋮ ]
𝑥𝑛
where 𝑥1 , 𝑥2 , … , 𝑥𝑛 are real numbers, which are called the components of X. The set of all
𝑛-vectors is denoted by ℝ𝑛 and is called 𝑛-space.
66
Definition 3.2.2: Let 𝑿 = (𝑥1 , 𝑥2 , … , 𝑥𝑛 ) and 𝒀 = (𝑦1 , 𝑦2 , … , 𝑦𝑛 ) be two vectors in ℝ𝑛 .
The sum of the vectors X and Y is the vector
(𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 , … , 𝑥𝑛 + 𝑦𝑛 )
and is denoted by X + Y.
Example 1: If X = (1, 2, -3) and Y = (0, 1, -2) are vectors in R3, then
X + Y = (1+0, 2+1, -3 + (-2) ) = ( 1, 3, -5 ).
Definition 3.2.3: If 𝑿 = (𝑥1 , 𝑥2 , … , 𝑥𝑛 ) is a vector in ℝ𝑛 and 𝑐 is a scalar, then the scalar
multiple 𝑐𝑿 of 𝑿 by 𝑐 is the vector
(𝑐𝑥1 , 𝑐𝑥2 , … , 𝑐𝑥𝑛 )
Example 2: If X = ( 2, 1, 5, -2 ) is a vector in ℝ4 and c = -3, then
cX = (-3) ( 2, 1, 5, -2 ) = (-6, -3, -15, 6 )
The operations of vector addition and scalar multiplication satisfy the following properties:
Theorem 3.2.1: Let X, Y, and Z be any vectors in ℝ𝑛 ; let 𝑐 and d be any scalars. Then
I. X + Y is a vector in ℝ𝑛 (that is, ℝ𝑛 is closed under the operation of vector addition).
a. X + Y = Y + X
b. X + ( Y + Z ) = ( X + Y ) +Z
c. There is a unique vector in ℝ𝑛 , 0 = (0, 0, …, 0) such that X + 0 = 0 + X = X.
d. There is a unique vector – X,
-X = (- x1 , - x2 , … , - xn ) such that X + (- X) = 0
67
II. cX is a vector in ℝ𝑛
a. ( X + Y ) = cX + cY
b. ( c + d )X = cX + dX
c. c(dX) = (cd)X
d. 1X = X
3.3 Vector Spaces and Subspaces
Definition 3.3.1: A vector space V over a field 𝐹 is a nonempty set of elements, called
vectors, together with two operations called vector addition and scalar multiplication
satisfying the following properties:
[A1] (Closure under addition)
To every pair of vectors X, Y ∈ V, then X + Y ∈ V.
[A2] (Commutative law of vector addition)
X+ Y= Y+X
∀ X , Y  V.
[A3] (Associative law of vector addition)
X + ( Y + Z ) = ( X + Y ) + Z ∀ X, Y, and Z  V.
[A4] (Zero Element)
∃𝑶 ∈ V such that
X + 𝑶 = 𝑶 + X = X ∀ X  V.
[A5] (Negative Elements)
∀ X  V, ∃ X ∈ V such that
X + ( X) = 𝑶
[M1] (Closure under scalar multiplication)
To every X  V and 𝑐 ∈ 𝐹, then 𝑐 X  V.
[M2] (Distributive Law)
c ( X + Y ) = c X + c Y, ∀𝑐 ∈ 𝐹 , ∀ X, Y  V.
68
[M3] (Distributive Law)
(𝑐 + 𝑑 ) X = 𝑐 X + 𝑑 X, ∀𝑐, 𝑑 ∈ 𝐹, ∀ X  V.
[M4] (Associative Law of Scalar Multiplication)
(𝑐𝑑 )X = 𝑐(𝑑 X ), ∀𝑐, 𝑑 ∈ 𝐹, ∀ X  V.
[M5] (Identity Element)
1 X = X, ∀ X  V.
Example 1. Consider the set 𝑃𝑛 of all polynomials of degree ≤ 𝑛 together with the zero
polynomial. Show that 𝑃𝑛 is a vector space.
Solution: We have to show that all properties are satisfied. A polynomial in 𝑃𝑛 is expressible
as
𝑝(𝑥) = 𝑝0 𝑥 𝑛 + 𝑝1 𝑥 𝑛−1 + … + 𝑝𝑛−1 𝑥 + 𝑝𝑛
where 𝑝0 , 𝑝1 , … , 𝑝𝑛 are real numbers. Let 𝑝(𝑥), 𝑞(𝑥), and 𝑟(𝑥) ∈ 𝑃𝑛 and 𝑐, 𝑑 ∈ ℝ. Then
[A1] 𝑝(𝑥) + 𝑞(𝑥) = (𝑝0 + 𝑞0 )𝑥 𝑛 + (𝑝1 + 𝑞1 )𝑥 𝑛−1 + ⋯ + (𝑝𝑛−1 + 𝑞𝑛−1 )𝑥 + (𝑝𝑛 + 𝑞𝑛 )
Clearly, the sum of two polynomials of degree ≤ 𝑛 is another polynomial with degree ≤ 𝑛.
Hence 𝑃𝑛 is closed under addition.
[A2] 𝑝(𝑥) + 𝑞(𝑥) = (𝑝0 + 𝑞0 )𝑥 𝑛 + (𝑝1 + 𝑞1 )𝑥 𝑛−1 + ⋯ + (𝑝𝑛−1 + 𝑞𝑛−1 )𝑥 + (𝑝𝑛 + 𝑞𝑛 )
= (𝑞0 + 𝑝0 )𝑥 𝑛 + (𝑞1 + 𝑝1 )𝑥 𝑛−1 + ⋯ + (𝑞𝑛−1 + 𝑝𝑛−1 )𝑥 + (𝑞𝑛 + 𝑝𝑛 )
= 𝑞(𝑥) + 𝑝(𝑥)
Thus addition is commutative.
[A3] [𝑝(𝑥) + 𝑞(𝑥)] + 𝑟(𝑥) = [(𝑝0 + 𝑞0 )𝑥 𝑛 + (𝑝1 + 𝑞1 )𝑥 𝑛−1 + ⋯ + (𝑝𝑛−1 + 𝑞𝑛−1 )𝑥
+(𝑝𝑛 + 𝑞𝑛 )] + 𝑟0 𝑥 𝑛 + 𝑟1 𝑥 𝑛−1 + … + 𝑟𝑛−1 𝑥 + 𝑟𝑛
= [(𝑝0 + 𝑞0 ) + 𝑟0 ]𝑥 𝑛 + [(𝑝1 + 𝑞1 )+𝑟1 ]𝑥 𝑛−1 + ⋯ +
[(𝑝𝑛−1 + 𝑞𝑛−1 ) + 𝑟𝑛−1 ]𝑥 +[(𝑝𝑛 + 𝑞𝑛 ) + 𝑟𝑛 ]
= [𝑝0 + (𝑞0 + 𝑟0 )]𝑥 𝑛 + [𝑝1 + (𝑞1 + 𝑟1 )]𝑥 𝑛−1 + ⋯ +
[𝑝𝑛−1 + (𝑞𝑛−1 + 𝑟𝑛−1 )]𝑥 + [𝑝𝑛 + (𝑞𝑛 + 𝑟𝑛 )]
69
= (𝑝0 𝑥 𝑛 + 𝑝1 𝑥 𝑛−1 + … + 𝑝𝑛−1 𝑥 + 𝑝𝑛 ) +
[(𝑞0 + 𝑟0 )𝑥 𝑛 + (𝑞1 + 𝑟1 )𝑥 𝑛−1 + ⋯ + (𝑞𝑛−1 + 𝑟𝑛−1 )𝑥 +
(𝑞𝑛 + 𝑟𝑛 )]
= 𝑝(𝑥) + [𝑞(𝑥) + 𝑟(𝑥)]
Thus vector addition is associative.
[A4] Let 𝑂 = 0𝑥 𝑛 + 0𝑥 𝑛−1 + … + 0𝑥 + 0 be the zero polynomial then
𝑝(𝑥) + 𝑂 = 𝑂 + 𝑝(𝑥) = 𝑝(𝑥)
[A5] Let – 𝑝(𝑥) = −𝑝0 𝑥 𝑛 − 𝑝1 𝑥 𝑛−1 − ⋯ − 𝑝𝑛−1 𝑥 − 𝑝𝑛 , then
𝑝(𝑥) + (−𝑝(𝑥)) = 𝑂
[M1]
𝑐𝑝(𝑥) = 𝑐𝑝0 𝑥 𝑛 + 𝑐𝑝1 𝑥 𝑛−1 + … + 𝑐𝑝𝑛−1 𝑥 + 𝑐𝑝𝑛
where 𝑐𝑝0 , 𝑐𝑝1 , … , 𝑐𝑝𝑛 ∈ ℝ . Clearly , the product of a real number and a polynomial is also
a polynomial. Hence 𝑃𝑛 is closed under scalar multiplication.
[M2] 𝑐[𝑝(𝑥) + 𝑞(𝑥)] = 𝑐[(𝑝0 + 𝑞0 )𝑥 𝑛 + (𝑝1 + 𝑞1 )𝑥 𝑛−1 + ⋯ + (𝑝𝑛−1 + 𝑞𝑛−1 )𝑥
+(𝑝𝑛 + 𝑞𝑛 )]
= 𝑐(𝑝0 + 𝑞0 )𝑥 𝑛 + 𝑐(𝑝1 + 𝑞1 )𝑥 𝑛−1 + ⋯ + 𝑐(𝑝𝑛−1 + 𝑞𝑛−1 )𝑥
+ 𝑐(𝑝𝑛 + 𝑞𝑛 )
= (𝑐𝑝0 + 𝑐𝑞0 )𝑥 𝑛 + (𝑐𝑝1 + 𝑐𝑞1 )𝑥 𝑛−1 + ⋯ + (𝑐𝑝𝑛−1 + 𝑐𝑞𝑛−1 )𝑥
+ (𝑐𝑝𝑛 + 𝑐𝑞𝑛 )
= (𝑐𝑝0 𝑥 𝑛 + 𝑐𝑝1 𝑥 𝑛−1 + … + 𝑐𝑝𝑛−1 𝑥 + 𝑐𝑝𝑛 ) +
(𝑐𝑞0 𝑥 𝑛 + 𝑐𝑞1 𝑥 𝑛−1 + … + 𝑐𝑞𝑛−1 𝑥 + 𝑐𝑞𝑛 )
= 𝑐𝑝(𝑥) + 𝑐𝑞(𝑥)
[M3] (𝑐 + 𝑑)𝑝(𝑥) = (𝑐 + 𝑑)(𝑝0 𝑥 𝑛 + 𝑝1 𝑥 𝑛−1 + … + 𝑝𝑛−1 𝑥 + 𝑝𝑛 )
= (𝑐 + 𝑑)𝑝0 𝑥 𝑛 + (𝑐 + 𝑑)𝑝1 𝑥 𝑛−1 + … + (𝑐 + 𝑑)𝑝𝑛−1 𝑥 + (𝑐 + 𝑑)𝑝𝑛
70
= (𝑐𝑝0 + 𝑑𝑝0 )𝑥 𝑛 + (𝑐𝑝1 + 𝑑𝑝1 )𝑥 𝑛−1 + ⋯ + (𝑐𝑝𝑛−1 + 𝑑𝑝𝑛−1 )𝑥
+ (𝑐𝑝𝑛 + 𝑑𝑝𝑛 )
= (𝑐𝑝0 𝑥 𝑛 + 𝑐𝑝1 𝑥 𝑛−1 + … + 𝑐𝑝𝑛−1 𝑥 + 𝑐𝑝𝑛 ) +
𝑑𝑝0 𝑥 𝑛 + 𝑑𝑝1 𝑥 𝑛−1 + … + 𝑑𝑝𝑛−1 𝑥 + 𝑑𝑝𝑛
= 𝑐𝑝(𝑥) + 𝑑𝑝(𝑥)
[M4] (𝑐𝑑)𝑝(𝑥) = (𝑐𝑑)𝑝0 𝑥 𝑛 + (𝑐𝑑)𝑝1 𝑥 𝑛−1 + … + (𝑐𝑑)𝑝𝑛−1 𝑥 + (𝑐𝑑)𝑝𝑛
= 𝑐(𝑑𝑝0 )𝑥 𝑛 + 𝑐(𝑑𝑝1 )𝑥 𝑛−1 + … + 𝑐(𝑑𝑝𝑛−1 )𝑥 + 𝑐(𝑑𝑝𝑛 )
= 𝑐(𝑑𝑝0 𝑥 𝑛 + 𝑑𝑝1 𝑥 𝑛−1 + … + 𝑑𝑝𝑛−1 𝑥 + 𝑑𝑝𝑛 )
= 𝑐(𝑑𝑝(𝑥))
[M5] 1𝑝(𝑥) = 1𝑝0 𝑥 𝑛 + 1𝑝1 𝑥 𝑛−1 + … + 1𝑝𝑛−1 𝑥 + 1𝑝𝑛
= 𝑝0 𝑥 𝑛 + 𝑝1 𝑥 𝑛−1 + … + 𝑝𝑛−1 𝑥 + 𝑝𝑛
= 𝑝(𝑥)
Since all ten properties are satisfied, then 𝑃𝑛 is a vector space.
Example 2. Let V be the set of all ordered triples of real numbers (𝑥, 𝑦, 𝑧) with the
operations
(𝑥, 𝑦, 𝑧) + (𝑥 ′ , 𝑦 ′ , 𝑧 ′ ) = (𝑥 + 𝑥 ′ , 𝑦 + 𝑦 ′ , 𝑧 + 𝑧 ′ )
𝑐(𝑥, 𝑦, 𝑧) = (𝑥, 1, 𝑧)
Determine if V is a vector space.
Solution: To prove that V is a vector space, we have to show that all ten properties are
satisfied.
Let 𝑿 = (𝑥, 𝑦, 𝑧), 𝒀 = (𝑥 ′ , 𝑦 ′ , 𝑧 ′ ) and 𝒁 = (𝑥 ′′ , 𝑦 ′′ , 𝑧 ′′ )
[A1]
V and 𝑐, 𝑑 ∈ ℝ. Then
𝑿 + 𝒀 = (𝑥 + 𝑥 ′ , 𝑦 + 𝑦 ′ , 𝑧 + 𝑧 ′ )
Since 𝑥 + 𝑥 ′ , 𝑦 + 𝑦 ′ , and 𝑧 + 𝑧 ′
addition.
∈ ℝ, then 𝑿 + 𝒀 ∈ V.
Hence V is closed under
71
[A2] 𝑿 + 𝒀 = (𝑥 + 𝑥 ′ , 𝑦 + 𝑦 ′ , 𝑧 + 𝑧 ′ )
= (𝑥 ′ + 𝑥, 𝑦 ′ + 𝑦, 𝑧 ′ + 𝑧)
=𝒀+𝑿
Hence vector addition is commutative.
[A3] (𝑿 + 𝒀) + 𝒁 = (𝑥 + 𝑥 ′ , 𝑦 + 𝑦 ′ , 𝑧 + 𝑧 ′ ) + (𝑥 ′′ , 𝑦 ′′ , 𝑧 ′′ )
= ((𝑥 + 𝑥 ′ ) + 𝑥 ′′ , (𝑦 + 𝑦 ′ ) + 𝑦 ′′ , (𝑧 + 𝑧 ′ ) + 𝑧′′)
= (𝑥 + (𝑥 ′ + 𝑥 ′′ ), 𝑦 + (𝑦 ′ + 𝑦 ′′ ), 𝑧 + (𝑧 ′ + 𝑧 ′′ ))
= (𝑥, 𝑦, 𝑧) + (𝑥 ′ + 𝑥 ′′ , 𝑦′ + 𝑦 ′′ , 𝑧 ′ + 𝑧 ′′ )
= 𝑿 + (𝒀 + 𝒁)
Thus vector addition is associative.
[A4] Let 𝑶 = (0, 0, 0) ∈ V be the zero vector then
𝑿 + 𝑶 = (𝑥, 𝑦, 𝑧) + (0, 0, 0)
= (𝑥, 𝑦, 𝑧)
=𝑿
[A5] Let – 𝑿 = (−𝑥, −𝑦, −𝑧) then
𝑿 + (−𝑿) = (𝑥, 𝑦, 𝑧) + (−𝑥, −𝑦, −𝑧)
= (0, 0, 0)
=𝑶
[M1] 𝑐𝑿 = 𝑐(𝑥, 𝑦, 𝑧) = (𝑥, 1, 𝑧)
Since 𝑥 and 𝑧 are real numbers then 𝑐𝑿 ∈ V. Hence V is closed under scalar multiplication.
72
[M2] 𝑐(𝑿 + 𝒀) = 𝑐(𝑥 + 𝑥 ′ , 𝑦 + 𝑦 ′ , 𝑧 + 𝑧 ′ )
= (𝑥 + 𝑥 ′ , 1, 𝑧 + 𝑧 ′ )
Also,
𝑐𝑿 + 𝑐𝒀 = 𝑐(𝑥, 𝑦, 𝑧) + 𝑐(𝑥 ′ , 𝑦 ′ , 𝑧 ′ )
= (𝑥, 1, 𝑧) + (𝑥 ′ , 1, 𝑧 ′ )
= (𝑥 + 𝑥 ′ , 2, 𝑧 + 𝑧 ′ )
Since 𝑐(𝑿 + 𝒀) ≠ 𝑐𝑿 + 𝑐𝒀 then M2 is not satisfied. Hence V under the prescribed
operations is not a vector space.
Theorem 3.3.1: If V is a vector space, then:
a) 0𝑿 = 0, for every 𝑿 in V.
b) 𝑐0 = 0, for every scalar 𝑐.
c) If 𝑐𝑿 = 0, then 𝑐 = 0 or 𝑿 = 0
d) (-1)𝑿 = −𝑿, for every 𝑿 in V.
SAQ 3-1
Consider the set 𝑀3,3 (ℝ) of all 3 x 3 matrices with entries
in under the usual operations of matrix addition and
scalar multiplication. Show that 𝑀3,3 (ℝ) is a vector space.
ASAQ 3-1
Let 𝐴, 𝐵, and 𝐶 ∈ 𝑀3,3 (ℝ) and 𝑐, 𝑑 ∈ ℝ. Then
[A1] For every 𝐴, 𝐵 𝑀3,3 (ℝ), we have 𝐴 + 𝐵 ∈ 𝑀3,3 (ℝ). Hence 𝑀3,3 (ℝ) is closed
under addition.
[A2] For every 𝐴, 𝐵
commutative.
𝑀3,3 (ℝ), we have 𝐴 + 𝐵 = 𝐵 + 𝐴, that is matrix addition is
73
[A3] For every 𝐴, 𝐵, 𝐶 𝑀3,3 (ℝ), we have (𝐴 + 𝐵) + 𝐶 = 𝐴 + (𝐵 + 𝐶), that is matrix
addition is associative.
[A4] Let 𝑂 be the 3 x 3 zero matrix, then for every 𝐴
𝐴+𝑂 =𝑂+𝐴=𝐴
[A5] For every 𝐴
𝑀3,3 (ℝ), we have
𝑀3,3 (ℝ), we have 𝐴 + (−𝐴) = 𝑂.
[M1] For every 𝑐 ∈ ℝ and 𝐴 𝑀3,3 (ℝ), we have 𝑐𝐴
closed under scalar multiplication.
𝑀3,3 (ℝ). Hence 𝑀3,3 (ℝ) is
[M2] For every 𝑐 ∈ ℝ and 𝐴, 𝐵
𝑀3,3 (ℝ), we have 𝑐(𝐴 + 𝐵) = 𝑐𝐴 + 𝑐𝐵.
[M3] For every 𝑐, 𝑑 ∈ ℝ and 𝐴
𝑀3,3 (ℝ), we have (𝑐 + 𝑑)𝐴 = 𝑐𝐴 + 𝑑𝐴.
[M4] For every 𝑐, 𝑑 ∈ ℝ and 𝐴
𝑀3,3 (ℝ), we have (𝑐𝑑)𝐴 = 𝑐(𝑑𝐴).
[M5] For every 𝐴
𝑀3,3 (ℝ),we have 1𝐴 = 𝐴.
Hence 𝑀3,3 (ℝ) under the usual operations of matrix addition and scalar multiplication is a
vector space.
Definition 3.3.2: Let V be a vector space and W a nonempty subset of V. If W is a vector
space under the operations of addition and scalar multiplication defined on V, then W is
called a subspace of V.
Theorem 3.3.2: Let V be a vector space under the operations addition and scalar
multiplication and let W be a nonempty subset of V. Then W is a subspace of V if and only if
the following conditions hold:
a) The zero vector 𝑶 belongs to W;
b) For every vector 𝑿, 𝒀  W, 𝑐  ℝ:
i. The sum 𝑿 + 𝒀  W
ii. The multiple 𝑐𝑿  W.
Property (i) in (b) states that W is closed under vector addition, and property (ii) in
(b) states that W is closed under scalar multiplication. Both properties may be combined
into the following equivalent single statement:
(b’) For every 𝑿, 𝒀  W, 𝑎, 𝑏 
, the linear combination 𝑎𝑿 + 𝑏𝒀  W.
74
REMARK: If V is any vector space, then V automatically contains two subspaces, the set {0}
consisting of the zero vector alone and the whole space V itself. These are called the trivial
subspaces of V.
Example 3: Consider the vector space ℝ3 . Let U consists of all vectors in ℝ3 whose entries
are equal; that is U = { (a, b, c) : a = b = c }. Show that U is a subspace of ℝ3 .
Solution:
(a) The zero vector ( 0, 0, 0 ) belongs to U.
(b) Let 𝑿 = (𝑎, 𝑎, 𝑎) and 𝒀 = (𝑏, 𝑏, 𝑏) be vectors in U and let 𝑘, 𝑑 
.
𝑘𝑿 + 𝑑𝒀 = (𝑘𝑎, 𝑘𝑎, 𝑘𝑎) + (𝑑𝑏, 𝑑𝑏, 𝑑𝑏)
= (𝑘𝑎 + 𝑑𝑏, 𝑘𝑎 + 𝑑𝑏, 𝑘𝑎 + 𝑑𝑏)
Clearly, 𝑘𝑿 + 𝑑𝒀
U hence U is a subspace of ℝ3 .
Example 4: Let V = ℝ3 and let W = { (a, b, c ): a  0}. Determine if W is a subspace of V.
Solution:
(a) The zero vector ( 0, 0, 0 )  W.
(b) Let 𝑿 = (𝑎, 𝑏, 𝑐) and 𝒀 = (𝑎′ , 𝑏 ′ , 𝑐 ′ ) be vectors in W.
𝑋 + 𝑌 = (𝑎 + 𝑎′ , 𝑏 + 𝑏 ′ , 𝑐 + 𝑐 ′ )
Since 𝑎  0 and 𝑎′  0 then 𝑎 + 𝑎′  0. Hence 𝑿 + 𝒀  W.
(c) Let 𝑋 = (1, 2, 3) and 𝑘 = −2 then
𝑘𝑿 = (−2, −4, −6)  W since −2 < 0.
Therefore, W is not a subspace of V.
75
ACTIVITY
I. In problems 1-4, determine whether the given set together with the given operations is a
vector space.
1. The set of ordered pairs (𝑎, 𝑏) of real numbers with the operations
(𝑎, 𝑏) + (𝑐, 𝑑) = (𝑎 + 𝑐, 𝑏 + 𝑑) and
𝑘(𝑎, 𝑏) = (𝑘𝑎, 0)
2. The set of all ordered triples of real numbers of the form (0,0, 𝑧) with the operations
( 0, 0, 𝑧 ) + ( 0, 0, 𝑧′ ) = ( 0, 0, 𝑧 + 𝑧′ ) and
𝑐 ( 0, 0, 𝑧 ) = ( 0, 0, 𝑐𝑧 ).
3. The set of polynomials (in 𝑥) of degree  𝑛 together with the zero polynomial with
positive constant term.
4. The set of all ordered pairs (𝑎, 𝑏) of real numbers with addition and scalar multiplication
in V defined by
(𝑎, 𝑏) + (𝑐, 𝑑) = (𝑎𝑐, 𝑏𝑑) and 𝑘(𝑎, 𝑏) = (𝑘𝑎, 𝑘𝑏)
5. Which of the following subsets of ℝ3 are subspaces of ℝ3 ? The set of all vectors of the
form
(a) (𝑎, 𝑏, 𝑐), where 𝑎 = 𝑐 = 0
(b) (𝑎, 𝑏, 𝑐),, where 𝑎 = −𝑐
(c) (𝑎, 𝑏, 𝑐),, where 𝑏 = 2𝑎 + 1
6. Let V be the set of all 2 x 3 matrices under the usual operations of matrix addition and
scalar multiplication. Which of the following subsets of V are subspaces? The set of all
matrices of the form
a
(a) 
d
b c
, where 𝑏 = 𝑎 + 𝑐.
0 0 
a
(b) 
d
b c
, where 𝑐 > 0
0 0 
76
3.4 Linear Combinations and Spanning Sets
Definition 3.4.1: A vector 𝑿  V is said to be a linear combination of the set of vectors
{ 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒌 }  V if there exists scalars 𝑐1 , 𝑐2 , ⋯ , 𝑐𝑘 such that
𝑿 = 𝑐1 𝑿𝟏 + 𝑐2 𝑿𝟐 + ⋯ + 𝑐𝑘 𝑿𝒌
We also say that 𝑿 is linearly dependent on the 𝑿𝒊 .
Example 1. In ℝ3 , (−15, −4, 0) is a linear combination of (−3, −1, 4) and (3, 2, 8) since
2(−3, −1, 4) − 3(3, 2, 8) = (−15, −4, 0).
−1
Example 2. [
−1
6
7
−1 0 3
1 3 −1
] is a linear combination of [
] and [
] since
15 16
1 5 6
−2 0 −1
−1 0
3[
1 5
3
1 3
] + 2[
6
−2 0
−1
−1 6
]=[
−1
−1 15
7
]
16
Example 3. In ℝ3 , let X1 = (4, 2, 3), X2 = (2, 1, 2), and X3 = ( 2, 1, 0). Determine if
(4, 2, 6) is a linear combination of X1, X2, and X3.
𝑿=
Solution: X is a linear combination of X1, X2, and X3 if we can find 𝑐1 , 𝑐2, and 𝑐3 
that
such
𝑐1(4, 2, 3) + 𝑐2 (2, 1, 2) + 𝑐3 ( 2, 1, 0)
(4, 2, 6)
Multiplying and adding yields the following linear system.
4𝑐1 + 2𝑐2
2𝑐3 = 4
2𝑐1 + 𝑐2 + 𝑐3 = 2
3𝑐1
2𝑐2
= 6
Row reducing the augmented matrix we have
4
2 −2
[2
1
1
−3 −2 0
4
2]
−6
1
4
𝑅1
1 1/2 −1/2
𝑅1 [ 2
1
1
−3 −2
0
1
2]
−6
77
−2𝑅1 + 𝑅2
3𝑅1 + 𝑅3
1
𝑅2
−2𝑅3
2
𝑅2
1 1/2
𝑅2
0
[0
𝑅3
0 −1/2
1 1/2
𝑅2
[0
0
𝑅3 0
1
𝑅3
−1/2
2
−3/2
1
0]
−3
−1/2
1
3
1
0]
6
1 1/2 −1/2
[0
1
3
0
0
1
1
6]
0
This implies that 𝑐1 = 2, 𝑐2 = 6, and 𝑐3 = 0. It can be verified that
2(4, 2, 3) + 6(2, 1, 2) + 0( 2, 1, 0) = (4, 2, 6).
Thus X is a linear combination of X1 = (4, 2, 3), X2 = (2, 1, 2), and X3 = ( 2, 1, 0).
SAQ 3-2
In 𝑃3 , let 𝑝(𝑥) = 𝑥 2 − 2𝑥, 𝑞(𝑥) = 5𝑥 − 2, and
𝑟(𝑥) = 𝑥 2 + 3. Determine if 𝑓(𝑥) = 2𝑥 2 + 4𝑥 − 7 is a
linear combination of 𝑝(𝑥), 𝑞(𝑥), and 𝑟(𝑥).
ASAQ 3-2
𝑓(𝑥) is a linear combination of 𝑝(𝑥), 𝑞(𝑥), and 𝑟(𝑥) if we can find 𝑐1 , 𝑐2 , and 𝑐3 ∈ ℝ such
that
𝑐1 (𝑥 2 − 2𝑥) + 𝑐2 (5𝑥 − 2) + 𝑐3 (𝑥 2 + 3) = 2𝑥 2 + 4𝑥 − 7
Equating the coefficients of similar terms we obtain the system
𝑐1
+ 𝑐3 = 2
−2𝑐1 + 5𝑐2
=4
−2𝑐2 + 3𝑐3 = −7
Row reducing the augmented matrix we obtain (you must do the calculation here)
78
1 0
[0 1
0 0
1
2/5
1
2
8/5]
−1
Thus the solution to the given system is 𝑐1 = 3, 𝑐2 = 2, 𝑐3 = −1. It can be verified that
3(𝑥 2 − 2𝑥) + 2(5𝑥 − 2) − (𝑥 2 + 3) = 2𝑥 2 + 4𝑥 − 7
Thus 𝑓(𝑥) is a linear combination of 𝑝(𝑥), 𝑞(𝑥), and 𝑟(𝑥).
Definition 3.4.2: Let S = { 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 } be a set of vectors in a vector space V. The set S
spans V or V is spanned by S, if every vector in V is a linear combination of the vector in S.
That is for every 𝑿 ∈ V there are scalars 𝑐1 , 𝑐2 , ⋯ , 𝑐𝑛 such that
𝑿 = 𝑐1 𝑿𝟏 + 𝑐2 𝑿𝟐 + ⋯ + 𝑐𝑛 𝑿𝒏
Example 4. The vectors 𝑿1 = (1, 0) and 𝑿𝟐 = (0, 1) span ℝ2 since every vector (𝑎, 𝑏) in ℝ2
can be expressed as a linear combination of 𝑿1 and 𝑿𝟐 . That is
(𝑎, 𝑏) = 𝑎(1, 0) + 𝑏(0, 1) for every 𝑎, 𝑏 ∈ ℝ.
Example 5. Every 2 x 2 matrix can be written as a linear combination of the four matrices
1 0
0 1
0 0
0 0
𝐸1 = [
], 𝐸2 = [
], 𝐸3 = [
], and 𝐸4 = [
]. That is
0 0
0 0
1 0
0 1
𝑎
[
𝑐
1 0
0
𝑏
] = 𝑎[
]+𝑏[
0 0
0
𝑑
1
0
]+𝑐[
0
1
0
0 0
]+𝑑[
]
0
0 1
for every 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℝ. Thus 𝐸1 , 𝐸2 , 𝐸3 , and 𝐸4 span 𝑀2,2 .
Example 6. Let V be the vector space ℝ3 and let S = { X1 = (1, 2, 3), X2 = (−1, −2, 1),
X3 = (0, 1, 0) }. Does S span ℝ3 ?
Solution.
Choose an arbitrary vector X = (𝑎, 𝑏, 𝑐) in ℝ3 . The set of vectors S = {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 } spans ℝ3 if
we can find scalars 𝑐1 , 𝑐2 , and 𝑐3 such that
𝑐1(1, 2, 3) + 𝑐2 (−1, −2, 1) + 𝑐3 (0, 1, 0) = (𝑎, 𝑏, 𝑐)
79
Multiplying and adding we obtain the linear system
𝑐1 − 𝑐2 −
= 𝑎
2𝑐1 − 2𝑐2 + 𝑐3 = 𝑏
3𝑐1 + 𝑐2
= 𝑐
The augmented matrix
1 −1 0
[2 −2 1
3 1 0
𝑎
𝑏 ] is row equivalent to
𝑐
1 0 0
[0 1 0
0 0 1
(𝑎 + 𝑐)/4
(−3𝑎 + 𝑐)/4] (verify)
−2𝑎 + 𝑏
The solution to the given system is
𝑐1 =
𝑎+𝑐
4
, 𝑐2 =
−3𝑎+𝑐
4
, and 𝑐3 = −2𝑎 + 𝑏 .
Since there exists 𝑐1 , 𝑐2, and 𝑐3 for every choice of 𝑎, 𝑏, and 𝑐, we say that every vector in
ℝ3 is a linear combination of X1 = (1, 2, 3), X2 = (−1, −2, 1), and X3 = (0, 1, 0).
For example, if 𝑿 = (2, −1, 6) then𝑐1 = 2, 𝑐2 = 0 and 𝑐3 = −5. It can be verified that
(2, −1, 6) = 2(1, 2, 3) + 0(−1, −2, 1) − 5(0, 1, 0)
Thus S = {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 } spans ℝ3 .
Example 7: Does 𝑿 = {−1, 4, 2, 2} belongs to span {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 } where X1 = (1, 0, 0, 1), X2 =
(1, −1, 0, 0), and X3 = (0, 1, 2, 1).
Solution: Every vector in span {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 } is of the form
𝑐1 𝑿𝟏 + 𝑐2 𝑿𝟐 + 𝑐3 𝑿𝟑
Thus 𝑿 belongs to span {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 } if we can write it as a linear combination of
𝑿𝟏 , 𝑿𝟐 , and 𝑿𝟑 . Let 𝑐1 , 𝑐2 , and 𝑐3 be scalars such that
𝑐1 (1, 0, 0, 1) + 𝑐2 (1, −1, 0, 0) + 𝑐3 (0, 1, 2, 1) = (−1, 4, 2, 2)
80
Multiplying and adding we obtain the linear system
𝑐1 + 𝑐2
−𝑐2 + 𝑐3
2𝑐3
𝑐1
+ 𝑐3
=
=
=
=
−1
4
2
2
The third equation implies that 𝑐3 = 1. Substituting it to the second and fourth equations,
we get 𝑐1 = 1 and 𝑐2 = −3. However,
𝑐1 + 𝑐2 = 1 + (−3) = −2 ≠ −1
Thus 𝑿 does not belong to span {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 }.
SAQ 3-3
Let S = {X1 = (1, 1, 0), X2 = (1, 3, 2), X3 = (4, 9, 5) } be
vectors in ℝ3 . Does S span ℝ3 .
ASAQ 3-3
Choose an arbitrary vector X = (𝑎, 𝑏, 𝑐) in ℝ3 . The set of vectors S = {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 } spans ℝ3 if
we can find scalars 𝑐1 , 𝑐2 , and 𝑐3 such that
𝑐1 (1, 1, 0) + 𝑐2 (1, 3, 2) + 𝑐3 (4, 9, 5) = (𝑎, 𝑏, 𝑐)
Multiplying and adding we obtain the linear system
𝑐1 + 𝑐2 + 4𝑐3 = 𝑎
𝑐1 + 3𝑐2 + 9𝑐3 = 𝑏
2𝑐2 + 5𝑐3 = 𝑐
The augmented matrix
1 1
[1 3
0 2
4
9
5
𝑎
1 1
𝑏 ] is row equivalent to [0 2
𝑐
0 0
4
5
0
𝑎
−1 + 𝑏 ](verify)
𝑎−𝑏+𝑐
81
The system has no solution if 𝑎 − 𝑏 + 𝑐 ≠ 0. Thus S = {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 } does not span ℝ3 . For
example, the vector (3, 1, 0) cannot be written as a linear combination of 𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 .
ACTIVITY
1. Determine whether 𝑓(𝑥) = 3𝑥 2 − 3𝑥 + 1 belongs to span {𝑝(𝑥), 𝑞(𝑥), 𝑟(𝑥)} where
𝑝(𝑥) = 𝑥 2 − 𝑥, 𝑞(𝑥) = 𝑥 2 − 2𝑥 + 1 and 𝑟(𝑥) = −𝑥 2 + 1.
2. Let S = {X1 = ( 6, 4, -2, 4 ), X2 = ( 2, 0, 0, 1 ), X3 = ( 3, 2, -1, 2 ), X4 = ( 5, 6, -3, 2 ),
X5 = ( 0, 4, -2, -1 ) }. Does S spans ℝ4 ?
82
MODULE 4
LINEAR INDEPENDENCE
Introduction
In this chapter we will discuss linear dependence and independence of a given set of
vectors, the basis and dimension of a vector space V, the rank of a matrix and how it can be
used to determine whether a matrix is singular or nonsingular and whether a homogeneous
system of equations has a trivial or nontrivial solution.
Objectives
1.
2.
3.
4.
5.
Define linear dependence and independence.
Determine whether a given set of vectors is linearly independent or dependent.
Define and explain basis and dimension of a given vector.
Find a basis for a vector space spanned by a given set of vectors.
Recognize the rank of a matrix and used this information to determine whether a
homogeneous system has a nontrivial solution.
4.1 Definition and Examples
Definition 4.1.1: A set of vectors { 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒌 }  V is said to be linearly dependent if
there exists scalars 𝑐1 , 𝑐2 , ⋯ , 𝑐𝑘 not all of which are zeros such that
𝑐1 𝑿𝟏 + 𝑐2 𝑿𝟐 + ⋯ + 𝑐𝑘 𝑿𝒌 = 𝟎
Otherwise, the set is said to be linearly independent. Meaning the set of vectors
{𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒌 } is linearly independent if 𝑐1 𝑿𝟏 + 𝑐2 𝑿𝟐 + ⋯ + 𝑐𝑘 𝑿𝒌 = 𝟎 is true if and only
if 𝑐1 = 𝑐2 = ⋯ = 𝑐3 = 0.
Example 1. Let S ={X1 = (1, -1), X2 = (1, 1) } be vectors in ℝ2 . Determine whether
S = {X1, X2 } is linearly dependent or linearly independent.
Solution: Let 𝑐1 and 𝑐2 be scalars. Next we form the equation
𝑐1(1, -1) + 𝑐2 (1, 1) = (0, 0)
Multiplying and adding we get the homogeneous system
𝑐1 + 𝑐2 = 0
−𝑐1 + 𝑐2 = 0
Since the only solution to this system is 𝑐1 = 𝑐2 = 0 then S = {X1, X2 } is linearly independent.
83
13
Example 2. Let S = {𝑿𝟏 = (1, 2 3), 𝑿𝟐 = (2, −1, 4), 𝑿𝟑 = (3, − 2 , 4)} be vectors in ℝ3 .
Determine whether S = {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 } is linearly dependent or linearly independent.
Solution: Let 𝑐1 , 𝑐2 , and 𝑐3 be scalars. Next we form the equation
13
𝑐1 (1, 2, 3) + 𝑐2 (2, −1, 4) + 𝑐3 (3, − 2 , 4) = (0, 0, 0)
Multiplying and adding we obtain the homogeneous system
𝑐1 + 2𝑐2 + 3𝑐3 = 0
13
2𝑐1 − 𝑐2 − 2 𝑐3 = 0
3𝑐1 + 4𝑐2 + 4𝑐3 = 0
Row reducing the augmented matrix we have
1 2
3
[2 −1 −13/2
3 4
4
0 −2𝑅 + 𝑅
0] −3𝑅 1 + 𝑅 2
1
3
0
−15𝑅2
2𝑅2 + 𝑅3
3
𝑅2 1 2
[0 −5 −25/2
𝑅3
0 −2
−5
1 2
𝑅2 [0 1
0 −2
1 2
𝑅3 [0 1
0 0
3
5/2
−5
3
5/2
0
0
0]
0
0
0]
0
0
0]
0
Since column 3 has no pivot then 𝑐3 is a free variable. This implies that the system has
infinitely many solutions. In particular if we let 𝑐3 = 2 then 𝑐1 = 4 and 𝑐2 = −5. It can be
verified that
4(1, 2, 3) − 5(2, −1, 4) + 2 (3, −
13
, 4) = (0, 0, 0)
2
Hence 𝑿𝟏 , 𝑿𝟐 , and 𝑿𝟑 are linearly dependent.
REMARKS:
1. Any set of vectors that includes the zero vector is linearly dependent.
Proof:
Let S = {𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 , 𝑶}. Then
0 ∙ 𝑿𝟏 + 0 ∙ 𝑿𝟐 + ⋯ + 0 ∙ 𝑿𝒏 + 2 ∙ 𝑶 = 0
Hence S is linearly dependent.
84
2. If X ≠ 0, { X } is linearly independent
3. The zero vector is linearly dependent
SAQ 4-1
Determine whether the vectors 𝑿𝟏 = (1, −1, 2), 𝑿𝟐 = (4, 0, 0),
𝑿𝟑 = (−2, 3, 5) and 𝑿𝟒 = (7, 1, 2) are linearly dependent or
linearly independent.
ASAQ 4-1
Let 𝑐1 , 𝑐2, 𝑐3 and 𝑐4 be scalars. Then we form the equation
𝑐1 (1, −1, 2) + 𝑐2 (4, 0, 0) + 𝑐3 (−2, 3, 5) + 𝑐4 (7, 1, 2) = (0, 0, 0)
Multiplying and adding we obtain the homogeneous system
𝑐1 + 4𝑐2 − 2𝑐3 + 7𝑐4 = 0
−𝑐1
+ 3𝑐3 + 𝑐4 = 0
2𝑐1
+ 5𝑐3 + 2𝑐4 = 0
1 4 −2 7
The augmented matrix [−1 0 3 1
2 0 5 2
0
0] is row equivalent to
0
1 4 −2
7
0
2
0] (you must do the calculation here)
[0 1 1/4
0 0
1
4/11 0
The last equation implies that 𝑐4 is a free variable. Hence
𝑐1 = −4𝑐2 + 2𝑐3 − 7𝑐4
1
𝑐2 = −4𝑐3 − 2𝑐4
4
𝑐3 = −11𝑐4
85
If we let 𝑐4 = 11 then 𝑐1 = −1, 𝑐2 = −21, and 𝑐3 = −4. It can be verified that
−1(1, −1, 2) − 21(4, 0, 0) − 4(−2, 3, 5) + 11(7, 1, 2) = (0, 0, 0)
Hence the given vectors are linearly dependent.
A very important theorem about linear dependence or independence of vectors can
now be stated.
Theorem 4.1.1. A set of 𝑛 vectors in ℝ𝑚 is always linearly dependent if 𝑛 > 𝑚.
Corollary 4.1.1. A set of linearly independent vectors in ℝ𝑛 contains at most 𝑛 vectors.
Corollary 4.1.1 can be interpreted in this way: if we have 𝑛 linearly independent
vectors, then adding more vectors will make the set of vectors linearly dependent. For
example, the set of vectors {𝑿𝟏 = (1, -1), 𝑿𝟐 = (1, 1)} of Example 1 is linearly independent.
Then by Corollary 4.1.1, the set of vectors {𝑿𝟏 = (1, -1), 𝑿𝟐 = (1, 1), 𝑿𝟑 = (2, 3)} is linearly
dependent.
Theorem 4.1.2: Let S = { 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒌 } be a set of nonzero vectors in a vector space V.
Then S is linearly dependent if and only if one of the vectors 𝑿𝒋 is a linear combination of
the preceding vectors in S.
Example 3. Let S = {X1 = (1, 1, 0), X2 = (0, 2, 3), X3 = (1, 2, 3), X4 = (3, 6, 6) } be a set of vectors
in ℝ3 . Determine if V is linearly dependent. If linearly dependent, express one of the vectors
as a linear combination of the rest.
Solution: By Corollary 4.1.1, we know that S is linearly dependent. Now we form the
equation
𝑐1(1, 1, 0) + 𝑐2 (0, 2, 3) + 𝑐3 (1, 2, 3) + 𝑐4 (3, 6, 6) = (0, 0, 0)
Multiplying and adding we get the homogeneous system
𝑐1
+ 𝑐3 + 3𝑐4 = 0
𝑐1 + 2𝑐2 + 2𝑐3 + 6𝑐4 = 0
3𝑐2 + 3𝑐3 + 6𝑐4 = 0
Row reducing the augmented matrix we obtain
86
1 0 1
[0 1 1
0 0 1
3 0
2 0]
1 0
This implies that 𝑐4 is a free variable. If we let 𝑐4 = 𝑟 ∈ ℝ and 𝑟 ≠ 0 then
𝑐3 = −𝑐4
= −𝑟
𝑐2 = −𝑐3 − 2𝑐4
= 𝑟 − 2𝑟
= −𝑟
𝑐1 = −𝑐3 − 3𝑐4
= 𝑟 − 3𝑟
= −2𝑟
Thus the given vectors are linearly dependent and it can be verified that
−2𝑟(1, 1, 0) − 𝑟(0, 2, 3) − 𝑟(1, 2, 3) + 𝑟(3, 6, 6) = (0, 0, 0)
By Theorem 4.1.2, we can write one of the vectors as a linear combination of the rest.
Transposing the first three terms to the right side we have
𝑟(3, 6, 6) = 2𝑟(1, 1, 0) + 𝑟(0, 2, 3) + 𝑟(1, 2, 3)
(3, 6, 6) = 2(1, 1, 0) + (0, 2, 3) + (1, 2, 3)
(dividing both sides by 𝑟)
Clearly, (3, 6, 6) is a linear combination of the preceding vectors.
NOTE: Any vector in the given set can be expressed as a linear combination of the other
vectors.
4.2 Basis and Dimension
Definition 4.2.1: A set of vectors S = {𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 } in a vector space V is called a basis for
V if S spans V and S is linearly independent.
Example 1. Show that the set S = {X1 = (3, 2, 2), X2 = (-1, 2, 1), X3 = (0, 1, 0) } is a basis for ℝ3 .
Solution:
First we have to show that S is linearly independent. Forming the equation
𝑐1(3, 2, 2) + 𝑐2 (−1, 2, 1)+ 𝑐3 (0, 1, 0) = (0, 0, 0)
we obtain the linear system
3𝑐1 − 𝑐2
= 0
2𝑐1 + 2𝑐2 + 𝑐3 = 0
2𝑐1 + 𝑐2
= 0
87
Solving the given system we obtain only the zero solution, 𝑐1 = 𝑐2 = 𝑐3 = 0. This implies
that S is linearly independent. Next we have to show that S spans ℝ3 . To show that S spans
ℝ3 , let 𝑿 = (𝑎, 𝑏, 𝑐) be any vector in ℝ3 . Then we form the equation
𝑐1(3, 2, 2) + 𝑐2 (−1, 2, 1)+ 𝑐3 (0, 1, 0) = (𝑎, 𝑏, 𝑐)
Multiplying and adding we obtain the linear system
3𝑐1 − 𝑐2
= 𝑎
2𝑐1 + 2𝑐2 + 𝑐3 = 𝑏
2𝑐1 + 𝑐2
= 𝑐
ac
3c  2a
2a  5b  8c
, c2 
, and c3 
. Since the above
5
5
5
system is consistent for any choice of 𝑎, 𝑏, and 𝑐, then S spans ℝ3 .
Solving the system we get c1 
Thus S = { (3, 2, 2), (−1, 2, 1), (0, 1, 0) } } is a basis for ℝ3 .
A basis for a vector space V is not unique. In fact it can be easily verified that the set
of vectors {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is also a basis for ℝ3 . This is called the natural basis
for ℝ3 . Likewise the set of vectors {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)} forms a
natural basis for ℝ4 . In general, the vectors 𝑋1 = (1, 0, ⋯ , 0), 𝑋2 = (0, 1, ⋯ , 0), ⋯ , 𝑋𝑛 =
(0, 0, ⋯ , 1) constitute a basis for ℝ𝑛 .
Example 2. The monomials {𝑥 3 , 𝑥 2 , 𝑥, 1} is a natural basis for 𝑃3 . In general the monomials
{𝑥 𝑛 , 𝑥 𝑛−1 , ⋯ , 𝑥, 1} constitute a basis for 𝑃𝑛 .
Example 3. The 2 x 2 matrices {[
1 0 0
],[
0 0 0
1 0 0 0 0
],[
],[
]} is a natural basis for 𝑀22 .
0 1 0 0 1
SAQ 4-2
Determine whether A = {𝑥 2 − 1, 𝑥 2 − 2, 𝑥 2 − 3} is a basis
for 𝑃2 .
88
ASAQ 4-2
First we have to determine if A is linearly independent. Let 𝑐1 , 𝑐2, and 𝑐3 be scalars. Then
we form the equation
𝑐1 (𝑥 2 − 1) + 𝑐2 (𝑥 2 − 2) + 𝑐3 (𝑥 2 − 3) = 0
Equating the coefficients of similar terms we obtain the system
𝑐1 + 𝑐2 + 𝑐3 = 0
−𝑐1 − 2 𝑐2 − 3𝑐3 = 0
The augmented matrix in reduced row echelon form is (verify)
[
1 0
0 1
−1
2
0
]
0
This implies that A is linearly dependent. Thus A is not a basis for 𝑃2 .
Theorem 4.2.1: If S = { 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 } is a basis for a vector space V, then every vector in V
can be written in one and only one way as a linear combination of the vectors in S.
Definition 4.2.2: If the vector space V has a finite basis, then the dimension of V is the
number of vectors in every basis and V is called a finite dimensional vector space.
Otherwise V is called an infinite dimensional vector space. If V = {0}, then V is said to be
zero dimensional. We often write dimV for the dimension of V.
Example 4. Since a basis for ℝ3 consists of 3 linearly independent vectors then dimℝ3 = 3.
Likewise dimℝ4 = 4, dimℝ5 = 5 and so on. In general dimℝ𝑛 = 𝑛 since 𝑛 linearly
independent vectors constitute a basis for ℝ𝑛 .
Example 5. By Example 2, we see that dim𝑃3 = 4 and in general the dim𝑃𝑛 = 𝑛 + 1.
Theorem 4.2.2: Let S = { 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 } be a set of nonzero vectors in a vector space V and
let W = spans S. Then some subset of S is a basis for W.
Theorem 4.2.2 means that if W consists of all vectors that can be expressed as a
linear combination of the vectors in S (W=spanS) then some subset of S forms a basis for W.
Thus if S is linearly independent then { 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 } is a basis for W. If S is linearly
independent there exists a proper subset of S which forms a basis for W. Let us consider the
next example:
89
Example 6: Let S = {X1 = (1, 2, 2), X2 = (3, 2, 1), X3 = (11, 10, 7), X4 = (7, 6, 4) }. Find a basis for
the subspace W = span S of ℝ3 .
Solution: W is a set of all vectors in ℝ3 which can be expressed as a linear combination of
the vectors in S. By Theorem 4.1.1 we know that S is linearly dependent. Hence S is not a
basis for W but by Theorem 4.2.2 S contains a proper subset of linearly independent vectors
which forms a basis for W. Next we form the equation
𝑐1(1, 2, 2) + 𝑐2 (3, 2, 1) + 𝑐3 (11, 10, 7) + 𝑐4 (7, 6, 4) = (0, 0, 0)
Equating the corresponding components, we obtain the homogeneous system
𝑐1 + 3𝑐2 + 11𝑐3 + 7𝑐4 = 0
2𝑐1 + 2𝑐2 + 10𝑐3 + 6𝑐4 = 0
2𝑐1 + 𝑐2 + 7𝑐3 + 4𝑐4 = 0
The augmented matrix in reduced row echelon form is (verify)
1 0
[0 1
0 0
2 1 0
3 2 0]
0 0 0
The leading 1s appear in columns 1 and 2, thus {X1, X2} is a basis for W = span S. It means
that {X1, X2} is the smallest set possible that could span W. By Theorem 4.1.2, we can write
𝑿𝟑 and 𝑿𝟒 as a linear combination of the two vectors X1 and X2. Thus 𝑿𝟑 and 𝑿𝟒 can be
discarded and the remaining vectors still span W. That is
W = span{ 𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 , 𝑿𝟒 } = span{X1, X2}.
Theorem 4.2.3: Suppose that dimV = 𝑛. If T = { Y1, Y2, …, Yr } is a linearly independent set of
vectors in V, then 𝑟 ≤ 𝑛.
Proof: Let 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 be a basis for V. If 𝑟 > 𝑛 then we can find scalars 𝑐1 , 𝑐2 , ⋯ , 𝑐𝑟 not
all zero such that
𝑐1 𝒀𝟏 + 𝑐2 𝒀𝟐 + ⋯ + 𝑐𝑟 𝒀𝒓 = 𝑶
is satisfied. This will contradict the linear independence of the 𝒀𝒊 ′s. Thus 𝑟 ≤ 𝑛.
Corollary 4.2.1: If S = { 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 } and T = { Y1, Y2, …, Ym } are bases for a vector space,
then 𝑛 = 𝑚. (If a vector space has one basis with a finite number of elements, then all other
bases are finite and have the same number of elements)
90
Theorem 4.2.4: If S is linearly independent set of vectors in a finite-dimensional vector
space V, then there is a basis T for V, which contains S.
Example 7: In V = ℝ4 , the set A = {(1, 1, 0, 0), (0, 0, 1, 1), (1, 0, 1, 0), (0, 1, 0, −1)} is a basis,
and B = { (1, 2, −1, 1), (0, 1, 2, −1) } is linearly independent. Extend B into a basis for V using
A.
Solution:
A  B = {(1, 2, −1, 1), (0, 1, 2, −1), (1, 1, 0, 0), (0, 0, 1, 1), (1, 0, 1, 0), (0, 1, 0, −1) }
Delete vectors in A which are linear combination of the preceding vectors.
a. Try (1, 1, 0, 0)
𝑐1(1, 2, −1, 1) + 𝑐2 (0, 1, 2, −1) = (1, 1, 0, 0)
Then
𝑐1
=1
2𝑐1 + 𝑐2 = 1
−𝑐1 + 2𝑐2 = 0
𝑐1 – 𝑐2 = 0
If 𝑐1 = 1 then 𝑐2 = 1 but 2𝑐1 + 𝑐2 = 3  1. The system of equations has no solution so
(1, 1, 0, 0) is not a linear combination of the preceding vectors. Thus we retain (1, 1, 0, 0).
b. Try (0, 0, 1, 1)
𝑐1 (1, 2, -1, 1) + 𝑐2 (0, 1, 2, -1) = (0, 0, 1, 1)
Then
𝑐1
=0
2𝑐1 + 𝑐2 = 0
−𝑐1 + 2𝑐2 = 1
𝑐1 – 𝑐2 = 1
If 𝑐1 = 0 then 𝑐2 = 0 but –𝑐1 + 2𝑐2 = 0  1. The system has no solution so (0, 0, 1, 1) is not a
linear combination of the preceding vectors. Thus we retain (0, 0, 1, 1).
Hence, {(1, 2, −1, 1), (0, 1, 2, −1), (1, 1, 0, 0), (0, 0, 1, 1)} is a basis for V.
Theorem 4.2.5: Let V be an 𝑛-dimensional vector space, and let S = { 𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 } be a
set of 𝑛 vectors in V.
a. If S is linearly independent, then it is a basis for V.
b. If S spans V, then it is a basis for V.
91
From our previous lesson, we see that the set of all solutions to the homogeneous
system AX = 0, where A is 𝑚 𝑥 𝑛 , is a subspace of ℝ𝑛 . To find a basis for this solution space
we consider the following example.
Example 8: Find a basis for the solution space W of
 x1 
1 2 1 2 1    0 
1 2 2 1 2   x2  0 

 x    
 2 4 3 3 3   3  0 

  x4   
 0 0 1 1 1   0 
 x5 
Solution:
Transform the augmented matrix to reduced row-echelon form using the Gauss-Jordan
reduction method. The augmented matrix in reduced row-echelon form is (verify)
1
0
[
0
0
2
0
0
0
0 3 0
1 −1 0
0 0 1
0 0 0
0
0
]
0
0
The solution is 𝑐1 = −2𝑠 – 3𝑡, 𝑐2 = 𝑠, 𝑐3 = 𝑡, 𝑐4 = 𝑡, and 𝑐5 = 0 where 𝑠 and 𝑡 are any real
numbers. Thus every solution is of the form
 2 s  3t 
 s 


 , where 𝑠 and 𝑡 are real numbers.
X  t


 t

 0 
Since W is a solution space then every vector in W can be written in the form of
 2 
 3 
1
0
 
 
X  s 0  + t 1 
 
 
0
1 
 0 
 0 
Since 𝑠 and 𝑡 can take on any values, we first let 𝑠 = 1, 𝑡 = 0 and let 𝑠 = 0, 𝑡 = 1, obtaining as
solutions
92
 2 
1
 
X1   0 
 
0
 0 
 3 
0
 
X2   1 
 
1 
 0 
and
Thus S = {X1, X2} belongs to W. Since any vector in W can be written as a linear combination
of the vectors in S then S spans W.
We have to show that S = {X1, X2} is linearly independent. We form the equation
𝑐1(−2, 1, 0, 0, 0 ) + 𝑐2 (−3, 0, 1, 1, 0 ) = (0, 0, 0, 0, 0)
Then
−2𝑐1 − 3𝑐2 = 0
𝑐1
= 0
𝑐2 = 0
The only solution to the above system is 𝑐1 = 𝑐2 = 0, hence S is linearly independent and is
a basis for W. Thus the dimension of W is 2.
SAQ 4-3
Find a basis for the solution space W of the homogeneous
system
𝑥1 + 2𝑥2 + 2𝑥3 − 𝑥4 + 𝑥5 = 0
2𝑥2 + 2𝑥3 − 2𝑥4 − 𝑥5 = 0
2𝑥1 + 6𝑥2 + 2𝑥3 − 4𝑥4 + 𝑥5 = 0
𝑥1 + 4𝑥2
− 3𝑥4
=0
What is the dimension of W?
ASAQ 4-3
1
0
The augmented matrix [
2
1
2
2
6
4
2
2
2
0
−1 1 0
−2 −1 0
] is row equivalent (verify) to
−4 1 0
−3 0 0
93
1
0
[
0
0
0
1
0
0
0 1
2
0 −1 −1/2
1 0
0
0 0
0
0
0
] which is in reduced row echelon form.
0
0
Since columns 4 and 5 have no pivots then 𝑥4 and 𝑥5 are free variables. If we let 𝑥4 = 𝑠 and
𝑥5 = 𝑡 where 𝑠 and 𝑡 are real numbers then
𝑥1 = −𝑠 − 2𝑡
1
𝑥2 = 𝑠 + 2 𝑡
𝑥3 = 0
𝑥4 = 𝑠
𝑥5 = 𝑡
where 𝑠 and 𝑡 are real numbers.
Thus all solutions are of the form
−𝑠 − 2𝑡
1
𝑠 +2𝑡
𝑋=
0
𝑠
[ 𝑡
]
−2
−1
1/2
1
=𝑠 0 +𝑡 0
1
0
[0]
[ 1 ]
Since 𝑠 and 𝑡 can be any real number, we first let 𝑠 = 1, 𝑡 = 0 and then let 𝑠 = 0, 𝑡 = 2 to
obtain the solution
−1
−4
1
1
𝑋1 = 0 and 𝑋2 = 0
1
0
[2]
[0]
which span W. It can be easily verified that 𝑋1 and 𝑋2 are linearly independent because one
vector is not a multiple of the other. Thus 𝑋1 and 𝑋2 form a basis for W and dimW = 2. Note
that you may obtain a different basis since 𝑠 and 𝑡 can take on any values.
94
ACTIVITY
1. Let
a) X1 = ( 4, 2, 1 ), X2 = ( 2, 6, -5 ), X3 = ( 1, -2, 3 )
b) X1 = ( 1, 2, 3 ), X2 = ( 1, 1, 1 ), X3 = ( 1, 0, 1 )
c) X1 = ( 1, 1, 0 ), X2 = ( 0, 2, 3 ), X3 = ( 1, 2, 3 ), X4 = ( 3, 6, 6 )
Which of the given set of vectors in ℝ3 is linearly dependent? For those that are, express
one vector as a linear combination of the rest.
3. Determine whether the set
 1
S  
 0
1 0 0 1 0 0 1 
,
,
,

0 1 1  0 1  1 1 
is a basis for the vector space V of all 2 x 2 matrices.
4. Find a basis of ℝ4 containing the vector (1, 2, 3, 4).
4.3 The Rank of a Matrix
Definition 4.3.1: Let
 a11 a12 ... a1n 
a
a22 ... a2 n 
21

A




 am1 am 2 ... amn 
be an 𝑚 x 𝑛 matrix. The rows of A
X 1  (a11 , a12 ,..., a1n )
X 2  (a 21 , a 22 ,..., a 2 n )
⋮
X m  (a m1 , a m 2 ,..., a mn )
considered as vectors in ℝ𝑛 , span a subspace of ℝ𝑛 , called the row space of A. Similarly,
the columns of A,
 a11 
 a12 
 a1n 
a 
a 
a 
21 
22 
2n


Y1 
, Y2 
, … , Yn   
 
 
 
 
 
 
 am1 
 am 2 
 amn 
95
considered as vectors in ℝ𝑚 , span a subspace of ℝ𝑚 called the column space of A.
1 2 0 1 
Example 1. Let A   2 6 3 2  . The rows of A,
3 10 6 5 
X1 = (1, 2, 0, 1), X2 = (2, 6, −3, −2), and X3 = (3, 10, −6, −5) are vectors in ℝ4 , and these
vectors span a subspace of ℝ4 called the row space of A. That is
Row space of A = span { 𝑋1 , 𝑋2 , 𝑋3 }
Similarly, the columns of A, Y1 = (1, 2, 3), Y2 = (2, 6, 10), Y3 = (0, −3, −6) and Y4 = (1, −2, −5)
are vectors in ℝ3 , and these vectors span a subspace of ℝ3 called the column space of A.
That is
Column space of A = span { 𝑌1 , 𝑌2 , 𝑌3 , 𝑌4 }
Theorem 4.3.1: If A and B are two 𝑚 x 𝑛 row equivalent matrices, then the row spaces of A
and B are equal.
Example 2: Let S = {𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 , 𝑿𝟒 } where 𝑿𝟏 = (1, 2, −1), 𝑿𝟐 = (6, 3, 0), 𝑿𝟑 = (4, −1, 2),
and 𝑿𝟒 = (2, −5, 4). Find a basis for the subspace V = spanS of ℝ3 .
Solution: V is the row space of the matrix A whose rows are the given vectors
1 2 −1
6 3
0
A=[
]
4 −1 2
2 −5 4
Applying the Gauss-Jordan elimination we obtain the matrix B
1
0
𝐵=[
0
0
0
1
0
0
1/3
−2/3
]
0
0
which is row equivalent to A. By Theorem 4.3.1, the row spaces of A and B are equal. Since
the nonzero rows of B are linearly independent then
(1, 0, 1/3) and (0, 1, −2/3)
96
form a basis for V. It means that all vectors in V can be expressed as a linear combination of
these two vectors. Note that the basis for V is not a subset of S. However, expressing any
vector in V as a linear combination of the basis obtained by this procedure is very simple.
For example, the vectors (2, −5, 4) and (−1, 4, −3) are in V. Since the leading 1s appear in
columns 1 and 2 then
(2, −5, 4) = 2(1, 0, 1/3) – 5(0, 1, −2/3) and
(−1, 4, −3) = −1(1, 0, 1/3) + 4(0, 1, −2/3)
Also note that the dimV = 2 ≠ 3 hence V ≠ ℝ3 . Since V ≠ ℝ3 then not all vectors in ℝ3 can
be expressed as a linear combination of (1, 0, 1/3) and (0, 1, −2/3). It is very easy to see
from our example above that all vectors of the form (2, −5, 𝑐), 𝑐 ≠ 4, are not in V.
Definition 4.3.2: The dimension of the row space of A is called the row rank of A, and the
dimension of the column space of A is called the column rank of A.
1 2
Example 3. Let 𝐴 = [2 6
3 10
0
1
−3 −3]. Find the row and column ranks of A.
−6 −7
Solution:
1 0
A is row equivalent (verify) to 𝐵 = [0 1
0 0
3
6
−3/2 −5/2] which is in reduced row
0
0
echelon form. The vectors (1, 0, 3, 6) and (0, 1, −3/2, −5/2) form a basis for the row space
of A. Thus the row rank of A is 2.
To find the column rank of A, we form the matrix
1
2
𝐴𝑇 = [
0
1
2
3
6 10
]
−3 −6
−3 −7
which is row equivalent (verify) to the matrix
1
0
𝐶=[
0
0
0 −1
1 2
]
0 0
0 0
97
The vectors (1, 0 −1) and (0, 1, 2) form a basis for the row space of 𝐴𝑇 . Thus
1
0
[ 0 ] and [1]
−1
2
form a basis for the column space of A. Hence the column rank of A is 2. Note that the row
rank and column rank of a matrix are equal. This is stated in the next theorem.
Theorem 4.3.2. The row and column ranks of the (𝑚 x 𝑛) matrix A =  aij  are equal.
The next example will show us how we can apply the method used in Example 3 in
finding a basis for a subspace when the vectors are given in column form.
1 0 2 3
5
2 2 1 2
0
Example 4: Let S = {[ ] , [ ] , [ ] , [ ] , [ ]} . Find a basis for the subspace V = spanS of
1 1 3 1
0
1 2 1 4 −1
ℝ4 .
1
2
Solution: Let 𝐴 = [
1
1
0
2
1
2
1
0
𝐴𝑇 = 2
3
[5
2
1
3
1
2
2
1
2
0
3 5
2 0
]. Next we form the matrix
1 0
4 −1
1 1
1 2
3 1
1 4
0 −1]
Applying elementary row operations we obtain (verify) the matrix
1
0
𝐵𝑇 = 0
0
[0
0
1
0
0
0
0
0
1
0
0
−1
3/5
4/5
0
0 ]
The basis for V are the nonzero rows of 𝐵 𝑇 written as columns. Thus
98
0
0
1
1
0
0
[ ], [
], and [
]
0
1
0
3/5
4/5
−1
form a basis for V.
Theorem 4.3.3. An 𝑛 x 𝑛 matrix is nonsingular if and only if rank A = 𝑛.
1
Example 5. Find the rank of 𝐴 = [1
1
2 0
1 −3].
3 3
Solution:
Transforming A to reduced row echelon form we obtain (verify)
1 0
[0 1
0 0
−6
3]
0
Since rank of A = 2 < 3 then A is singular. We know from our previous lesson that a matrix is
singular if and only if |𝐴| = 0. It can be verified that
1 2
|1 1
1 3
0
−3| = 0
3
This result is stated in the following corollary and can be used in determining the
rank of an 𝑛 x 𝑛 matrix.
Corollary 4.3.1. If A is an 𝑛 x 𝑛 matrix, then rank A = 𝑛 if and only if A  0.
The next corollary gives another method of testing whether the homogeneous
system 𝐴𝑋 = 0 has a trivial or nontrivial solution.
Corollary 4.3.2. The homogeneous system AX = 0 of 𝑛 linear equations in 𝑛 unknowns has a
nontrivial solution if and only if rank A < 𝑛.
99
Example 6: The 3 x 3 matrix A of Example 5 is singular. Hence the homogeneous system
𝑥1 + 2𝑥2
= 0
𝑥1 + 𝑥2 − 3𝑥3 = 0
𝑥1 + 3𝑥2 + 3𝑥3 = 0
has a nontrivial solution.
Corollary 4.3.3. Let S = {𝑿𝟏 , 𝑿𝟐 , ⋯ , 𝑿𝒏 } be a set of vectors in ℝ𝑛 and let A be the matrix
whose rows (columns) are the vectors in S. Then S is linearly independent if and only if
|𝐴| ≠ 0.
Example 7: The vectors (1, −2, 3), (2, 4, 7) and (0, −1, 5) are linearly independent because
1
[2
0
−2 3
4 7] = 41 ≠ 0
−1 5
By Theorem 4.2.5, (1, −2, 3), (2, 4, 7) and (0, −1, 5) form a basis for ℝ3 .
ACTIVITY
1. Find a basis for the row space and column space of the following matrices:
1 2 3
a. [4 5 6]
7 8 9
2 1 3
b. [2 −1 5
1 1 1
−2
2]
1
2. Let V = P1 and let A = { x + 1, x – 1, 2x + 3 }
a. Show that S spans P1.
b. Find a subset of A which is a basis for P1.
3. Consider the following subset of the vector space of all real-valued functions
S = { cos2t, sin2t, cos2t }
Find a basis for the subspace W = spanS. What is the dimension of W?
100
MODULE 5
LINEAR TRANSFORMATIONS AND MATRICES
Introduction
In this chapter we will discuss a function mapping one vector space to another
vector space, one-to-one and onto linear transformations, the kernel of a linear
transformation and how it can be used to determine if the linear transformation is one-toone, the range of a linear transformation and how it can be used to determine whether the
linear transformation is onto and the matrix of a linear transformation.
Objectives
After going through this chapter, you are expected to be able to do the following:
1. Define and explain linear transformation or linear mapping.
2. Determine whether a function mapping one vector space to another is a linear
transformation.
3. Differentiate between one-to-one linear transformation and onto linear transformation.
4. Define the kernel and range of a linear transformation.
5. Using kernel, distinguish between one-to-one linear transformation and onto linear
transformation.
6. Find the matrix of a linear transformation with respect to some bases.
5.1 Linear Transformations (Linear Mappings)
Definition 5.1.1: Let V and W be vector spaces. A linear transformation L of V into W is a
function assigning a unique vector L(X) in W to each X in V such that:
a. L ( X + Y ) = L (X) + L(Y), for every vector X and Y in V,
b. L ( 𝑐X ) = 𝑐L(X), for every vector X in V and every scalar 𝑐.
Example 1: Let L: ℝ3 → ℝ3 be defined by
L(x, y, z) = (x, y, 0)
Determine whether L is a linear transformation or not.
Solution: Let X = (𝑥1 , 𝑦1 , 𝑧1 ) and Y = (𝑥2 , 𝑦2 , 𝑧2 )  ℝ3 then
𝑋 + 𝑌 = (𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝑧1 + 𝑧2 ).
101
a) We have to show that L(X + Y) = L(X) + L(Y)
L(X + Y) = L(𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝑧1 + 𝑧2 )
= (𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 0)
= (𝑥1 , 𝑦1 , 0) + ( 𝑥2 , 𝑦2 , 0)
= L(X) + L(Y)
Hence L(X + Y) = L(X) + L(Y)
b) Let 𝑐𝑿 = (𝑐𝑥1 , 𝑐𝑦1 , 𝑐𝑧1 ). We have to show that L(𝑐X) = 𝑐L(X).
L(𝑐𝑿) = (𝑐𝑥1 , 𝑐𝑦1 , 0)
= 𝑐(𝑥1 , 𝑦1 , 0)
= 𝑐L(𝑿)
Hence L(cX) = cL(X). Since the two conditions are satisfied then L is a linear transformation.
SAQ 5-1
Let L: ℝ3
ℝ3 be defined by
L(𝑥, 𝑦, 𝑧) = (𝑥 − 𝑦, 𝑥 2 , 2𝑧)
Determine whether L is a linear transformation or not.
102
ASAQ 5-1
Let 𝑿 = (𝑥1 , 𝑦1 , 𝑧1 ) and Y = (𝑥2 , 𝑦2 , 𝑧2 )  ℝ3 then 𝑋 + 𝑌 = (𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝑧1 + 𝑧2 ).
L(𝑿 + 𝒀) = ((𝑥1 + 𝑥2 ) − (𝑦1 + 𝑦2 ), (𝑥1 + 𝑥2 )2 , 2(𝑧1 + 𝑧2 ))
= (𝑥1 + 𝑥2 − 𝑦1 − 𝑦2 , 𝑥1 2 + 2𝑥1 𝑥2 + 𝑥2 2 , 2𝑧1 + 2𝑧2 )
and L(X) + L(Y) = (𝑥1 − 𝑦1 , 𝑥1 2 , 2𝑧1 ) + (𝑥2 − 𝑦2 , 𝑥2 2 , 2𝑧2 )
= ((𝑥1 + 𝑥2 − 𝑦1 − 𝑦2 , 𝑥1 2 + 𝑥2 2 , 2𝑧1 + 2𝑧2 )
Since L(𝑿 + 𝒀) ≠ L(𝑿) + L(𝒀) then L is not a linear transformation.
Theorem 5.1.1: If L: V→ W is a linear transformation, then
L(c1X1 + c2X2 + … + ckXk ) = c1L(X1) + c2L(X2) + … + ckL(Xk)
for any vectors X1, X2, … , Xk in V and any scalars c1, c2, … , ck.
Theorem 5.1.2: Let L: ℝ𝑛 → ℝ𝑚 be a linear transformation. Then
a. L(0) = 0
b. L(−𝑿) = − L(𝑿) for every 𝑿 ∈ ℝ𝑛
c. L ( X – Y ) = L(X) – L(Y)
Proof of Theorem 5.1.2.a.
a.
L(0) = 0
L(0) = L (0 + 0)
L(0) = L(0) + L(0)
L(0) – L(0) = L(0) + L(0) – L(0)
0 = L(0)
103
Proof of Theorem 5.1.2.b.
b. L(−𝑿) = −L(𝑿)
L(−𝑿) = L(−1 ∙ 𝑿)
= −1L(𝑿)
= −L(𝑿)
Proof of Theorem 5.1.2.c.
c. L(𝑿 − 𝒀) = L(𝑿 + (−1)𝒀)
= L(𝑿) + L(−1 ∙ 𝒀)
= L(𝑿) − L(𝒀)
Example 2: Let L: ℝ3 → ℝ2 be defined by L(𝑥, 𝑦, 𝑧) = (𝑥, 𝑦) be a linear transformation
(verify). It can be verified that
L(0, 0, 0) = (0, 0)
That is, the zero vector in ℝ3 is mapped to the zero vector in ℝ2 .
Also, let X = (𝑥1 , 𝑦1 , 𝑧1 ) and Y = (𝑥2 , 𝑦2 , 𝑧2 ) be any vectors in ℝ3 then
𝑿 − 𝒀 = (𝑥1 − 𝑥2 , 𝑦1 − 𝑦2 , 𝑧1 − 𝑧2 ). Thus
L(X – Y) = (𝑥1 − 𝑥2 , 𝑦1 − 𝑦2 )
= (𝑥1 , 𝑦1 ) − (𝑥2 , 𝑦2 )
= L(X) – L(Y)
Example 3: Let T: ℝ2 → ℝ2 be the “translation mapping” defined by
T(𝑥, 𝑦) = (𝑥 + 1, 𝑦 + 2)
By Theorem 5.1.2 letter a, T is not a linear transformation since
T(0, 0) = (0 + 1, 0 + 2) = (1, 2) ≠ (0, 0)
(The zero vector is not mapped into the zero vector).
104
Theorem 5.1.3: Let L: V→ W be a linear transformation of an n-dimensional vector space V
into a vector space W. Also let S = { X1 , X2, … , Xn } be a basis for V. If X is any vector in V,
then L(X) is completely determined by {L(X1), L(X2), … , L(Xn) }.
Example 4: Let L: ℝ3 → ℝ3 be a linear transformation for which we know that
(2, −4) , L(0, 1, 0) = (3, −5) and L(0, 0, 1) = (2, 3).
a. What is L(1, −2, 3)?
b) What is L(𝑎, 𝑏, 𝑐)?
Solution: The set {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis for ℝ3 , and
(1, −2, 3) = 1(1, 0, 0) – 2(0, 1, 0) + 3(0, 0, 1)
By Theorem 5.1.3:
a)
L(1, −2, 3) = L1(1, 0, 0) – L2(0, 1, 0) + L3(0, 0, 1)
= L(1, 0, 0) – 2L(0, 1, 0) + 3L(0, 0, 1)
= (2, −4) – 2(3, −5) + 3(2, 3)
= (2, 23)
b) We see from part (a) that for any (𝑎, 𝑏, 𝑐)  ℝ3 , we have
L(𝑎, 𝑏, 𝑐) = 𝑎L(1, 0, 0) + 𝑏L(0, 1, 0) + 𝑐L(0, 0, 1)
= 𝑎(2, −4) + 𝑏(3, −5) + 𝑐(2, 3)
= (2𝑎 + 3𝑏 + 2𝑐, −4𝑎 −5𝑏 + 3𝑐)
SAQ 5-2
Let L: P2 → P3 be a linear transformation for which we know
that L(1) = 1, L(t) = t2 and L(t2) = t3 + t. Find
(a) L(2t2 – 5t + 3)
(b) L(at2 + bt + c)
L(1, 0,0) =
105
ASAQ2
The set {𝑡 2 , 𝑡, 1} is a basis for 𝑃2 . By Theorem 5.1.3
a. L(2𝑡 2 − 5𝑡 + 3) = L2(𝑡 2 ) + L(−5)(𝑡) + L3(1)
= 2L(𝑡 2 ) − 5L(𝑡) + 3L(1)
= 2(𝑡 3 + 𝑡) − 5(𝑡 2 ) + 3(1)
= 2𝑡 3 + 2𝑡 − 5𝑡 2 + 3
= 2𝑡 3 − 5𝑡 2 + 2𝑡 + 3
b. L(𝑎𝑡 2 + 𝑏𝑡 + 𝑐) = 𝑎L(𝑡 2 ) + 𝑏L(𝑡) + cL(1)
= 𝑎(𝑡 3 + 𝑡) + 𝑏(𝑡 2 ) + 𝑐
= 𝑎𝑡 3 + 𝑎𝑡 + 𝑏𝑡 2 + 𝑐
= 𝑎𝑡 3 + 𝑏𝑡 2 + 𝑎𝑡 + 𝑐
5.2 The Kernel and Range of a Linear Transformation
Definition 5.2.1: A linear transformation L: V→ W is said to be one-to-one if for all X1, X2 in
V, X1 ≠ X2 implies that L(X1) ≠ L(X2). An equivalent statement is that L is one-to-one if for all
X1, X2 in V, L(X1) = L(X2) implies that X1 = X2.
Example 1. Let L: ℝ2 → ℝ2 be a linear transformation defined by
L( 𝑥, 𝑦 ) = ( 𝑥 + 𝑦, 𝑥 )
Determine if L is one–to–one or not.
Solution:
Let X1 = (𝑥1 , 𝑦1 ) and X2 = (𝑥2 , 𝑦2 ) be vectors in ℝ2 . We have to show that if L(X1) = L(X2) then
X1 = X2.
L(X1) = L(X2)
(𝑥1 + 𝑦1 , 𝑥1 ) = (𝑥2 + 𝑦2 , 𝑥2 )
Equating the corresponding parts we have
𝑥1 + 𝑦1 = 𝑥2 + 𝑦2
𝑥1 = 𝑥2
106
If we subtract the second equation from the first, we get 𝑦1 = 𝑦2 which implies that 𝑿𝟏 =
𝑿𝟐 . Thus L is one-to-one.
Example 2. Let L: ℝ3 → ℝ3 be a linear transformation defined by
L(x, y, z) = (x, y, 0)
Determine if L is one-to-one or not.
Solution:
Let X1 = (4, 5, −3) and X2 = (4, 5, 2) be vectors in ℝ3 . We see that X1  X2 but
L(4, 5, −3) = L(4, 5, 2) = (4, 5, 0). Hence L is not one-to-one.
Definition 5.2.2: Let L: V→ W be a linear transformation. The kernel of L, kerL, is the subset
of V consisting of all vectors X such that L(X) = Ow.
Note that kerL is not empty since by Theorem 5.1.2 we know that if L: ℝ𝑛 → ℝ𝑚 is a linear
transformation then the zero vector in ℝ𝑛 is mapped to the zero vector in ℝ𝑚 . Thus 𝑂 ∈
kerL.
Example 3. Let L: ℝ4 → ℝ3 be a linear transformation defined by
L(𝑥, 𝑦, 𝑧, 𝑤) = (𝑥 + 𝑦, 𝑧 + 𝑤, 𝑥 + 𝑧)
The vector (1, −1, −1, 1) is in kerL since L(1, −1, −1, 1) = (0, 0, 0) while the vector
(1, 2, 3, −4) is not in kerL because L(1, 2, 3, −4) = ( 3, 1, 5 ) ≠ (0, 0, 0). Thus all vectors X in
ℝ4 such that L(X) = (0, 0, 0) are in kerL.
Example 4. Let L be the linear transformation of Example 3. The kerL consists of all vectors
X in ℝ4 such that L(X) = 0. If we let X = (𝑥, 𝑦, 𝑧, 𝑤) then
L(𝑥, 𝑦, 𝑧, 𝑤) = (0, 0, 0)
(𝑥 + 𝑦, 𝑧 + 𝑤, 𝑥 + 𝑧) = (0, 0, 0)
Equating the corresponding parts we obtain the homogeneous system
𝑥+𝑦
=0
𝑧 + 𝑤=0
𝑥 +𝑧
=0
107
The augmented matrix in reduced echelon form (verify) is
1
[0
0
0 0 −1 0
1 0 1 0]
0 1 1 0
This implies that 𝑤 is a free variable and can take on any value. Hence
𝑥=𝑟
𝑦 = −𝑟
𝑧 = −𝑟
𝑤 = 𝑟 where 𝑟 ∈ ℝ
Thus kerL consists of all vectors of the form (𝑟, −𝑟, −𝑟, 𝑟) where 𝑟 is any real number. That
is
ker L = { (𝑟, −𝑟, −𝑟, 𝑟): 𝑟  ℝ}
= { 𝑟(1, −1, −1, 1): 𝑟  ℝ}
Hence ker L = span { (1, −1, −1, 1)}
Since (1, −1, −1, 1) is linearly independent then it forms a basis for kerL. Thus
dim(kerL) = 1.
SAQ 5-3
Let L: ℝ4 → ℝ2 be defined by
x
 
y
x  y
L    
  z    z  w
   
  w 
Find kerL.
108
ASAQ 5-3
ker L = { (𝑥, 𝑦, 𝑧, 𝑤) : (𝑥 − 𝑦, 𝑧 − 𝑤) = (0, 0) }
Equating the corresponding parts we get:
𝑥 − 𝑦 = 0 which implies that 𝑥 = 𝑦
and
𝑧 − 𝑤 = 0 which implies that 𝑧 = 𝑤
If we let 𝑦 = 𝑟  ℝ and 𝑤 = 𝑠  ℝ then
kerL = { (𝑟, 𝑟, 𝑠, 𝑠) : 𝑟 and 𝑠  ℝ }
= { 𝑟 (1, 1, 0, 0) + 𝑠(0, 0, 1, 1) }
Hence kerL = span {(1, 1, 0, 0), (0, 0, 1, 1)}
Since {(1, 1, 0, 0), (0, 0, 1, 1)} is linearly independent (one vector is not a scalar multiple of
the other) then it forms a basis for kerL. Hence dim(kerL) = 2.
Theorem 5.2.1: If L: V→ W is a linear transformation, then kerL is a subspace of V.
Definition 5.2.3: If L: V→ W is a linear transformation, then the range of L, denoted by range
L, is the set of all vectors in W that are images, under L, of vectors in V. Thus a vector Y is in
range L if we can find some vector X in V such that L(X) = Y. If range L = W, we say that L is
onto.
Example 5: Let L: ℝ3 → ℝ2 be a linear transformation defined by
L(𝑥, 𝑦, 𝑧) = (𝑥 + 𝑦, 𝑦 − 𝑧)
The range or image of L (ImL) is:
ImL = { (𝑥 + 𝑦, 𝑦 − 𝑧) }
= { 𝑥 (1, 0) + 𝑦 (1, 1) + 𝑧 (0, -1) : 𝑥, 𝑦, 𝑧  ℝ }
Thus ImL = span {(1, 0), (1, 1), (0, −1)}. By Theorem 4.1.1 {(1, 0), (1, 1), (0, −1)} is linearly
dependent. Since (0, −1) is a linear combination of the other vectors such as
109
(0, −1) = 1 (1, 0) + (−1)(1, 1)
then we can delete (0, −1) from the set and the remaining vectors still span ImL. The set
{(1, 0), (1, 1)} is linearly independent therefore forms a basis for ImL. Hence dim(ImL) = 2.
Theorem 5.2.2: Let L: V→ W be a linear transformation
(i) L is a monomorphism (one-to-one) if and only if dim(kerL) = 0.
(ii) L is an epimorphism (onto) if and only if dim(ImL) = dim W.
Example 6. In ASAQ 5-3, since dim(kerL) = 2 then L is not a monomorphism (one-to-one)
and in Example 5, since dim(ImL) = 2 = dimℝ2 then L is an epimorphism (onto).
Theorem 5.2.3: If L: V→ W is a linear transformation, then range L is a subspace of W.
Theorem 5.2.4: If L: V→ W is a linear transformation, then
dim(kerL) + dim(range L) = dim V
Example 7: Verify Theorem 5.2.4 using the linear transformation of Example 5.
Solution:
kerL = { (𝑥, 𝑦, 𝑧) : (𝑥 + 𝑦, 𝑦 − 𝑧) = (0, 0) }
Equating the corresponding parts we get:
𝑥 + 𝑦 = 0 which implies that 𝑥 = −𝑦
and
𝑦 − 𝑧 = 0 which implies that 𝑧 = 𝑦 ;
If we let 𝑦 = 𝑟, 𝑟  ℝ, then
kerL = { (−𝑟, 𝑟, 𝑟) : 𝑟  ℝ }
= { 𝑟(−1, 1, 1) : 𝑟  ℝ }
Thus kerL = span { (−1, 1, 1) } which is linearly independent hence forms a basis for kerL.
Thus dim(kerL) = 1.
110
From the previous example, we know that dim(ImL) = 2. Therefore
dim(kerL) + dim(ImL) = dim ℝ3
1
+
2
= 3
Remark: Let L: V → W be a linear transformation. Then the rank of L is defined to be the
dimension of its image, and the nullity of L is defined to be the dimension of its kernel.
rank (L) = dim(ImL) and nullity (L) = dim(kerL)
Thus the preceding theorem yields the following formula for L when V has a finite
dimension:
rank(L) + nullity(L) = dim V.
Corollary 5.2.1: Let L: V→ W be a linear transformation and dim V = dim W,
a. If L is one-to-one, then it is onto.
b. If L is onto, then it is one-to-one.
SAQ 5-4
Let L: ℝ3 →ℝ4 be defined by
L(𝑥, 𝑦, 𝑧) = (𝑥 + 𝑧, 𝑦 − 𝑥, 𝑦 + 𝑧, 𝑥 + 𝑦 + 2𝑧)
Verify Theorem 5.2.7
111
ASAQ 5-4
L(𝑥, 𝑦, 𝑧) = (0, 0, 0, 0)
(𝑥 + 𝑧, 𝑦 − 𝑥, 𝑦 + 𝑧, 𝑥 + 𝑦 + 2𝑧) = (0, 0, 0, 0)
Equating the corresponding parts we obtain the homogeneous system
𝑥
+𝑧 =0
−𝑥 + 𝑦
=0
𝑦 +𝑧 =0
𝑥 + 𝑦 + 2𝑧 = 0
The augmented matrix in reduced row echelon form (verify) is
1
0
[
0
0
0
1
0
0
1
1
0
0
0
0
]
0
0
Thus 𝑥 = −𝑧, 𝑦 = −𝑧, where 𝑧 can be any real number. If we let 𝑧 = 𝑟 ∈ ℝ then
kerL = {(−𝑟, −𝑟, 𝑟): 𝑟 ∈ ℝ}
= {𝑟(−1, −1, 1): 𝑟 ∈ ℝ}
Hence kerL = span {(−1, −1, 1)}. Since (−1, −1, 1) is linearly independent then it is a basis
for kerL. Thus
nullity of L = dim(kerL) = 1.
Solving for the image of L, we have
Im(L) = {(𝑥 + 𝑧, 𝑦 − 𝑥, 𝑦 + 𝑧, 𝑥 + 𝑦 + 2𝑧)}
= {𝑥(1, −1, 0, 1) + 𝑦(0, 1, 1, 1) + 𝑧(1, 0, 1, 2): 𝑥, 𝑦, 𝑧 ∈ ℝ}
Since (1, 0, 1, 2) = (1, −1, 0, 1) + (0, 1, 1, 1) then (1, 0, 1, 2) is a linear combination of the
preceding vectors and may be deleted from the spanning set. Thus
Im(L) = span {(1, −1, 0, 1), (0, 1, 1, 1)}
Since {(1, −1, 0, 1), (0, 1, 1, 1)} is linearly independent (one vector is not a scalar multiple of
the other) then it is a basis for Im(L). Therefore
Rank of L = dim(ImL) = 2
112
Now it can be verified that
Nullity of L + rank of L = dim(ℝ3 )
1
+ 2
= 3
ACTIVITY
1. Which of the following are linear transformations?
(a) L(x, y) = (x2 + x, y – y2)
(b) L(x, y) = (x – y, 0, 2x + 3)
  x   2x  3 y 


(c) L   y    3 y  2 z 
  z    2z 

  
2. Let L: ℝ3 → ℝ4 be defined by
L(x, y, z) = (x + y + z, x + 2y – 3z, 2x + 3y – 2z, 3x + 4y – z)
(a) Find a basis for and the dim(kerL),
(b) Find a basis for and the dim(ImL),
(c) Verify Theorem 5.2.4.
3. Let L: ℝ3 → ℝ3 be defined by
L(x, y z) = (x + z, x + y + 2z, 2x + y + 3z )
(a) Is L one-to-one?
(b) Is L onto?
4. Let L: P2 → P2 be the linear transformation defined by L(at2 + bt + c) = (a + c)t2 +
(b + c)t.
(a) Is t2 – t – 1 in kerL?
(d) Find a basis for kerL.
(b) Is t2 + t -1 in kerL?
(e) Find a basis for ImL.
2
(c) Is 2t – t in range L?
113
5.3 The Matrix of a Linear Transformation
Coordinate Vectors
Definition 5.3.1: Let V be an n-dimensional vector space with basis S = { X1 , X2, … , Xn }. If
X = a1X1 + a2X2 + … + anXn
is any vector in V, then the vector
 a1 
a 
X

 S  2 
 
 an 
in ℝ𝑛 is called the coordinate vector of X with respect to the basis S. The components of
[ X ] S are called the coordinates of X with respect to S.
 1   2 
Example 1. Let S =   ,    be a basis for ℝ2 . Find the coordinate vectors of the
1  3 
following vectors with respect to S.
 3 
(a)  
 7 
12 
(b)  
13
Solution:
 3 
(a) Let X =   . To find [X]S we must find c1 and c2 such that
 7 
 3 
2
1
 7  = c1   + c2  3 
 
 
 1
Multiplying, adding, and equating the corresponding parts, we get
c1 + 2c2 = − 3
− c1 + 3c2 = − 7
The solution is c1 = 1 and c2 = −2. Thus the coordinate vector of X with respect to the
basis S is
1
 2 
 
114
Letter (b) is left as an exercise.

1 
0 
Example 2. Let S =  E1    , E2     be the natural basis for ℝ2 . Then
0
1  

 3 
1 
0 
 7  = −3  0  + (−7)  1 
 
 
 
 3 
Hence [X]S =   . Note that [X]S is the original vector itself when the natural basis is
 7 
used. This result is true in general.
Theorem 5.3.1: Let L: V→ W be a linear transformation of an 𝑛-dimensional vector space V
into an 𝑚-dimensional vector space W ( 𝑛 ≠ 0 and 𝑚 ≠ 0 ) and let S = { X1 , X2, … , Xn} and T =
{Y1, Y2, …, Ym } be bases for V and W, respectively. Then the 𝑚 x 𝑛 matrix A, whose jth
column is the coordinate vector [L(Xj)]T of L(Xj) with respect to T, is associated with L and
has the following property:
If Y = L(X) for some X in V, then [Y]T = A[X]S ,
where [X]S and [Y]T are the coordinate vectors of X and Y with respect to the respective
bases S and T. Moreover, A is the only matrix with this property.
Procedure for computing the matrix of a linear transformation L: V→ W with respect to the
bases S = { X1 , X2, … , Xn} and T = { Y1, Y2, …, Ym } for V and W, respectively:
STEP 1: Compute L(Xj) for j = 1, 2, … , 𝑛.
STEP 2: Find the coordinate vector [L(Xj)]T of L(Xj) with respect to the basis T. This means
that we have to express L(Xj) as a linear combination of the vectors in T.
STEP 3: The matrix A of L with respect to S and T is formed by choosing [L(Xj)]T as the jth
column of A.
Definition 5.3.2: The matrix of Theorem 5.3.1 is called the matrix of L with respect to the
bases S and T. The equation [L(X)]T = A[X]S is called the representation of L with respect to S
and T. We also say that the said equation represents L with respect to S and T.
115
Example 3: Let L: ℝ2 → ℝ2 be defined by
  x 
x  2 y
L   = 

2x  y 
  y 

1 
0 
Let S =  X1    , X 2     be the natural basis for ℝ2 and let
0
1  


 1
 2 
T = Y1    , Y2     be another basis for ℝ2 . Find the matrix of L with respect to S and
2
0 

 1  
T. Compute L     using the matrix of L.
 2 
Solution:
 1  
1 
 1
 2
L     =    a1    a2  
2
2
0
 0  
1 = a1  2a2
2 = 2a1
1
The solution is a1  1 and a2  1, so  L( X1 )T =  
1
 0  
2
 1
 2
L     =    a1    a2  
 1
2
0
 1  
2 = a1  2a2
−1 = 2a1
 1/ 2 
The solution is a1  1/ 2 and a2  3/ 4 , so  L( X 2 )T = 
.
 3/ 4 
Thus the matrix A of L with respect to S and T is
1 1/ 2 
A

1 3 / 4 
  1   
 1  
To compute for  L      using A, we first compute for     . Since S is the natural
   2  T
 2  S
basis for ℝ2 then
116
 1  
1 
  =  
2
 2  S
So,
  1   
1 1/ 2 
 L     = 

1 3 / 4 
   2  T
1 
 0 
 2  = 5 / 2 


 
ACTIVITY
  1   0  1  


1. Let S =   1 ,  2  ,  0   be a basis for ℝ3 . Find the coordinate vector of each of the
  2  1  0 
      
following vectors with respect to S.
1 
(a)  4 
 2 
3
(b)  4 
 3 
1
(c)  2 
 4 
x  2 y


x


2. Let L: ℝ2 → ℝ3 be defined by L       2 x  y  . Let S and T be the natural bases for
  y   x  y 


  1  0  
ℝ and ℝ , respectively. Also, let S’ =   ,    and T’ =
1 1 
2
3
 1   0   1  
      
 1  , 1  ,  1  be bases for
 0  1   1  
      
ℝ2 and ℝ3 , respectively. Find the matrix of L with respect to
(a) S and T
(b) S’ and T’
3. Let L: P1 → P3 be defined by L(p(t)) = t2p(t). Let S = { t, t + 1 } and
T = { t3 , t2 – 1, t, t + 1 } be bases for P1 and P3, respectively. Find the matrix of L
with respect to S and T. Compute [ L(–3t + 3) ]T using the matrix of L.
117
MODULE 6
EIGENVALUES AND EIGENVECTORS
Introduction
In this section we will discuss the concepts of eigenvalues, eigenvectors, and
eigenspaces, algebraic and geometric multiplicity of an eigenvalue, Hamilton-Cayley
theorem, similar matrices and diagonalizable matrices.
Objectives
At the end of this chapter, you are expected to be able to do the following:
1. Define eigenvalues and eigenvectors.
2. Compute for the eigenvalues and the associated eigenvectors of a matrix representing a
linear transformation.
3. Determine the algebraic and geometric multiplicities of the eigenvalues.
3. Identify the properties of a diagonalizable matrix.
4. Determine the matrix of transition P so that 𝑃 −1 𝐴𝑃 is a diagonal matrix, given a
matrix A representing a linear transformation of a vector space into itself.
6.1 Characteristic Polynomial
Definition 6.1.1: If A  [ a ij ] is an 𝑛 x 𝑛 matrix, the polynomial matrix
𝑥𝐼 − 𝐴 = 𝐶
is called the characteristic matrix of 𝐴. The determinant of 𝐶 is called the characteristic
polynomial of 𝐴. The equation det𝐶 = 0 is called the characteristic equation of 𝐴.
1
Example 1: Let 𝐴 = [
4
−3
]. Then
−2
𝑥𝐼2 − 𝐴 = 𝐶 = [
= [
is called the characteristic matrix of 𝐴.
𝑥
0
0
1
]−[
𝑥
4
𝑥−1
−4
3
]
𝑥+2
−3
]
−2
118
The determinant of 𝐶
𝑥−1
|
−4
3
| = (𝑥 − 1)(𝑥 + 2) − (−12)
𝑥+2
= 𝑥 2 + 𝑥 + 10
is called the characteristic polynomial of 𝐴 and the equation
𝑥 2 + 𝑥 + 10 = 0
is called the characteristic equation of 𝐴.
6.2 Hamilton-Cayley Theorem
Let 𝑓(𝑥) = 𝑎0 𝑥 𝑛 + 𝑎1 𝑥 𝑛−1 + … + 𝑎𝑛−1 𝑥 + 𝑎𝑛 be a polynomial in 𝑥 with real
coefficients. If 𝐴 is an 𝑛 x 𝑛 matrix then 𝑓(𝐴) is equal to the matrix
𝑎0 𝐴𝑛 + 𝑎1 𝐴𝑛−1 + … + 𝑎𝑛−1 𝐴 + 𝑎𝑛 𝐼𝑛
Note that we replace the constant term by 𝑎𝑛 𝐼𝑛 so that each term of 𝑓(𝐴) is a matrix.
Theorem 6.2.1. (Hamilton-Cayley Theorem) If 𝐴 is an 𝑛 x 𝑛 matrix and 𝑓(𝑥) is its
characteristic polynomial, then 𝑓(𝐴) = 0.
2 3
Example 1: Let 𝐴 = [
]. The characteristic matrix of 𝐴 is
−1 4
𝑥 − 2 −3
𝐶=[
]
1
𝑥−4
Next we compute for the characteristic polynomial of 𝐴.
𝑓(𝑥) = det𝐶
𝑥−2
𝑓(𝑥) = |
1
−3
|
𝑥−4
= (𝑥 − 2)(𝑥 − 4) − (−3)
= 𝑥 2 − 6𝑥 + 11
It can be verified that the
constant term is equal to
the determinant of 𝐴.
119
Computing for 𝑓(𝐴) we have
𝑓(𝐴) = 𝐴2 − 6𝐴 + 11𝐼2
2 3 2
=[
][
−1 4 −1
1
3
2 3
] − 6[
] + 11 [
0
4
−1 4
1 18
12 18
11
= [
]−[
]+[
−6 13
−6 24
0
0
= [
0
0
]
1
0
]
11
0
]
0
Hence 𝐴 satisfies its characteristic equation, that is 𝑓(𝐴) = 0. Since det𝐴 = 11 ≠ 0 then 𝐴
is nonsingular so 𝐴−1 exists. By the Hamilton-Cayley Theorem, we form the equation
𝐴2 − 6𝐴 + 11𝐼2 = 0
11𝐼2 = 6𝐴 − 𝐴2
𝐼𝟐 =
𝐼𝟐 =
1
11
(6𝐴 − 𝐴2 )
1
11
(𝐴)(6𝐼 − 𝐴)
Therefore,
1
𝐴−1 = 11 (6𝐼 − 𝐴)
4 −3
]
1 2
4/11 −3/11
= [
]
1/11 2/11
1
= 11 [
120
SAQ 6-1
Find the characteristic polynomial for the matrix
2 −2 3
[1 1
1]
1 3 −1
and
a) Show by direct substitution that this matrix
satisfies its characteristic equation.
b) Find 𝐴−1 .
ASAQ 6-1
The characteristic matrix of 𝐴 is
𝑥−2
[ −1
−1
2
−3
𝑥−1
−1 ]
−3 𝑥 + 1
and its characteristic polynomial is
𝑓(𝑥) = (𝑥 − 2)(𝑥 − 1)(𝑥 + 1) − 7 − [3(𝑥 − 1) − 2(𝑥 + 1) + 3(𝑥 − 2)]
= 𝑥 3 − 2𝑥 2 − 5𝑥 + 6
Next we form the equation
𝑓(𝐴) = 𝐴3 − 2𝐴2 − 5𝐴 + 6𝐼3 = 0
(1)
where
2 −2 3 2 −2 3
5
𝐴2 = 𝐴 ∙ 𝐴 = [1 1
1 ] [1 1
1 ] = [4
1 3 −1 1 3 −1
4
5 3 1 2 −2
𝐴3 = 𝐴2 ∙ 𝐴 = [4 2 3] [1 1
4 −2 7 1 3
Substituting these to (1), we have
3 1
2 3] and
−2 7
3
14 −4
1 ] = [13 3
−1
13 11
17
11]
3
121
14
𝑓(𝐴) = [13
13
0 0
= [0 0
0 0
−4 17
2 −2 3
6 0
5 3 1
]
−
2
[
]
−
5
[
]
+
[
3 11
1 1
1
0 6
4 2 3
11 3
1 3 −1
0 0
4 −2 7
0
0]
6
0
0]
0
Thus 𝐴 satisfies its characteristic equation.
b. To solve for 𝐴−1 , we form the equation
𝐴3 − 2𝐴2 − 5𝐴 + 6𝐼3 = 0
6𝐼3 = −𝐴3 + 2𝐴2 + 5𝐴
1
𝐼3 = 6 (−𝐴3 + 2𝐴2 + 5𝐴)
1
𝐼3 = 6 (𝐴)(−𝐴2 + 2𝐴 + 5𝐼3 )
Therefore
1
𝐴−1 = 6 (−𝐴2 + 2𝐴 + 5𝐼3 )
2
1 −5 −3 −1
= [[−4 −2 −3] + 2 [1
6
1
−4 2 −7
−2 3
5 0 0
]
+
[
1
1
0 5 0]]
3 −1
0 0 5
4 −7 5
= 6 [−2 5 −1]
−2 8 −4
2/3 −7/6 5/6
−1/3
5/6 −1/6]
=[
−1/3 4/3 −2/3
1
6.3 Eigenvalues, Eigenvectors, and Eigenspaces
Definition 6.3.1: Let 𝐴 be an 𝑛 x 𝑛 matrix. The real number 𝜆 is called an eigenvalue of 𝐴 if
there exists a nonzero vector 𝑿 in ℝ𝑛 such that
𝐴𝑿 = 𝜆𝑿
(1)
122
Every nonzero vector 𝑿 satisfying (1) is called an eigenvector of 𝐴 associated with the
eigenvalue 𝜆.
Example 1: Let 𝐴 = [
[
2 −1
]. Since
−2 3
2 −1 1
1
][ ] = 4[ ]
−2 3 −2
−2
then 4 is an eigenvalue of 𝐴 and [
4.
1
] is the eigenvector associated with the eigenvalue 𝜆 =
−2
Theorem 6.3.1: The eigenvalues of 𝐴 are the real roots of the characteristic polynomial of
A.
1 0
Example 2. Let 𝐴 = [−1 3
3 2
𝑥−1
| 1
−3
0
0 ]. The characteristic polynomial of 𝐴 is
−2
0
𝑥−3
−2
0
0 | = (𝑥 − 1)(𝑥 − 3)(𝑥 + 2)
𝑥+2
By Theorem 6.3.1, the eigenvalues of 𝐴 are the roots of the characteristic equation
(𝑥 − 1)(𝑥 − 3)(𝑥 + 2) = 0
Thus 𝜆1 = 1, 𝜆2 = 3, and 𝜆3 = −2 are the eigenvalues of 𝐴.
To find the eigenvector associated with 𝜆1 = 1, we form the system (1𝐼3 − 𝐴)𝑋 = 0:
0
0 0 𝑥1
0
𝑥
[ 1 −2 0] [ 2 ] = [0]
−3 −2 3 𝑥3
0
6
A solution to this system is {(4 𝑟, 8 𝑟, 𝑟) : 𝑟 ∈ ℝ}. Thus if 𝑟 = 8 then 𝑋1 = [3] is the
8
eigenvector associated with 𝜆1 = 1.
3
3
123
To find the eigenvector associated with 𝜆2 = 3, we form the system (3𝐼3 − 𝐴)𝑋 = 0:
2
0 0 𝑥1
0
𝑥
[1
0 0] [ 2 ] = [0]
−3 −2 5 𝑥3
0
0
A solution to this system is {(0, 2 𝑟, 𝑟) : 𝑟 ∈ ℝ}. Thus if 𝑟 = 2 then 𝑋2 = [5] is the
2
eigenvector associated with 𝜆2 = 3.
5
To find the eigenvector associated with 𝜆3 = −2, we form the system (−2𝐼3 − 𝐴)𝑋 = 0:
−3 0 0 𝑥1
0
𝑥
[ 1 −5 0] [ 2 ] = [0]
−3 −2 0 𝑥3
0
0
A solution to this system is {(0, 0, 𝑟): 𝑟 ∈ ℝ}. Thus 𝑋2 = [0] is the eigenvector associated
1
with 𝜆3 = −2.
 2 1
Example 3: Let A  
.
 1 3
The characteristic polynomial of 𝐴 is
𝑥−2
|
1
−1
| = (𝑥 − 2)(𝑥 − 3) − (−1)
𝑥−3
= 𝑥 2 − 5𝑥 + 7
Since 𝑥 2 − 5𝑥 + 7 = 0 has no real roots then A has no eigenvalues.
Definition 6.3.2: Let L: V→ V be a linear transformation and let  be an eigenvalue of L.
Denote by S(  ) the set of all eigenvectors associated with  , together with the zero vector.
Then S(  ) is called the eigenspace of L associated with  .
Definition 6.3.3: Let  be an eigenvalue of L.
(a) The geometric multiplicity of  is the dimension of S(  ).
(b) The algebraic multiplicity of  is its multiplicity as a root of the characteristic polynomial
f(  ).
124
Example 4. Let L be a linear transformation represented by the matrix
2 0
0
𝐴 = [3 −1 0 ]
0 4 −1
Determine the geometric and algebraic multiplicities of its eigenvalues.
Solution: The characteristic polynomial of A is
𝑥−2
0
0
𝑓(𝑥) = | −3 𝑥 + 1
0 |
0
−4
𝑥+1
= (𝑥 − 2)(𝑥 + 1)(𝑥 + 1)
= (𝑥 − 2)(𝑥 + 1)2
Thus the eigenvalues of A are 𝜆1 = 2 with algebraic multiplicity 1 and 𝜆2 = 𝜆3 = −1 with
algebraic multiplicity 2.
To find the geometric multiplicity of 𝜆1 = 2, we have to determine the dimension of the
eigenspace S(2) associated with 𝜆1 = 2. Substituting 2 for x in the characteristic matrix of A
we obtain
0
0 0
[−3 3 0]
0 −4 3
This matrix in reduced row echelon form is
1
[0
0
3
The solution is {(4 𝑟,
3
4
0 −3/4
1 −3/4]
0
0
𝑟, 𝑟) : 𝑟 ∈ ℝ}. Thus
3
S(2) = {(4 𝑟,
3
4
𝑟, 𝑟) : 𝑟 ∈ ℝ}
= span {(3, 3, 4)}
Hence 𝜆1 = 2 has geometric multiplicity 1.
125
Similarly, substituting −1 for x in the characteristic matrix we obtain
−3 0 0
[−3 0 0]
0 −4 0
This matrix in reduced row echelon form is
1 0
[0 1
0 0
0
0]
0
Thus
S(−1) = {(0, 0, 𝑟): 𝑟 ∈ ℝ}
= span{(0, 0, 1)}
Hence 𝜆2 = 𝜆3 = −1 has geometric multiplicity 1.
Theorem 6.3.2. Let 𝜆 be an eigenvalue. Then the geometric multiplicity of 𝜆 does not
exceed its algebraic multiplicity.
Theorem 6.3.3. If the eigenvalues 𝜆1 , … , 𝜆𝑛 are all different and {𝜉1 , … , 𝜉2 } is a set of
eigenvectors, 𝜉𝑖 corresponding to 𝜆𝑖 , then the set {𝜉1 , … , 𝜉2 } is linearly independent.
Example 5. Let A be the matrix of Example 2. The three distinct eigenvalues of A are
6
0
0
𝜆1 = 1, 𝜆2 = 3, and 𝜆3 = −2 and the corresponding eigenvectors are [3], [5], and [0]. It
8
2
1
6
0
0
can be verified that [3], [5], and [0] are linearly independent.
8
2
1
126
SAQ 6-2
3
2
Let 𝐴 = [ 1
4
−2 −4
2
1 ].
−1
Determine the algebraic and geometric multiplicities of its
eigenvalues.
ASAQ 6-2
The characteristic polynomial of A is
𝑥−3
−2
−2
𝑓(𝑥) = | −1 𝑥 − 4 −1 |
2
4
𝑥+1
= 𝑥 3 − 6𝑥 2 + 11𝑥 − 6
= (𝑥 − 1)(𝑥 − 2)(𝑥 − 3)
Thus the eigenvalues of A are 1, 2, and 3 all of which have algebraic multiplicity 1.
To solve for the geometric multiplicity of 𝜆 = 1, we substitute 1 for x in the characteristic
matrix obtaining the matrix
−2 −2 −2
[−1 −3 −1]
2
4
2
This matrix in reduced row echelon form is
1 0
[0 1
0 0
Thus
S(1) = {(−𝑟, 0, 𝑟): 𝑟 ∈ ℝ
= span{(−1, 0, 1)}
Hence 𝜆 = 1 has geometric multiplicity 1.
1
0]
0
127
To solve for the geometric multiplicity of 𝜆 = 2, we substitute 2 for x in the characteristic
matrix obtaining the matrix
−1 −2 −2
[−1 −2 −1]
2
4
3
This matrix in reduced row echelon form is equal to
1 2
[0 0
0 0
0
1]
0
Thus
S(2) = {(−2𝑟, 𝑟, 0): 𝑟 ∈ ℝ
= span{(−2, 1, 0)}
Hence 𝜆 = 2 has geometric multiplicity 1.
To solve for the geometric multiplicity of 𝜆 = 3, we substitute 3 for x in the characteristic
matrix obtaining the matrix
0 −2 −2
[−1 −1 −1]
2
4
4
This matrix in reduced row echelon form is
1 0
[0 1
0 0
Thus
S(3) = {(0, −𝑟, 𝑟): 𝑟 ∈ ℝ
= span{(0, −1, 1)}
Hence 𝜆 = 3 has geometric multiplicity 1.
0
1]
0
128
6.4 Diagonalization
Definition 6.4.1: A matrix B is similar to a matrix A if there is a nonsingular matrix P such
that
B = P-1AP
Example 1: Let 𝐴 = [
𝑃−1 𝐴𝑃 we have
1 −1
−1 −1
1
] and 𝑃 = [
]. Then 𝑃−1 = [
2 4
2
1
−2
𝐵=[
= [
1
]. If we let 𝐵 =
−1
1
1 1 −1 −1 −1
][
][
]
−2 −1 2 4
2
1
3 0
]
0 2
By Definition 6.4.1, B is similar to A.
REMARKS:
1. A is similar to A.
2. If B is similar to A, then A is similar to B.
3. If A is similar to B and B is similar to C, then A is similar to C.
Theorem 6.4.1. Similar matrices have the same characteristic polynomial.
Theorem 6.4.2. Similar matrices have the same eigenvalues and eigenvectors.
Example 2. Let A and B be the matrices of Example 1. The characteristic polynomial of A is
𝑥−1
1
𝑓(𝑥) = |
| = 𝑥 2 − 5𝑥 + 6
−2 𝑥 − 4
and the characteristic polynomial of B is
𝑥−3
0
𝑓(𝑥) = |
| = 𝑥 2 − 5𝑥 + 6
0
𝑥−2
Thus A and B have the same characteristic polynomial. Consequently, A and B also have the
same eigenvalues and eigenvectors.
129
Definition 6.4.2: We shall say that the matrix A is diagonalizable if it is similar to a diagonal
matrix. In this case we also say that A can be diagonalized.
Example 3. Let A be the matrix of Example 1. Since A is similar to a diagonal matrix then A is
diagonalizable.
Theorem 6.4.3: An 𝑛 x 𝑛 matrix A is diagonalizable if and only if it has 𝑛 linearly independent
eigenvectors. In this case A is similar to a diagonal matrix D, with P-1AP = D, whose diagonal
elements are the eigenvalues of A, while P is a matrix whose columns are 𝑛 linearly
independent eigenvectors of A.
Theorem 6.4.4: A matrix A is diagonalizable if all the roots of its characteristics polynomial
are real and distinct.
Example 4. Let 𝐴 = [
1 4
]. The characteristic polynomial of A is
1 −2
𝑥−1
𝑓(𝑥) = |
−1
−4
| = 𝑥 2 + 𝑥 − 6 = (𝑥 + 3)(𝑥 − 2)
𝑥+2
Since the eigenvalues are real and distinct then A is diagonalizable.
Example 5. Let 𝐴 be the matrix of SAQ6-2. The distinct eigenvalues of A are 1, 2, and 3
−1
−2
hence A is diagonalizable. The linearly independent eigenvectors of A are [ 0 ] , [ 1 ] ,
1
0
0
and [−1]. Thus by Theorem 6.4.3
1
−1 −2 0
1
2
2
𝑃=[ 0
1 −1] and 𝑃−1 = [−1 −1 −1] (verify)
1
0
1
−1 −2 −1
Then
𝐷 = 𝑃−1 𝐴𝑃
1
= [−1
−1
2
2
3
2
2 −1 −2 0
−1 −1] [ 1
4
1 ][ 0
1 −1]
−2 −1 −2 −4 −1 1
0
1
1
2
2 −1 −2 0
= [−2 −2 −2] [ 0
1 −1]
−3 −6 −3 1
0
1
130
1 0
= [0 2
0 0
0
0]
3
Note that D is a diagonal matrix whose diagonal elements are the eigenvalues of A.
ACTIVITY
2  2 3 
1. Let A  0 3  2 . Find the characteristic polynomial, eigenvalues, and
0  1 2 
eigenvectors of A.
3  2 1 
2. Let A  0 2 0 . Find a nonsingular matrix P such that P-1AP is diagonal.
0 0 0
3. Let L be a linear transformation represented by the matrix
0 0 0 
A  0 1 0 
1 0 1 
Determine the geometric and algebraic multiplicities of its eigenvalues.
131
MODULE 7
INNER PRODUCT SPACES
Introduction
In this section we will discuss inner product, inner product spaces, orthonormal
bases in ℝ𝑛 , the Gram-Schmidth orthogonalization process, diagonalization of symmetric
matrix, and quadratic forms.
Objectives
After studying this module, you should be able to:
1. Explain the steps on the diagonalization of the matrix.
2. Enumerate the properties of a diagonizable symmetric matrix.
3. Perform diagonalization of a symmetric matrix by orthogonalization.
7.1 Inner Product in ℝ𝒏
Definition 7.1.1. The length or magnitude or norm of the vector 𝑋 = (𝑥1 , 𝑥2 , … , 𝑥𝑛 ) in ℝ𝑛
is
‖𝑋‖ = √𝑥1 2 + 𝑥2 2 + … + 𝑥𝑛 2
The above formula is also use to define the distance from the point (𝑥1 , 𝑥2 , … , 𝑥𝑛 ) to the
origin.
Example 1. Let 𝑋 = (1, 2, −3) be a vector in ℝ3 . The length of 𝑋 is
‖𝑋‖ = √12 + 22 + (−3)2 = √14
Definition 7.1.2. If 𝑋 = (𝑥1 , 𝑥2 , … , 𝑥𝑛 ) and 𝑌 = (𝑦1 , 𝑦2 , … , 𝑦𝑛 ) are vectors in ℝ𝑛 , then their
inner product (also called dot product) is given by
𝑋 ∙ 𝑌 = 𝑥1 𝑦1 + 𝑥2 𝑦2 + ⋯ 𝑥𝑛 𝑦𝑛 .
Example 2. Let 𝑋 = (1, 0, 1) and 𝑌 = (2, 3, −1). Then their inner product is
𝑋 ∙ 𝑌 = 1 ∙ 2 + 0 ∙ 3 + 1(−1) = 1
132
Theorem 7.1.1. (Properties of Inner Product) If 𝑋, 𝑌 and 𝑍 are vectors in ℝ𝑛 and 𝑐 is a
scalar, then:
a. 𝑋 ∙ 𝑋 = ‖𝑋‖2 ≥ 0, with equality if and only if 𝑋 = 0.
b. 𝑋 ∙ 𝑌 = 𝑌 ∙ 𝑋
c. (𝑋 + 𝑌) ∙ 𝑍 = 𝑋 ∙ 𝑍 + 𝑌 ∙ 𝑍
d. (𝑐𝑋) ∙ 𝑌 = 𝑋 ∙ (𝑐𝑌) = 𝑐(𝑋 ∙ 𝑌)
NOTE: A vector space V along with an inner product or scalar product operation satisfying
all these properties is called an inner product space.
Example 3.
product.
ℝ𝑛 is an inner product space since it satisfies all the properties of inner
Definition 7.1.3. The angle between two nonzero vectors 𝑋 and 𝑌 is defined as the unique
number 𝜃, 0 ≤ 𝜃 ≤ 𝜋, such that
𝑐𝑜𝑠𝜃 =
𝑋∙𝑌
‖𝑋‖‖𝑌‖
Example 4. Let 𝑋 = (0, 0, 1, 1) and 𝑌 = (1, 0, 1, 0). Then
‖𝑋‖ = √2, ‖𝑌‖ = √2 and 𝑋 ∙ 𝑌 = 1
Thus
𝑐𝑜𝑠𝜃 =
1
(√2)(√2)
1
𝑐𝑜𝑠𝜃 = 2
1
𝜃 = cos−1 (2)
𝜃 = 60𝑜
Definition 7.1.4. Two nonzero vectors 𝑋 and 𝑌 in ℝ𝑛 are said to be orthogonal if 𝑋 ∙ 𝑌 = 0.
If one of the vectors is the zero vector, we agree to say that the vectors are orthogonal.
They are said to be parallel if |𝑋 ∙ 𝑌| = ‖𝑋‖‖𝑌‖. They are in the same direction if 𝑋 ∙ 𝑌 =
‖𝑋‖‖𝑌‖. That is, they are orthogonal if 𝑐𝑜𝑠𝜃 = 0, parallel if 𝑐𝑜𝑠𝜃 = ±1 and in the same
direction if 𝑐𝑜𝑠𝜃 = 1.
133
Example 5. Let 𝑋1 = (4, 2, 6, −8), 𝑋2 = (−2, 3, −1, −1) and 𝑋3 = (−2, −1, −3, 4). Then
𝑋1 ∙ 𝑋2 = 4(−2) + 2 ∙ 3 + 6(−1) + (−8)(−1) = 0,
𝑋2 ∙ 𝑋3 = (−2)(−2) + 3(−1) + (−1)(−3) + (−1)(4) = 0 and
𝑋1 ∙ 𝑋3 = 4(−2) + 2(−1) + 6(−3) + (−8)(4) = −60
This shows that 𝑋1 and 𝑋2 are orthogonal and 𝑋2 and 𝑋3 are also orthogonal.
Moreover, ‖𝑋1 ‖ = 2√30 and ‖𝑋3 ‖ = √30 then
|𝑋1 ∙ 𝑋3 | = ‖𝑋1 ‖‖𝑋3 ‖
|−60| = (2√30)(√30)
60 = 60
This implies that 𝑋1 and 𝑋3 are parallel but not in the same direction.
Definition 7.1.5. A unit vector 𝑈 in ℝ𝑛 is a vector of unit length. If 𝑋 is a nonzero vector,
then the vector
𝑈= [
1
‖𝑋‖
]𝑋
is a unit vector in the direction of 𝑋.
Example 6. Consider the vector 𝑋 = (0, 4, 2, 3). Since ‖𝑋‖ = √29 then the vector
𝑈=
1
√29
(0, 4, 2, 3) = (0,
Is a unit vector in the direction of 𝑋.
4
√29
,
2
√29
,
3
)
√29
134
72. Orthonormal Bases in ℝ𝒏
Definition 7.2.1. A set 𝑆 = {𝑥1 , 𝑥2 , … , 𝑥𝑘 } in ℝ𝑛 is called orthogonal if any two distinct
vectors in S are orthogonal. An orthonormal set of vectors is an orthogonal set of unit
vectors.
Example 1. Let 𝑋1 = (1, 2, −1, 1), 𝑋2 = (0, −1, −2, 0) and 𝑋3 = (1, 0, 0, −1). Since
𝑋1 ∙ 𝑋2 = 0, 𝑋1 ∙ 𝑋3 = 0 and 𝑋2 ∙ 𝑋3 = 0 then the set {𝑋1, 𝑋2 , 𝑋3 } is orthogonal. The
vectors
𝑌1 = (
1
,
2
√7 √7
,−
1
,
1
) , 𝑌2 = (0, −
√7 √7
1
√5
,−
2
√5
, 0) and 𝑌3 = (
1
√2
, 0, 0, −
1
)
√2
are unit vectors in the direction of 𝑋1 , 𝑋2 and 𝑋3 , respectively. Thus {𝑌1 , 𝑌2 , 𝑌3 } is an
orthonormal set and span the same subspace as {𝑋1 , 𝑋2 , 𝑋3 }, that is, span{𝑋1 , 𝑋2 , 𝑋3 } =
span{𝑌1 , 𝑌2 , 𝑌3 }.
Notice that an orthonormal set is just a set of orthogonal vectors in which each
vector has been normalized to unit length.
Theorem 7.2.1. Let 𝑆 = {𝑥1 , 𝑥2 , … , 𝑥𝑘 } be an orthogonal set of nonzero vectors in ℝ𝑛 . Then
S is linearly independent.
Example 2. The vectors 𝑋1 , 𝑋2 and 𝑋3 of Example 1 are orthogonal hence they are linearly
independent.
Corollary 7.2.1. An orthonormal set of vectors in ℝ𝑛 is linearly independent.
Example 3. The set of vectors {𝑌1 , 𝑌2 , 𝑌3 } of Example 1 is orthonormal hence it is linearly
independent.
Gram-Schmidt Orthogonalization Process
Gram-Schmidth process is a procedure by which an orthonormal set of vectors is
obtained from a linearly independent set of vectors in an inner product space.
135
Definition 7.2.2. (Gram-Schmidt Process) Let W be a nonzero subspace of ℝ𝑛 with basis 𝑆 =
{𝑋1 , 𝑋2 , … , 𝑋𝑚 }. Then there exists an orthonormal basis 𝑇 = {𝑍1 , 𝑍2 , … , 𝑍𝑚 } for W.
Steps in the Gram-Schmidth process:
STEP 1. Let 𝑌1 = 𝑋1.
STEP 2. Compute the vectors 𝑌2 , 𝑌3 , … , 𝑌𝑚 , by the formula
𝑋𝑘 ∙𝑌𝑗
𝑌𝑘 = 𝑋𝑘 − ∑𝑘−1
𝑗=1 ( 𝑌
𝑗 ∙𝑌𝑗
) 𝑌𝑗 for 2 ≤ 𝑘 ≤ 𝑚
The set of vectors 𝑆 ′ = {𝑌1 , 𝑌2 , … , 𝑌𝑚 } is an orthogonal set.
STEP 3. Normalize each vector in 𝑆′ to obtain an orthonormal basis for W.
Example 4. Use the Gram-Schmidth process to transform the basis {(1, 1, 1), (0, 1, 1),
(1, 2, 3)} for ℝ3 into an orthonormal basis for ℝ3 .
Solution: We apply the Gram-Schmidth process to obtain 3 vectors 𝑍1 , 𝑍2 , 𝑍3 which also
span ℝ3 and orthogonal to each other.
STEP 1. Let 𝑌1 = 𝑋1 = (1, 1, 1).
STEP 2. We now compute for 𝑌2 and 𝑌3 :
𝑋2 ∙ 𝑌1
𝑌2 = 𝑋2 − (
)𝑌
𝑌1 ∙ 𝑌1 1
2
= (0, 1, 1) − ( ) (1, 1, 1)
3
2 1 1
= (− , , )
3 3 3
and
𝑋3 ∙ 𝑌1
𝑋3 ∙ 𝑌2
𝑌3 = 𝑋3 − (
) 𝑌1 − (
)𝑌
𝑌1 ∙ 𝑌1
𝑌2 ∙ 𝑌2 2
6
1
2 1 1
= (1, 2, 3) − ( ) (1, 1, 1) − (
) (− , , )
3
6/9
3 3 3
136
3
2 1 1
= (1, 2, 3) − (2, 2, 2) − (− , , )
2
3 3 3
1 1
= (0, − , )
2 2
2 1 1
1 1
The set of vectors 𝑆 ′ = {(1, 1, 1), (− 3 , 3 , 3) , (0, − 2 , 2)} is orthogonal.
STEP 3. We normalize each vector found in STEP 2.
Let
𝑍1 =
𝑌1
‖𝑌1 ‖
1
𝑍2 =
𝑌2
1
=
‖𝑌2 ‖
√6/9
=
=
√3
3
𝑌3
‖𝑌3 ‖
=
=
1
,
1
,
1
√3 √3 √3
) , (−
1
,
1
)
2 1 1
2 1 1
√6
2
√2
,
,
1
)
1 1
√2/4
2
1
,
√6 √6 √6
1
2
,
√3 √3 √3
(− 3 , 3 , 3)
(0, − 2 , 2)
1 1
(0, − 2 , 2)
= (0, −
Then 𝑇 = {(
1
(− 3 , 3 , 3)
= (−
𝑍3 =
(1, 1, 1) = (
1
1
,
1
)
√2 √2
1
1
1
, ) , (0, − √2 , √2)} is an orthonormal basis for ℝ3 .
√6 √6 √6
137
SAQ 7-1
Use the Gram-Schmidt process to find an orthonormal basis for the
subspace W of ℝ4 with basis {(1, 1, -1, 0), (0, 2, 0, 1), (-1, 0, 0, 1)}.
ASAQ 7-1
STEP 1. Let 𝑌1 = 𝑋1 = (1, 1, −1, 0)
STEP 2. We now solve for 𝑌2 and 𝑌3 :
𝑋2 ∙ 𝑌1
𝑌2 = 𝑋2 − (
)𝑌
𝑌1 ∙ 𝑌1 1
2
= (0, 2, 0, 1) − (1, 1, −1, 0)
3
2 4 2
= (− , , , 1)
3 3 3
𝑋3 ∙ 𝑌1
𝑋3 ∙ 𝑌2
𝑌3 = 𝑋3 − (
) 𝑌1 − (
)𝑌
𝑌1 ∙ 𝑌1
𝑌2 ∙ 𝑌2 2
1
5/3
2 4 2
= (−1, 0, 0,1) − (− ) (1, 1, −1, 0) −
(− , , , 1)
3
11/3
3 3 3
= (−
4
3
7 6
,− ,− , )
11 11 11 11
2 4 2
4
3
7
6
Thus 𝑆 ′ = {(1, 1, −1, 0), (− 3 , 3 , 3 , 1) , (− 11 , − 11 , − 11 , 11)} is an orthogonal basis for a
subspace W of ℝ4 .
138
STEP 3. We normalize each vector in 𝑆′.
Let 𝑍1 =
𝑍2 =
𝑌1
‖𝑌1 ‖
1
=
(1, 1, -1, 0) = (
√3
,
1
√3 √3
,−
1
√3
,0 )
𝑌2
1
2 4 2
=
(− 3 , 3 , 3 , 1)
‖𝑌2 ‖
√33/9
=
3
2 4 2
𝑌3
‖𝑌3 ‖
1
=
11
1 1
3 3
4
,
,
2
,
3
)
√33 √33 √33 √33
3
7
6
(− 11 , − 11 , − 11 , 11)
4
√110
= (−
2
4
√110/121
=
(− 3 , 3 , 3 , 1)
√33
= (−
𝑍3 =
1
3
7
6
(− 11 , − 11 , − 11 , 11)
4
√110
,−
3
√110
1
3
,−
7
,
6
√110 √110
)
2
4
2
3
4
3
7
6
, , , ) , (− √110 , − √110 , − √110 , √110)} is
33 √33 √33 √33
Thus 𝑇 = {(√ , √ , − √ , 0 ) , (− √
an orthonormal basis for the subspace W of ℝ4
ACTIVITY
1. Which of the following are orthonormal sets of vectors?
1
1
1 1 1
a. ( , 0, − ) , ( , , ) , (0, 1, 0)
√2
√2
√3 √3 √3
b. (0, 2, 2, 1), (−1, 1, 2, 2), (0, 1, −2, 1)
1
c. (
√6
,−
2
√6
, 0,
1
) , (−
√6
1
√3
,−
1
√3
, 0, −
1
√3
) , (0, 0, 1, 0)
139
2. Use the Gram-Schmidt process to find an orthonormal basis for the subspace of ℝ4
with basis {(1, -1, 0, 1), (2, 0, 0, -1), (0, 0, 1, 0)}.
7.3 Diagonalization of Symmetric Matrix
Theorem 7.3.1. All roots of the characteristic polynomial of a symmetric matrix are real
numbers.
−1 0
Example 1. Let 𝐴 = [ 0 2
0 0
0
0]. The characteristic polynomial of A is
3
𝑥+1
𝑓(𝑥) = | 0
0
0
0
𝑥−2
0 |
0
𝑥−3
= (𝑥 + 1)(𝑥 − 2)(𝑥 − 3)
Clearly the roots of 𝑓(𝑥) are all real numbers.
Corollary 7.3.1. If A is a symmetric matrix all of whose eigenvalues are distinct, then A is
diagonalizable.
Example 2. Let A be the matrix of Example 1. The eigenvalues 𝜆1 = −1, 𝜆2 = 2 and 𝜆3 = 3
are real and distinct. Hence A is diagonalizable.
Theorem 7.3.2. If A is a symmetric matrix, then the eigenvectors that belong to distinct
eigenvalues of A are orthogonal.
Example 3. Let A be the matrix of Example 1. To find the associated eigenvectors, we form
the system (𝜆𝐼3 − 𝐴)𝑋 = 0 and solve for x. Thus
1
0
0
𝑋1 = [0], 𝑋2 = [1] and X 3 = [0]
0
0
1
140
are the eigenvectors associated with 𝜆1 = −1, 𝜆2 = 2 and 𝜆3 = 3, respectively. It can be
verified that {𝑋1 , 𝑋2 , 𝑋3 } is an orthogonal set of vectors in ℝ3 2.
Definition 7.3.1. A nonsingular matrix A is called orthogonal if 𝐴−1 = 𝐴𝑇 .
1
0
0
Example 4. Let 𝐴 = [0 1/√2 −1/√2].
0 −1/√2 −1/√2
−1
Since 𝐴
1
= [0
0
0
0
1/√2 −1/√2] = 𝐴𝑇 then A is an orthogonal matrix.
−1/√2 −1/√2
Theorem 7.3.3. The 𝑛 x 𝑛 matrix A is orthogonal if and only if the columns (and rows) of A
form an orthonormal set of vectors in ℝ𝑛 .
Example 5. Let A be the matrix of Example 4. The columns (and rows) of A are of unit length
and are mutually orthogonal. Thus A is orthogonal.
Theorem 7.3.4. If A is symmetric 𝑛 x 𝑛 matrix , then there exists an orthogonal matrix P
such that 𝑃−1 𝐴𝑃 = 𝑃𝑇 𝐴𝑃 = 𝐷, a diagonal matrix, where the columns of P consist of the
linearly independent eigenvectors of A and the diagonal elements of D are the eigenvalues
of A associated with these eigenvectors.
0 −1 −1
Example 6. Let 𝐴 = [−1 0 −1]. Find an orthogonal matrix P and a diagonal matrix D
−1 −1 0
such that 𝐷 = 𝑃−1 𝐴𝑃 = 𝑃𝑇 𝐴𝑃.
Solution: The characteristic polynomial of A is
𝑥
𝑓(𝑥) = |1
1
1 1
𝑥 1| = (𝑥 − 1)2 (𝑥 + 2)
1 𝑥
141
The eigenvalues of A are 𝜆1 = 1, 𝜆2 = 1 and 𝜆3 = −2. To find the eigenvector associated
with 𝜆1 = 1, we solve for the homogeneous system (𝜆1 𝐼3 − 𝐴)𝑋 = 0:
1 1 1 𝑥1
0
[1 1 1] [𝑥2 ] = [0]
1 1 1 𝑥3
0
(1)
The solution to this system is (verify)
−𝑟 − 𝑠
[ 𝑟 ] where 𝑟, 𝑠 ∈ ℝ
𝑠
Thus a basis for the solution space of (1) consists of the eigenvectors
−1
−1
𝑋1 = [ 1 ] and 𝑋2 = [ 0 ]
0
1
However, the two vectors are not orthogonal so we apply the Gram-Schmidt process to
obtain an orthonormal basis for the solution space of (1).
Let 𝑌1 = 𝑋1 = (−1, 1, 0). Then
𝑋2 ∙ 𝑌1
𝑌2 = 𝑋2 − (
)𝑌
𝑌1 ∙ 𝑌1 1
1
= (−1, 0, 1) − (−1, 1, 0)
2
1 1
= (− , − , 1)
2 2
Normalizing these two vectors we get
𝑍1 =
𝑋1
‖𝑋1 ‖
=
1
√2
(−1, 1, 0)
2
1
1
𝑋
𝑍2 = ‖𝑋2 ‖ =
(− 2 , − 2 , 1)
√6
2
= (−
1
√6
,−
1
,
2
)
√6 √6
142
Thus {𝑍1 , 𝑍2 } is an orthonormal basis of eigenvectors for the solution space of (1).
Next we find the eigenvector associated with 𝜆3 = −2 by solving the homogeneous system
(−2𝐼3 − 𝐴)𝑋 = 0:
−2 1
1 𝑥1
0
[ 1 −2 1 ] [𝑥2 ] = [0]
1
1 −2 𝑥3
0
(2)
The solution to this system is (verify)
𝑟
[𝑟] where 𝑟 ∈ ℝ
𝑟
Thus a basis for the solution space of (2) consists of the eigenvector
1
𝑋3 = [1]
1
Normalizing this we get
𝑋3
‖𝑋3 ‖
𝑍3 =
=
1
√3
=(
1
(1, 1, 1)
,
1
,
1
)
√3 √3 √3
Note that 𝑍3 is orthogonal to both 𝑍1 and 𝑍2 thus {𝑍1 , 𝑍2 , 𝑍3 } is an orthonormal basis of
ℝ3 consisting of eigenvectors of A. By Theorem 7.3.4,
−
𝑃=
1
√2
1
√2
[ 0
1
1
√6
1
√3
1
√6
2
1
−
−
√6
1
− √2
1
and 𝑃 −1 = 𝑃𝑇 = −
√6
√3
√3]
1
[ √3
1
√2
1
− √6
1
√3
0
2
√6
1
√ 3]
143
Hence
𝐷 = 𝑃𝑇 𝐴𝑃
1
− √2
1
6
= −
√
1
[ √3
−
1
0
√2
1
− √6
1
√3
1
1
√2
1
√2
1
= − √6 − √6
2
0
2
[−1
√6
1 −1
√3]
2
0
2
√6
2
− √1
2
−1 −1
0 −1]
−1 0
1
√2
[
−
1
√2
1
√2
[− √3 − √3 − √3] [ 0
0
1
1
√6
1
√3
1
√6
2
√3
1
−
−
√6
− √1
6
1
−√
6
2
√6
1
√3
1
√3
1
√3]
√3]
1 0 0
= [0 1 0 ]
0 0 −2
Note that D is a diagonal matrix whose main diagonal consists of the eigenvalues of A.
SAQ 7-2
1
Let 𝐴 = [0
0
0
0
3 −2]. Find an orthogonal matrix P and a
−2 3
diagonal matrix D such that 𝐷 = 𝑃−1 𝐴𝑃 = 𝑃𝑇 𝐴𝑃.
144
ASAQ 7-2
The characteristic polynomial of A is
𝑥−1
0
0
𝑓(𝑥) = | 0
𝑥−3
2 |
0
2
𝑥−3
= (𝑥 − 1)(𝑥 − 3)(𝑥 − 3) − 4(𝑥 − 1)
= (𝑥 − 1)2 (𝑥 − 5)
The eigenvalues of A are 𝜆1 = 1, 𝜆2 = 1 and 𝜆3 = 5. To find the eigenvector associated
with 𝜆1 = 1, we solve for the homogeneous system (𝜆1 𝐼3 − 𝐴)𝑋 = 0:
0 0
0 𝑥1
0
[0 −2 2 ] [𝑥2 ] = [0]
0 2 −2 𝑥3
0
(1)
The solution to this system is (verify)
𝑟
[𝑠] where 𝑟, 𝑠 ∈ ℝ
𝑠
Thus a basis for the solution space of (1) consists of the eigenvectors
1
0
𝑋1 = [0] and 𝑋2 = [1]
0
1
Note that 𝑋1 and 𝑋2 are already orthogonal. Next we normalize 𝑋2:
0
𝑋2
𝑍2 =
= [1/√2]
‖𝑋2 ‖
1/√2
Thus {𝑋1 , 𝑍2 } is an orthonormal basis of eigenvectors for the solution space of (1).
Next we find the eigenvector associated with 𝜆3 = 5 by solving the homogeneous system
(5𝐼3 − 𝐴)𝑋 = 0:
145
4 0 0 𝑥1
0
[0 2 2] [𝑥2 ] = [0]
0 2 2 𝑥3
0
(2)
The solution to this system is (verify)
0
[−𝑟] 𝑟 ∈ ℝ
𝑟
Thus a basis for the solution space of (2) consists of the eigenvector
0
𝑋3 = [−1]
1
Normalizing 𝑋3 we obtain
0
𝑋3
𝑍3 =
= [−1/√2]
‖𝑋3 ‖
1/√2
Since 𝑍3 is orthogonal to both 𝑋1 and 𝑍2 then {𝑋1 , 𝑍2 , 𝑍3 } is an orthonormal basis of ℝ3
consisting of the eigenvectors of A. Thus
1
0
0
1
0
0
−1
𝑇
𝑃 = [0 1/√2 −1/√2] and 𝑃 = 𝑃 = [0 1/√2 1/√2]
0 1/√2 1/√2
0 −1/√2 1/√2
Hence
1
0
0
1
𝐷 = [0 1/√2 1/√2] [0
0 −1/√2 1/√2 0
1
= [0
0
0
0
1
1/√2 1/√2] [0
−5/√2 5/√2 0
1 0 0
= [0 1 0 ]
0 0 5
0
0 1
3 −2] [0
−2 3 0
0
0
1/√2 −1/√2]
1/√2 1/√2
0
0
1/√2 −1/√2]
1/√2 1/√2
146
ACTIVITY
−3 0
1. Let 𝐴 = [ 0 −2
−1 0
−1
0 ]. Find an orthogonal matrix P and a diagonal matrix D such
−3
that 𝐷 = 𝑃 −1 𝐴𝑃 = 𝑃𝑇 𝐴𝑃.
REFERENCES:
1. Kolman. Bernard. Introductory Linear Algebra; 4th edition, New York: Mcmillan
Publishing Company, 1988.
2. Lipschutz, Seymour. Linear Algebra; SI (Metric Edition), Singapore: McGraw-Hill
International Book Company, 1981.
3. Johnson and Riess. Introduction to Linear Algebra; Phil: Addison-Wesley Publishing
Company, Inc. 1981.
4. Perry, William L. Elementary Linear Algebra; USA: McGraw-Hill, Inc. 1988.