Chapter 2Matrices

Linear Algebra

Ch2_2

2.1 Addition, Scalar Multiplication, and Multiplication of Matrices

DefinitionTwo matrices are equal if they are of the same size and if their corresponding elements are equal.

• aij: the element of matrix A in row i and column j.• For a square nn matrix A, the main diagonal is:

nnnn

n

n

aaa

aaaaaa

A

21

22221

11211

Thus A = B if aij = bij i, j. ( for every, for all)

Ch2_3

Addition of MatricesDefinitionLet A and B be matrices of the same size. Their sum A + B is the matrix obtained by adding together the corresponding elements of A and B. The matrix A + B will be of the same size as A and B. If A and B are not of the same size, they cannot be added, and we say that the sum does not exist.

. then , if Thus i,jbacBAC ijijij

Ch2_4

Example 2.72

45 and ,813652 ,320

741Let

CBA

Determine A + B and A + C, if the sum exist.

Solution

.1113193

831230675421

813652

320741 )1(

BA

(2) Because A is 2 3 matrix and C is a 2 2 matrix, there are not of the same size, A + C does not exist.

Ch2_5

Scalar Multiplication of matricesDefinitionLet A be a matrix and c be a scalar. The scalar multiple of A by c, denoted cA, is the matrix obtained by multiplying every element of A by c. The matrix cA will be the same size as A.

Example 3

.027421Let

A

.09211263

03)3(37343)2(3133

A

Observe that A and 3A are both 2 3 matrices.

., then , if Thus jicabcAB ijij

Ch2_6

Negation and SubtractionDefinitionWe now define subtraction of matrices in such a way that makes it compatible with addition, scalar multiplication, and negative. Let

A – B = A + (–1)B

Example

.640182 and 563

205 Suppose

BA

.1123183

654603)1(28025

BA

Ch2_7

njinjiji

nj

j

j

iniiij bababa

b

bb

aaac

22112

1

21

Multiplication of MatricesDefinitionLet the number of columns in a matrix A be the same as the number of rows in a matrix B. The product AB then exists.

If the number of columns in A does not equal the number of row B,

we say that the product does not exist.

Let A: mn matrix, B: nk matrix,The product matrix C=AB has elements

C is a mk matrix.

Ch2_8

Example 4

exist. products theif , and , , Determine

.526 and ,623105

,0231

Let

ACBAAB

CBA

623

1050231AB

.201091614

)60()12())2(0()02()30()52()63()11())2(3()01()33()51(

61

0220

0235

02

61

3120

3135

31

BA and AC do not exist.

Sol.

Note. In general, ABBA.

Ch2_9 2)14()23(1

243

Example 5

.5301 and

230712

Let

BA

50233

123

50073

107

50123

112

5301

230712

AB

1030751

1006300075032

23c

Determine AB.

105

237 and 4312 BA

Example 6 Let C = AB, Determine c23.

Ch2_10

Size of a Product MatrixIf A is an m r matrix and B is an r n matrix, then AB will be an m n matrix.

A

m r

Br n

= ABm n

Example 7

If A is a 5 6 matrix and B is an 6 7 matrix. Because A has six columns and B has six rows. Thus AB exits.And AB will be a 5 7 matrix.

Ch2_11

DefinitionA zero matrix is a matrix in which all the elements are zeros. A diagonal matrix is a square matrix in which all the elements not on the main diagonal are zeros. An identity matrix is a diagonal matrix in which every diagonal element is 1.

matrix zero

000

000000

mn0

Amatrix diaginal

00

0000

22

11

nna

aa

A

matrixidentity

100

010001

nI

Special Matrices

Ch2_12

Theorem 2.1Let A be m n matrix and Omn be the zero m n matrix. Let B be an n n square matrix. On and In be the zero and identity n n matrices. Then

A + Omn = Omn + A = ABOn = OnB = On

BIn = InB = BExample 8

.4312 and 854

312Let

BA

AOA

854312

000000

854312

23

22 0000

0000

3312

OOB

BBI

4312

1001

4312

2

Ch2_13

Homework

Exercise 11

Let A be a matrix whose third row is all zeros. Let B be any matrix such that the product AB exists. Prove that the third row of AB is all zeros.

Solution

. ,0 ]0 0 0[ ] [)( 2

1

2

1

332313 i

b

bb

b

bb

aaaAB

ni

i

i

ni

i

i

ni

Ch2_14

2.2 Algebraic Properties of Matrix Operations

Theorem 2.2 -1

Let A, B, and C be matrices and a, b, and c be scalars. Assume that the size of the matrices are such that the operations can be performed.Properties of Matrix Addition and scalar Multiplication1. A + B = B + A Commutative property of addition2. A + (B + C) = (A + B) + C Associative property of addition3. A + O = O + A = A (where O is the appropriate zero matrix)4. c(A + B) = cA + cB Distributive property of addition5. (a + b)C = aC + bC Distributive property of addition6. (ab)C = a(bC)

Ch2_15

Let A, B, and C be matrices and a, b, and c be scalars. Assume that the size of the matrices are such that the operations can be performed.Properties of Matrix Multiplication1. A(BC) = (AB)C Associative property of multiplication2. A(B + C) = AB + AC Distributive property of multiplication3. (A + B)C = AC + BC Distributive property of multiplication4. AIn = InA = A (where In is the appropriate zero matrix)

5. c(AB) = (cA)B = A(cB)Note: AB BA in general. Multiplication of matrices is not

commutative.

Theorem 2.2 -2

Ch2_16

15

201873

5431

.5964

115584273031

CBA

.1520 and ,18

73 ,5431Let

CBA

Example 1

Proof of Thm 2.2 (A+B=B+A)

.)()( ijijijijijij ABabbaBA Consider the (i,j)th elements of matrices A+B and B+A:

A+B=B+A

Ch2_17

Example 2 and 3

.1131112

201310

1321

AB

.19

014

1131112)(

CAB

.014

and ,201310 ,13

21Let

CBA Compute ABC.

Sol. (1) (AB)C

(2) A(BC)

4

1014

201310BC

.19

41

1321)(

BCA

Count the number of multiplications.

Which method is better?

26+32=12+6=18

32+22=6+4=10

A(BC) is better.

Ch2_18

Arithmetic Operations

If A is an m r matrix and B is r n matrix, the number of scalar multiplications involved in computing the product AB is mrn.

A B C = D22 23 31 21

ProofConsider the (i,j)th element of matrix AB:

AB mn mrnrjirjijiij bababaAB 2211)(

(1) (AB)C: 223+ 231=18(2) A(BC): 231+ 221=10 better

Ch2_19

In algebra we know that the following cancellation laws apply. If ab = ac and a 0 then b = c. If pq = 0 then p = 0 or q = 0.

However the corresponding results are not true for matrices. AB = AC does not imply that B = C. PQ = O does not imply that P = O or Q = O.

Caution

Example

.but ,8643

that Observe

.23

83 and ,

1221

,4221

matrices heConsider t (1)

CBACAB

CBA

. and but , that Observe

.3162

and ,4221

matrices heConsider t (2)

OQOPOPQ

QP

Ch2_20

Powers of Matrices

Theorem 2.3

If A is an n n square matrix and r and s are nonnegative integers, then 1. ArAs = Ar+s.2. (Ar)s = Ars.3. A0 = In (by definition)

Definition

If A is a square matrix, then

timesk

k AAAA

Ch2_21

Example 4. compute ,

0121

If 4AA

2123

0121

01212A

.65

10112123

21234

A

Solution

2

2222

22

4635736257)2(3)2(

BBAABABBABBAABAABBABABBAA

Example 5 Simplify the following matrix expression.ABBABABBAA 57)2(3)2( 22

Solution

Ch2_22

Systems of Linear Equations

A system of m linear equations in n variables as follows

mnmnm

nn

bxaxa

bxaxa

11

11111

Let

mnmnm

n

b

bB

x

xX

aa

aaA

11

1

111 and , ,

We can write the system of equations in the matrix form

AX = B

Ch2_23

Idempotent and Nilpotent Matrices(Exercise 30~36)

Definition(1) A square matrix A is said to be idempotent if A2=A.(2) A square matrix A is said to nilpotent if there is a

positive integer p such that Ap=0. The least integer p such that Ap=0 is called the degree of nilpotency of the matrix.

.2163

,2163

(1) 2 AAA

Example

2 :nilpotency of degree The .0000

,3193

(2) 2

BB

Ch2_24

Homework

Exercise 2.2:6, 8, 11, 19, 27, 31, 33, 34, 35, 38

Exercise 8

A : m r matrix, B: r n matrix, C: n s matrix. Derive formulas for the number of scalar multiplications involved in computing (AB)C and A(BC).

Exercise 27

A, B: diagonal matrix of the same size, c: scalar Prove that A+B and cA are diagonal matrices.

Ch2_25

Homework

Exercise 31

Determine b, c, and d such that is idempotent.

Exercise 33

Prove that if A and B are idempotent and AB=BA, then AB is idempotent.

dcb1

Ch2_26

2.3 Symmetric Matrices

DefinitionThe transpose of a matrix A, denoted At, is the matrix whose columns are the rows of the given matrix A.

Example 1

.431 and ,654721 ,08

72

CBA

07

82tA

675241

tB .431

tC

jiAAmnAnmA jiijtt ., )( ,: : i.e.,

Ch2_27

Theorem 2.4 Properties of Transpose

Let A and B be matrices and c be a scalar. Assume that the sizes of the matrices are such that the operations can be performed.1. (A + B)t = At + Bt Transpose of a sum2. (cA)t = cAt Transpose of a scalar multiple3. (AB)t = BtAt Transpose of a product4. (At)t = A

Ch2_28

nijnijij

ni

i

i

jnjjjiijt

bababa

b

bb

aaaABAB

2211

2

1

21

)()(

Theorem 2.4 Properties of Transpose

Proof for 3. (AB)t = BtAt

nijnijij

jn

j

j

niii

ttttij

tt

bababa

a

aa

bbb

AjBiAjBiAB

22112

1

21

] of [row ] of column [] of [column ] of row[)(

Ch2_29

Symmetric Matrix

6394323293704201

384871410

4552

match

match

DefinitionA symmetric matrix is a matrix that is equal to its transpose.

jiaaAA jiijt , i.e., ,

Example

Ch2_30

If and only if

Let p and q be statements.Suppose that p implies q (if p then q), written p q,and that also q p, we say that

“p if and only if q” (iff )

Ch2_31

Example 3

*We have to show (a) AB is symmetric AB = BA, and the converse, (b) AB is symmetric AB = BA.

() Let AB be symmetric, thenAB= (AB)t by definition of symmetric matrix = BtAt by Thm 2.4 (3) = BA since A and B are symmetric

() Let AB = BA, then(AB)t = (BA)t

= AtBt by Thm 2.4 (3) = AB since A and B are symmetric

Proof

Let A and B be symmetric matrices of the same size. Prove that the product AB is symmetric if and only if AB = BA.

Ch2_32

Example 3

Proof

Let A be a symmetric matrix. Prove that A2 is symmetric.

tA )( 2 tAA)( )( tt AA 2AAA

Ch2_33

DefinitionLet A be a square matrix. The trace of A, denoted tr(A) is the sum of the diagonal elements of A. Thus if A is an n n matrix.

tr(A) = a11 + a22 + … + ann

Example 4

Determine the trace of the matrix .037652214

A

SolutionWe get

.10)5(4)( Atr

Ch2_34

Theorem 2.5 Properties of TraceLet A and B be matrices and c be a scalar. Assume that the sizes of the matrices are such that the operations can be performed. 1. tr(A + B) = tr(A) + tr(B) 2. tr(AB) = tr(BA) 3. tr(cA) = c tr (A) 4. tr(At) = tr(A)

Since the diagonal element of A + B are (a11+b11), (a22+b22), …, (ann+bnn), we get

tr(A + B) = (a11 + b11) + (a22 + b22) + …+ (ann + bnn)

= (a11 + a22 + … + ann) + (b11 + b22 + … + bnn)

= tr(A) + tr(B).

Proof of (1)

Ch2_35

Example of (2) tr(AB)=tr(BA)

101123

,21

0231

BA

52118

,121246

220BAAB

)( 3)( BAtrABtr

Ch2_36

Matrices with Complex ElementsThe element of a matrix may be complex numbers. A complex number is of the form

z = a + biWhere a and b are real numbers and a is called the real part and b the imaginary part of z.

.1i

The conjugate of a complex number z = a + bi is z = a bi

Ch2_37

Example 6.321

23 and 54232Let

ii

iBiiiA

Compute A + B, 2A, and AB.Solution

iii

iiiiii

iii

iiiBA

82535

3251422332

32123

54232

i

iiiiiA 108

46245423222

iiii

iii

iiiAB

iiiiiiiiiiii

181557910411

32123

54232

)32)(5()2(4)1)(5()3)(4()32)(23()2)(2()1)(23(3)2(

Ch2_38

Definition(i) The conjugate of a matrix A is denoted A and is obtained by taking the conjugate of each element of the matrix.(ii) The conjugate transpose of A is written and defined by A*=A

t.(iii) A square matrix C is said to be hermitian if C=C*.

iii

A76

4132

Example (i), (ii)

iii

A764132

iii

AAt

741632*

Example (iii)

*

643432

Ci

iC

Ch2_39

Homework

Exercise 2.3:2, 5, 11, 13, 15, 23

Exercise 11

A matrix A is said to be antisymmetric if A = At. (a) give an example of an antisymmetric matrix.(b) Prove that an antisymmetric matrix is a square matrix having diagonal elements zero.(c) Prove that the sum of two antisymmetric matrices of the same size is an antisymmetric matrix.

Ch2_40

Homework

Exercise 13

Prove that any square matrix A can be decomposed into the sum of a symmetric matrix B and an antisymmetric matrix C: A = B+C.(Find B: B = Bt, C: C = Ct, such that A = B + C.)

A = B + C

Solution

A = Bt Ct At = B C(1) (2)

(1)+(2): A + At = 2B B = ½(A + At )

(1)(2): A At = 2C C = ½(A At )

Ch2_41

2.4 The Inverse of a MatrixDefinitionLet A be an n n matrix. If a matrix B can be found such that AB = BA = In, then A is said to be invertible and B is called the inverse of A. If such a matrix B does not exist, then A has no inverse. (denote B = A1, and Ak=(A1)k )

43

21AExample 1Prove that the matrix has inverse .

12

2

1

2

3

B

Proof

221

23 10

01124321 IAB

221

23 10

01432112 IBA

Thus AB = BA = I2, proving that the matrix A has inverse B.

Ch2_42

Theorem 2.7The inverse of an invertible matrix is unique.

ProofLet B and C be inverses of A. Thus AB = BA = In, and AC = CA = In.

Multiply both sides of the equation AB = In by C.

C(AB) = CIn

(CA)B = CInB = C

B = CThus an invertible matrix has only one inverse.

Thm2.2

Ch2_43

A: an invertible nn matrixHow to find A-1?

. ,Let 21211

nnn CCCIXXXA

We shall find A1 by finding X1, X2, …, Xn.

Since AA1 =In, then .2121 nn CCCXXXA

.,, , i.e., 2211 nn CAXCAXCAX

Solve these systems by using Gauss-Jordan elimination: .:

::matrix augmented

21

21

nn

n

XXXICCCA

.:: 1 AIIA nn

Note. [In:B] ， A1 。

Ch2_44

Gauss-Jordan Elimination for finding the Inverse of a Matrix

Let A be an n n matrix.1. Adjoin the identity n n matrix In to A to form the matrix

[A : In].

2. Compute the reduced echelon form of [A : In].

If the reduced echelon form is of the type [In : B], then B is the inverse of A.

If the reduced echelon form is not of the type [In : B], in that the first n n submatrix is not In, then A has no inverse.

An n n matrix A is invertible if and only if its reduced echelon form is In.

Ch2_45

Example 2Determine the inverse of the matrix

531532211

A

Solution

100531010532001211

]:[ 3IA

101320012110001211

1RR1R3

2)(R2

101320012110001211

R2)1(

123100012110013101

R2)2(R3R2R1

123100135010010001

R3)1(R2R3R1

.123135110

Thus, 1

A 隨堂作業： 14

Ch2_46

135000011210012301

3R2R3R2)1(R1

Example 3Determine the inverse of the following matrix, if it exist.

412721511

A

Solution

100412010721001511

]:[ 3IA

102630011210001511

R1)2(R31)R1(R2

There is no need to proceed further. The reduced echelon form cannot have a one in the (3, 3) location. The reduced echelon form cannot be of the form [In : B]. Thus A–1 does not exist.

Ch2_47

Properties of Matrix InverseLet A and B be invertible matrices and c a nonzero scalar, Then

AA 11)( 1.11 1)( .2 A

ccA

111)( .3 ABAB

nn AA )()( .4 11 tt AA )()( .5 11

Proof1. By definition, AA1=A1A=I.

))(())(( .2 11 11 cAAIAcA cc

))(( )())(( .3 1111111 ABABIAAABBAABAB nn

nn

nn AAIAAAAAA )( )( .4 1

times

11

times

1

,)( )( ,

,)( )( , .5111

111

IAAAAIAA

IAAAAIAAttt

ttt

Ch2_48

Example 4

.)( compute n toinformatio

this Use.4311 shown that becan it then ,13

14 If1

1

tA

AA

Solution

.)()(4131

431111

ttt AA

Ch2_49

Theorem 2.8

Let AX = Y be a system of n linear equations in n variables. If A–1 exists, the solution is unique and is given by X = A–1Y.

Proof(X = A–1Y is a solution.)Substitute X = A–1Y into the matrix equation.

AX = A(A–1Y) = (AA–1)Y = InY = Y.

(The solution is unique.)Let X1 be any solution, thus AX1 = Y. Multiplying both sides of this equation by A–1 gives

A–1AX1= A–1Y InX1 = A–1YX1 = A–1Y.

Ch2_50

Example 5Solve the system of equations

253353212

321

321

321

xxxxxxxxx

SolutionThis system can be written in the following matrix form:

231

531532211

3

2

1

xxx

If the matrix of coefficients is invertible, the unique solution is

231

531532211 1

3

2

1

xxx

This inverse has already been found in Example 2. We get

121

231

123135110

3

2

1

xxx

.1 ,2 ,1 issolution unique The 321 xxx

Ch2_51

Elementary MatricesDefinition An elementary matrix is one that can be obtained from the identity matrix In through a single elementary row operation.

Example

100010001

3I

010100001

1ER2 R3

100050001

2E5R2

100012001

3ER2+ 2R1

Ch2_52

Elementary Matrices

ihgfedcba

A

AEAfedihgcba

1

010100001

R2 R3

AEAihgfed

cba

2

100050001

555

5R2

AEAihg

cfbeadcba

3

100012001

222

R2+ 2R1

elementary row operation, elementary matrix 。

Ch2_53

Notes for elementary matricesEach elementary matrix is invertible.

Example

If A and B are row equivalent matrices and A is invertible, then B is invertible.

ProofIf A … B, thenB=En … E2 E1 A for some elementary matrices En, … , E2 and E1. So B1 = (En … E2 E1A)1 =A1E1

1 E21 … En

1.

1221EI

RR

100010021

1E

100010001

I

100010021

2E

, 2211 IE

RR IEE 12 i.e.,

Ch2_54

Homework

Exercise 2.4:6, 14, 15, 16, 19, 21, 27

dcba

A

(section 2.5~2.7)

Exercise 7

If , show that .)(

11

acbd

bcadA

Ch2_55

Homework

Exercise 21

Prove that (AtBt)1 = (A1B1)t.

Exercise 27

True or False:(a) If A is invertible A1 is invertible.(b) If A is invertible A2 is invertible.(c) If A has a zero on the main diagonal it is not invertible.(d) If A is not invertible then AB is not invertible.(e) A1 is row equivalent to In.