chapter 2 matrices and linear transformations 大葉大學資訊工程系黃鈴玲 linear algebra

Chapter 2Matrices and Linear Transformations

大葉大學資訊工程系黃鈴玲

Linear Algebra

2.1 Addition, Scalar Multiplication, and Multiplication of Matrices

• aij: the element of matrix A in row i and column j.

Definition

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

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

Addition of Matrices

DefinitionLet 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

Example .72

45 and ,813652 ,320

741Let

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

Solution (自行練習 )

.1113193

831230675421

813652

320741 )1(

(2) A and C are not of the same size, A + C does not exist.

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

.027421Let

.09211263

03)3(37343)2(313

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

., then , if Thus jicabcAB ijij

Negation and SubtractionDefinitionThe matrix (1)C is written –C and is called the negative of C.

Example

.640182 and 563

205 Suppose

.1123183

654603)1(28025

We now define subtraction in terms of addition and scalar multiplication. Let

A – B = A + (–1)B

Example

701 then ,

., , then , if Thus jibacBAC ijijij

Matrix Multiplication

)54()23(

)52()21(

Examples

Example 1

exist.not does product that theShow

213Let

無法相乘 AB does not exist.

Note. In general, ABBA. 如此題之 BA 存在

njinjiji

iniiij bababa

DefinitionLet 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_10

Example 2

exist. products theif , and Determine

1050231AB .

Note. In general, ABBA.

BA does not exist.

2)14()23(1243

237 and 4312 BA

Example 3

Let C = AB, Determine c23.

隨堂作業： 11(d)

隨堂作業： 5(f)

Ch2_11

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.

Example

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.

隨堂作業： 8(f)

Ch2_12

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

000000

Amatrix diaginal

matrixidentity

010001

Special Matrices

Ch2_13

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 4

.4312 and 854

312Let

Ch2_14

(a) A: mn, B: nr Let the columns of B be the matrices B1, B2, …, Br. Write B=[B1 B2 … Br]. Thus AB=A[B1 B2 … Br]=[AB1 AB2 … ABr].

314 and

Matrix multiplication in terms of columns

Example

4321 BBB

628AB and

8321 ABABAB

Ch2_15

(b) A: mn, B: n1, where A=[A1 A2 … An] and .

We get .

and 584

Matrix multiplication in terms of columns

Example

2321 AAA

n AbAbAb

Ch2_16

Partitioning of MatricesA matrix can be subdivided into a number of submatrices.

Example

210where 15 and 2 ,

Example

Ch2_17

Example 5

隨堂作業： 21(a)

121 and

Consider the following partition of A.

Under this partition A is interpreted as a 22 matrix. For the product ABto be exist, B must be partitioned into a matrix having two rows.

Let .131

13141212

Ch2_18

Homework

Exercise 2.1:5, 8, 11, 17, 21

Exercise 17

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

332313 i

Ch2_19

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 Multiplication

1. A + B = B + A Commutative property of addition

2. A + (B + C) = (A + B) + C Associative property of addition

3. A + O = O + A = A (where O is the appropriate zero matrix)

4. c(A + B) = cA + cB Distributive property of addition

5. (a + b)C = aC + bC Distributive property of addition

6. (ab)C = a(bC)

Ch2_20

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 Multiplication

1. A(BC) = (AB)C Associative property of multiplication

2. A(B + C) = AB + AC Distributive property of multiplication

3. (A + B)C = AC + BC Distributive property of multiplication

4. 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_21

53101568

10216092

CBA 532

20 and ,

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_22

Example 2

.1131112

201310

1131112)(

and ,201310 ,13

CBA Compute ABC.

A B C = D22 23 31 21

ABC = (AB)C = A(BC)

Ch2_23

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 ,86

43 that Observe

83 and ,

21 matrices heConsider t (1)

CBACAB

. and but , that Observe

62 and ,

21 matrices heConsider t (2)

OQOPOPQ

Ch2_24

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_25

Example 3

. compute ,01

21 If 4AA

Solution

4635736257)2(3)2(

BBAABABBABBAABAABBABABBAA

注意這兩項不能合併！

Example 4 Simplify the following matrix expression.

ABBABABBAA 57)2(3)2( 22 Solution

Ch2_26

Systems of Linear Equations

A system of m linear equations in n variables as follows

and , ,

We can write the system of equations in the matrix form

AX = B

隨堂作業： 32(c)

Ch2_27

Solutions to Systems of Linear Equations

Consider a homogeneous system of linear equations AX=0. Let X1 and X2 be solutions. Then AX1=0 and AX2=0

A(X1 + X2) = AX1 + AX2 = 0 X1 + X2 is also a solution

Note. The set of solutions to a homogeneous system of linear equationsis closed under addition and under scalar multiplication. It is a subspace.

If c is a scalar, A(cX1) = cAX1 = 0 cX1 is also a solution

Ch2_28

Example 5

Consider the following homogeneous system of linear equations.

It can be shown that there are many solutions, x1=2r, x2=3r, x3 = r.

The solutions are vectors in R3 of the form (2r, 3r, r) or r(2, 3, 1). The solutions form a subspace of R3 of dimension 1.

Figure 2.4

Note. x1=0, …, xn = 0, is a solution to every homogeneous system. The set of solutions to every homogeneous system passes through the origin.

Ch2_29

Solutions to Nonhomogeneous systems

Let AX=Y be a nonhomogeneous system of linear equations, so Y0. Let X1 and X2 be two solutions. Then AX1=Y and AX2=Y

A(X1 + X2) = AX1 + AX2 = 2Y X1 + X2 is not a solution.

( 可省略 ) If c is a scalar, A(cX1) = cAX1 = cY cX1 is not a solution.

Exercise 41

Show that a set of solutions to a system of nonhomogeneous linear equations is not closed under addition or under scalar multiplication, and that it is thus not a subspace of Rn.

The set of solutions is not a subspace of Rn.

Example

Ch2_30

xxxConsider the following nonhomogeneous system of linear equations.

The general solution of this system is (2r+4, 3r+2, r).

(2r+4, 3r+2, r) = r(2, 3, 1)+(4, 2, 0)

Note that r(2, 3, 1) is the general solution of the corresponding homogeneous system. The vector (4, 2, 0) is the specific solution to the nonhomogeneous system corresponding to r=0.

Figure 2.5 隨堂作業： 40

Ch2_31

Idempotent and Nilpotent Matrices(Exercise 24~30)

Definition

(1) A square matrix A is said to be idempotent ( 等冪 ) if A2=A.(2) A square matrix A is said to nilpotent ( 冪零 ) 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.

63 (1) 2 AAA

Example

2 :nilpotency of degree The .00

93 (2) 2

隨堂作業： 28, 29(b)

Ch2_32

Homework

Exercise 2.2:6, 13, 21, 25, 27, 28, 29, 32, 36, 40, 41

Exercise 21

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

Ch2_33

Homework

Exercise 25

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

Exercise 27

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

Ch2_34

2.3 Symmetric Matrices

DefinitionThe transpose (轉置 ) 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

675241

tB .431

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

Ch2_35

Theorem 2.4 Properties of TransposeLet 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 sum

2. (cA)t = cAt Transpose of a scalar multiple

3. (AB)t = BtAt Transpose of a product

4. (At)t = A

Ch2_36

nijnijij

jnjjjiijt

bababa

aaaABAB

Theorem 2.4 Properties of Transpose

Proof for 3. (AB)t = BtAt

nijnijij

ttttij

bababa

AjBiAjBiAB

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

Ch2_37

Symmetric Matrix

6394323293704201

384871410

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

jiaaAA jiijt , i.e., ,

Example

隨堂作業： 2(c)

Ch2_38

Example 3

Let C = aA+bB, where a and b are scalars.

Ct = (aA+bB)t

= (aA)t + (bB)t by Thm 2.4 (1) = aAt + bBt by Thm 2.4 (2)

= aA + bB since A and B are symmetric

Thus C is symmetric.

Let A and B be symmetric matrices of the same size. Let C be a linear combination of A and B. Prove that the product C is symmetric.

Ch2_39

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” ( 若且唯若 ) ( 通常簡寫為 iff ) ( 也說成 : p is necessary and sufficient for q )

Ch2_40

Example 4

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

() Let AB be symmetric, then

AB= (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

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_41

Example 3 相關題目

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

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

Ch2_42

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 5

Determine the trace of the matrix .037652214

Solution

We get.10)5(4)( Atr

隨堂作業： 15(b)

Trace of a matrix

Ch2_43

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_44

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

)( 3)( BAtrABtr

Ch2_45

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

z = a + bi

Where a and b are real numbers and a is called the real part and b the imaginary part (虛部 ) of z.

The conjugate (共軛複數 ) of a complex number z = a + bi isdefined and written z = a bi.

Ch2_46

Example 7.321

23 and 54232Let

iBiiiA

Compute A + B, 2A, and AB.

Solution

iiiiiii

181557

910411

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

)32)(23()2)(2()1)(23(3)2(

Ch2_47

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

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

Example (i), (ii)

Example (iii)

隨堂作業： 23

Ch2_48

Homework

Exercise 2.3:2, 5, 11, 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_49

2.4 The Inverse of a Matrix

DefinitionLet 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 ( 可逆 ) and B is called the inverse ( 反矩陣 ) of A. If such a matrix B does not exist, then A has no inverse. (denote B = A1, and Ak=(A1)k )

21AExample 1Prove that the matrix has inverse .

01124321 IAB

01432112

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

Ch2_50

Theorem 2.7If a matrix has an inverse, that inverse is unique.

Let 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 = C

InB = C

Thus an invertible matrix has only one inverse.

Thm2.2 乘法結合律

Ch2_51

Let A be 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

.:: 1 AIIA nn

Note. 若是最後沒有變成 [In:B] 的形式，則 A1 不存在。

Ch2_52

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_53

Example 2

Determine the inverse of the matrix

531532211

Solution

100531010532001211

]:[ 3IA

101320012110001211

1RR1R3

101320012110001211

123100012110013101

R2)2(R3R2R1

.123135110

Thus, 1

隨堂作業： 4(a), 14

123100135010010001

R3)1(R2R3R1

Ch2_54

135000011210012301

3R2R3R2)1(R1

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

412721511

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. 隨堂作業： 4(c)

Ch2_55

Properties of Matrix Inverse

Let A and B be invertible matrices and c a nonzero scalar, Then

AA 11)( 1.11 1

)( .2 Ac

cA111)( .3 ABAB

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

1. By definition, AA1=A1A=I.

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

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

nn AAIAAAAAA )( )( .4 1

,)( )( ,

,)( )( , .5111

IAAAAIAA

IAAAAIAAttt

Ch2_56

Example 4

.)( compute n toinformatio

this Use.4311 shown that becan it then ,13

Solution

.)()(41

ttt AA

隨堂作業： 15,19

Ch2_57

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.

(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_58

Example 5

Solve the system of equations

253353212

xxxxxxxxx

SolutionThis system can be written in the following matrix form:

531532211

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

531532211

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

123135110

.1 ,2 ,1 issolution unique The 321 xxx

Ch2_59

Elementary Matrices

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

Example

1ER2 R3

3ER2+ 2R1

Ch2_60

Elementary Matrices

cfbead

R2+ 2R1

一個矩陣做 elementary row operation ，相當於在左邊乘一個對應的 elementary matrix 。

Ch2_61

Notes for elementary matricesEach elementary matrix is square and invertible.

Example

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

If 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

1221EI

1 IERR IEE 12 i.e.,

Ch2_62

Homework

Exercise 2.4:4, 7, 14, 15, 19, 21, 28

( 後面的 section 2.5~2.7 都跳過 )

Exercise 7

If , show that .)(

（求 22 之反矩陣，此公式較快）

Ch2_63

Homework

Exercise 21

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

Exercise 28

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.

2.5 Matrix Transformations, Rotations, and Dilations

Ch2_64

A function, or transformation, is a rule that assigns to each element of a set a unique element of another set.

We will be especially interested in linear transformations, which are transformations that preserve the mathematical structure of a vector space.

Consider the function f(x) =3x2+4.

domain ( 定義域 ) of the function: the set of all possible x

f(2)=16, we say that the image of 2 is 16.

Extend these ideas to functions between vector spaces

We usually use the term transformation rather than function in linear algebra.

Ch2_65

Consider the transformation T maps R3 into R2 defined byT(x, y, z) = (2x, yz)

The domain of T is R3 and we say the codomain ( 對應域 ) is R2.

T(1, 4, 2) = (2, 6) The image of (1, 4, 2) is (2, 6).

Convenient representations:

T 2 , and the image of .62 is

Ch2_66

DefinitionA transformation T of Rn into Rm is a rule that assigns to each vector u in Rn a unique vector v in Rm. Rn is called the domain of T and Rm is the codomain. We write T(u) = v; v is the image of u under T. The term mapping is also used for a transformation.

Dilation( 擴張 ) and Contraction( 收縮 )

Ch2_67

Consider the transformation , where r > 0.

Figure 2.11

This equation can be written as the following useful matrix form.

Reflection( 反射 )

Ch2_68

Consider the transformation . T is called a reflection.

This equation can be written as the following useful matrix form.

yxT 10

Figure 2.12

Matrix transformationsEvery matrix defines a transformation. Let A be a matrix and x be a column vector such that Ax exists. Then A defines the matrix transformation T(x) = Ax.

For example, defines

T 140235

Ch2_69Figure 2.14 Matrix transformations.

213, written as

431, written as

隨堂作業： 1

Definition

Ch2_70

Let A be an mn matrix. Let x be an element of Rn written in a column matrix form. A defines a matrix transformation T(x)=Ax of Rn into Rm. The vector Ax is the image of x. The domain of the transformation is Rn and codomain is Rm.

We write T: Rn Rm if T maps Rn into Rm.

Geometrical Properties:Matrix transformations maps line segments ( 線段 ) into line segments (or points). If the matrix is invertible ( 可逆 ) the transformation also maps parallel lines into parallel lines.

Example 1

Ch2_71

Consider the transformation T: R2 R2 defined by the matrix . Determine the image of the unit square under this

transformation.

The unit square is the square whose vertices are the points

Ch2_72

OOR'RQ'QP'PThe images are:

Figure 2.15

The square PQRO is mapped into the parallelogram P’Q’R’O.

Composition( 合成 ) of Transformations

Ch2_73

Figure 2.16

Consider the matrix transformations T1(x)=A1x and T2(x)=A2x. The composite transformation T=T2 T1 is given by T(x) = T2(T1(x)) = T2 (A1x) =A2A1xThus T is defined by the matrix product A2A1. T(x) = A2A1x

Ch2_74

Homework

Exercise 2.5:1, 10

2.6 Linear TransformationsA vector space has two operations: addition and scalar multiplication.Consider the matrix transformation T(u)=Au. T(u+v) = A(u+v) = Au+Av = T(u)+T(v)

Definition Let u and v be vectors in Rn and let c be a scalar. A transformation T: Rn → Rm is said to be a linear transformation if T(u + v) = T(u) + T(v) and T(cu) = cT(u).

uT T(u

u+v T(u+v) = T(u)+T(v)

u cu TT(cu) = cT(u)

T(cu) = A(cu) = cAu = cT(u)

Ch2_75

Example 1Prove that the following transformation T: R2 → R2 is linear.

T(x, y) = (x y, 3x)Solution“+”: Let (x1, y1) and (x2, y2) R2. Then

T((x1, y1) + (x2, y2)) = T(x1 + x2, y1 + y2) = (x1 + x2 y1 y2, 3x1 + 3x2) = (x1 y1, 3x1) + (x2 y2, 3x2) = T(x1, y1) + T(x2, y2)

“”: Let c be a scalar.T(c(x1, y1)) = T(cx1, cy1) = (cx1 cy1, 3cx1)

= c(x1 y1, 3x1) = cT(x1, y1)

Thus T is linear.Note A linear transformation T: U → U is called a linear operator.

Ch2_76

隨堂作業： 1

Example 2Show that the following transformation T: R3 → R2 is not linear.

T(x, y, z) = (xy, z)Solution“+”: Let (x1, y1, z1) and (x2, y2, z2) R3. Then T((x1, y1, z1) + (x2, y2, z2)) = T(x1 + x2 , y1 + y2 , z1 + z2) = ((x1 + x2)(y1 y2), z1 + z2)

T is not linear.

and T(x1, y1, z1) + T(x2, y2, z2) = (x1y1, z1) + (x2y2, z2)

Thus, in general T((x1, y1, z1) + (x2, y2, z2)) T(x1, y1, z1) + T(x2, y2, z2).

Ch2_77隨堂作業： 4(b)

Example 3Determine a matrix A that describes the linear transformation

SolutionIt can be shown that T is linear. The domain of T is R2.We find the effect of T on the standard basis of R2.

1(T and

T can be written as

yxT 30

Why? See the next page.

Ch2_78

Let T: Rn → Rm be a linear transformation. {e1, e2, …, en} be the standard basis of Rn, and u be an arbitrary vector in Rn.

Matrix Representation

Express u in terms of the basis, u = a1e1+a2e2+ … + anen.

Since T is a linear transformation,

T(u) = T(a1e1+ a2e2 + … + anen) = T(a1e1) + T(a2e2) + …+ T(anen) = a1T(e1) + a2T(e2) + …+ anT(en)

])()()([1

TTT eee

Ch2_79

Thus the linear transformation T is defined by the matrix

A = [ T(e1) … T(en) ].

A is called the standard matrix of T.

Ch2_80

Example 4The transformation defines a reflection in the line y = x.It can be shown that T is linear. Determine the standard matrix of this transformation. Find the image of .

SolutionWe find the effect of T on the standard basis.

1(T and

T can be written as

yxT 01

Ch2_81

0110)1

Ch2_82

Figure 2.22

隨堂作業： 12

Ch2_83

Homework

Exercise 2.6:1, 4, 12

(2.7~2.9 節跳過 )

chapter 2 matrices and linear transformations 大葉大學資訊工程系黃鈴玲 linear algebra

Documents

1 lecture 龚玲玲 accounting english topic 1 lecture...

graph theory chapter 8 hamiltonian graphs 大葉大學...

discrete mathematics chapter 1 the foundations : logic and...

lecturer: mu lingling （穆玲玲） email:...

discrete mathematics chapter 3 the fundamentals :...

chapter 6 connectivity and flow 大葉大學資訊工程系...

9. linear algebra computation -...

万以内数的大小比较教师：赵学玲

福井玲 fukui, -...

chapter 4 general vector spaces 大葉大學資訊工程系...

graph theory chapter 5 networks 大葉大學 (da-yeh univ.)...

浦口实小分校刘玲玲

chapter 3 trees and forests 大葉大學資訊工程系...

chapter 8 numerical technique 大葉大學資訊工程系...

graph theory chapter 9 planar graphs 大葉大學 (da-yeh...

数字电视王连兴熊熊沈玲玲 2010.11.9

discrete mathematics chapter 6 advanced counting techniques...

discrete mathematics section 3.7 applications of number...

linear slide hand lsh series 大切なワーク … products...

11108103 蔡依玲

chapter 2 matrices and linear transformations 大葉大學 資訊工程系 黃鈴玲 linear algebra

chapter 2 matrices and linear transformations 大葉大學資訊工程系黃鈴玲 linear algebra