Chap. 6 Linear Transformations Linear Algebra Ming-Feng Yeh Department of Electrical Engineering...

Chap. 6 Chap. 6 Linear TransformationsLinear Transformations

Linear AlgebraMing-Feng Yeh

Department of Electrical Engineering

Lunghwa University of Science and Technology

Ming-Feng Yeh Chapter 6 6-2

6.1 6.1 Introduction toIntroduction to Linear Transformation Linear Transformation Learn about functions that map a vector space V

into a vector space W --- T: V W

v

V: domain of T

w

W: codomain of T

range

T: V Wimage of v


MapMap If v is in V and w is in W s.t. T(v) = w, then

w is called the image of v under T. The set of all images of vectors in V is called the

range of T. The set of all v in V s.t. T(v) = w is called the

preimage of w.

Section 6-1


Ex. 1: A function from REx. 1: A function from R22 into R into R22

For any vector v = (v1, v2) in R2, and let T: R2 R2 be defined by T(v1, v2) = (v1 v2, v1 + 2v2)

Find the image of v = (1, 2) T(1, 2) =(1 2, 1+2·2) = (3, 3)

Find the preimage of w = (1, 11)

T(v1, v2) = (v1 v2, v1 + 2v2) = (1, 11) v1 v2 = 1; v1 + 2v2 = 11 v1 = 3; v2 = 4

Section 6-1


Linear TransformationLinear Transformation Let V and W be vector spaces. The function

T: V W is called a linear transformation of V into W if the following two properties are true for all u and v in V and for any scalar c.1. T(u + v) = T(u) + T(v)2. T(cu) = cT(u)

A linear transformation is said to be operation reserving (the operations of addition and scalar multiplication).

Section 6-1


Ex. 2: Verifying a linearEx. 2: Verifying a linear transformation from R transformation from R22 into R into R22

Show that the function T(v1, v2) = (v1 v2, v1 + 2v2) is a linear transformation from R2 into R2.

Let v = (v1, v2) and u = (u1, u2) Vector addition: v + u = (v1 + u1, v2 + u2)

T(v + u) = T(v1 + u1, v2 + u2) = ( (v1 + u1) (v2 + u2), (v1 + u1) + 2(v2 + u2) ) = (v1 v2 , v1 + 2v2) + (u1 u2, u1 +2u2) = T(v) + T(u)

Scalar multiplication: cv = c(v1, v2) = (cv1, cv2)T(cv) = (cv1 cv2, cv1+ 2cv2) = c(v1 v2, v1+ 2v2) = cT(v)

Therefore T is a linear transformation.

Section 6-1


Ex. 3: Not linear transformationEx. 3: Not linear transformation f(x) = sin(x)

In general, sin(x1 + x2) sin(x1) + sin(x2) f(x) = x2

In general, f(x) = x + 1

f(x1 + x2) = x1 + x2 + 1f(x1) + f(x2) = (x1 + 1) + (x2 + 1) = x1 + x2 + 2Thus, f(x1 + x2) f(x1) + f(x2)

22

21

221 )( xxxx

Section 6-1


Linear Operation &Linear Operation &Zero / Identity TransformationZero / Identity Transformation A linear transformation T: V V from a vector

space into itself is called a linear operator. Zero transformation (T: V W):

T(v) = 0, for all v Identity transformation (T: V V):

T(v) = v, for all v

Section 6-1


Thm 6.1: Linear transformationsThm 6.1: Linear transformations

Let T be a linear transformation from V into W, where u and v are in V. Then the following properties are true.

1. T(0) = 0

2. T(v) = T(v)

3. T(u v) = T(u) T(v)

4. If v = c1v1 + c2v2 + … + cnvn, thenT(v) = c1T(v1) + c2T(v2) + … + cnT(vn)

Section 6-1


Proof of Theorem 6.1Proof of Theorem 6.1

1. Note that 0v = 0. Then it follows thatT(0) = T(0v) = 0T(v) = 0

2. Follow from v = (1)v, which implies thatT(v) = T[(1)v] = (1)T(v) = T(v)

3. Follow from u v = u + (v), which implies thatT(u v) = T[u + (1)v] = T(u) + (1)T(v) = T(u) T(v)

4. Left to you

Section 6-1


Remark of Theorem 6.1Remark of Theorem 6.1 A linear transformation T: V W is determined

completely by its action on a basis of V.

If {v1, v2, …, vn} is a basis for the vector space V and if T(v1), T(v2), …, T(vn) are given, then T(v) is determined for any v in V.

Section 6-1


Ex 4: Linear transformationsEx 4: Linear transformations and bases and bases

Let T: R3 R3 be a linear transformation s.t.

T(1, 0, 0) = (2, 1, 4); T(0, 1, 0) = (1, 5, 2);

T(0, 0, 1) = (0, 3, 1). Find T(2, 3, 2). (2, 3, 2) = 2(1, 0, 0) + 3(0, 1, 0) 2(0, 0, 1)

T(2, 3, 2) = 2T(1, 0, 0) + 3T(0, 1, 0) 2T(0, 0, 1) = 2(2, 1, 4) + 3(1, 5, 2) 2(0, 3, 1) = (7, 7, 0)

Section 6-1


Ex 5: Linear transformationEx 5: Linear transformation defined by a matrix defined by a matrixThe function T: R2 R3 is defined as follows

Find T(v), where v = (2, 1)

Therefore, T(2, 1) = (6, 3, 0)

2

1

21

12

03

)(v

vAT vv

0

3

6

1

2

21

12

03

)( vv AT

Section 6-1


Example 5 (cont.)Example 5 (cont.) Show that T is a linear transformation from R2 to

R3.

1. For any u and v in R2, we haveT(u + v) = A(u + v) = Au +Av = T(u) +T(v)

2. For any u in R2 and any scalar c, we haveT(cu) = A(cu) = c(Au) = cT(u)

Therefore, T is a linear transformation from R2 to R3.

Section 6-1


Thm 6.2: Linear transformationThm 6.2: Linear transformation given by a matrix given by a matrix

Let A be an m n matrix. The function T defined byT(v) = Av

is a linear transformation from Rn into Rm.

In order to conform to matrix multiplication with an m n matrix, the vectors in Rn are represented by m 1 matrices and the vectors in Rm are represented by n 1 matrices.

Section 6-1


Remark of Theorem 6.2Remark of Theorem 6.2 The m n matrix zero matrix corresponds to the zero

transformation from Rn into Rm. The n n matrix identity matrix In corresponds to the

identity transformation from Rn into Rn. An m n matrix A defines a linear transformation from

Rn into Rm.

nmnmm

nn

nn

nmnmm

n

n

vavava

vavava

vavava

v

v

v

aaa

aaa

aaa

A

2211

2222121

1212111

2

1

21

22221

11211

v

nR mR

Section 6-1


Ex 7: Rotation in the planeEx 7: Rotation in the planeShow that the linear transformation T: R2 R2 given by the

matrix has the property that it rotates

every vector in counterclockwise about the origin through

the angle . Sol: Let

cossin

sincosA

),( yx

),( yxT

x

y)sin,cos(),( rryx v

)sin(

)cos(

sincoscossin

sinsincoscos

sin

cos

cossin

sincos)(

r

r

rr

rr

r

rAT vv

Section 6-1


Ex 8: A projection in REx 8: A projection in R33

The linear transformation T: R3 R3 given by

is called a projection in R3.

If v = (x, y, z) is a vector in R3,

then T(v) = (x, y, 0). In other words,

T maps every vector in R3 to its

orthogonal projection in the xy - plane.

000

010

001

A

xy

z

),,( zyx

)0,,(

),,(

yx

zyxT

Section 6-1


Ex 9: Linear transformationEx 9: Linear transformation from M from Mm,nm,n to M to Mn,mn,m

Let T: Mm,n Mn,m be the function that maps m n matrix A to its transpose. That is,Show that T is a linear transformation.

pf: Let A and B be m n matrix.

TAAT )(

mnnm

TT

TTT

MMT

AcTAccAcAT

BTATBABABAT

,, into formn nsformatiolinear tra a is

)()()()( and

)()()()(

Section 6-1


6.2 6.2 The Kernel and Range ofThe Kernel and Range of a Linear Transformation a Linear Transformation Definition of Kernel of a Linear Transformation

Let T:V W be a linear transformation. Then the set of all vectors v in V that satisfy T(v) = 0 is called the kernel of T and is denoted by ker(T).

The kernel of the zero transformation T: V W consists of all of V because T(v) = 0 for every v in V. That is,ker(T) = V.

The kernel of the identity transformation T: V V consists of the single element 0. That is, ker(T) = {0}.


Ex 3: Finding the kernelEx 3: Finding the kernel

Find the kernel of the projection T: R3 R3 given by T(x, y, z) = (x, y, 0).

Sol: This linear transformationprojects the vector (x, y, z)in R3 to the vector (x, y, 0)in xy-plane. Therefore,ker(T) = { (0, 0, z) : zR}

x y

z

)0,0,0(

),0,0( z

Section 6-2


Ex 4: Finding the kernelEx 4: Finding the kernel

Find the kernel of T: R2 R3 given by T(x1, x2) = (x1 2x2, 0, x1).

Sol: The kernel of T is the set of all x = (x1, x2) in R2 s.t. T(x1, x2) = (x1 2x2, 0, x1) = (0, 0, 0).Therefore, (x1, x2) = (0, 0). ker(T) = { (0, 0) } = { 0 }

Section 6-2


Ex 5: Finding the kernelEx 5: Finding the kernelFind the kernel of T: R3 R2 defined by T(x) = Ax, where

Sol: The kernel of T is the set of all x = (x1, x2, x3) in R3 s.t. T(x1, x2, x3) = (0, 0). That is,

Therefore, ker(T) = {t(1, 1,1): tR} = span{(1, 1,1) }

321

211A

Rt

t

t

t

x

x

x

x

x

x

,

0

0

321

211

3

2

1

3

2

1

Section 6-2


Thm 6.3: Kernel is a subspaceThm 6.3: Kernel is a subspace

The kernel of a linear transformation T: V W is a subspace of the domain V.

pf: 1. ker(T) is a nonempty subset of V.

2. Let u and v be vectors in ker(T). ThenT(u + v) = T(u) + T(v) = 0 + 0 = 0 (vector addition)Thus, u + v is in the kernel

3. If c is any scalar, then T(cu) = cT(u) = c0 = 0(scalar multiplication), Thus, cu is in the kernel.

The kernel of T sometimes called the nullspace of T.

Section 6-2


Ex 6: Finding a basis for kernelEx 6: Finding a basis for kernelLet T: R5 R4 be defined by T(x) = Ax, where x is in R5

and .

Find a basis for ker(T) as a subspace of R5.

82000

10201

01312

11021

A

Section 6-2


Example 6 (cont.)Example 6 (cont.)

Thus one basis for the kernel T is given by

B = { (2, 1, 1, 0, 0), (1, 2, 0, 4, 1) }

1

4

0

2

1

0

0

1

1

2

0

0

0

0

0

82000

10201

01312

11021

5

4

3

2

1

5

4

3

2

1

ts

x

x

x

x

x

x

x

x

x

x

Section 6-2


Solution SpaceSolution Space A basis for the kernel of a linear transformation

T(x) = Ax was found by solving the homogeneous system given by Ax = 0.

It is the same produce used to find the solution space of Ax = 0.

Section 6-2


Thm 6.4: Range is a subspaceThm 6.4: Range is a subspaceThe range of a linear transformation T: V W is a subspace

of the domain W. range(T) = { T(v): v is in V } ker(T) is a subspace of V.pf: 1. range(T) is a nonempty because T(0) = 0.2. Let T(u) and T(v) be vectors in range(T). Because u and v

are in V, it follows that u + v is also in V. Hence the sumT(u) + T(v) = T(u + v) is in the range of T. (vector addition)

3. Let T(u) be a vector in the range of T and let c be a scalar. Because u is in V, it follows that cu is also in V. Hence, cT(u) = T(cu) is in the range of T. (scalar multiplication)

Section 6-2

The Range of a Linear Transform


Figure 6.6Figure 6.6

T: V W

V

W0

Codomain

Range

Domain

Kernel ker(T) is a subspace of V

range(T) is a subspace of W

Section 6-2


Column SpaceColumn Space To find a basis for the range of a linear transformation

defined by T(x) = Ax, observe that the range consists of all vectors b such that the system Ax = b is consistent.

b is in the range of T if and only if b is a linear combination of the column vectors of A.

bx

mmn

n

n

n

mnmnmm

n

n

b

b

b

a

a

a

x

a

a

a

x

x

x

x

aaa

aaa

aaa

A

2

1

2

1

1

21

11

12

1

21

22221

11211

Section 6-2


Corollary of Theorems 6.3 & 6.4Corollary of Theorems 6.3 & 6.4

Let T: Rn Rm be the linear transformation given by T(x) = Ax.

[Theorem 6.3] The kernel of T is equal to the solution space of Ax = 0.

[Theorem 6.4] The column space of A is equal to the range of T.

Section 6-2


Ex 7: Finding a basis for rangeEx 7: Finding a basis for rangeLet T: R5 R4 be the linear transform given in Example 6.

Find a basis for the range of T.Sol: The row echelon of A:

One basis for the range of T isB = { (1, 2, 1, 0), (2, 1, 0, 0), (1, 1, 0, 2) }

00000

41000

20110

10201

82000

10201

01312

11021

A

Section 6-2


Rank and NullityRank and Nullity

Let T: V W be a linear transformation. The dimension of the kernel of T is called the

nullity of T and is denoted by nullity(T). The dimension of the range of T is called the

rank of T and is denoted by rank(T).

Section 6-2


Thm 6.5: Sum of rank and nullityThm 6.5: Sum of rank and nullity Let T: V W be a linear transformation from an

n-dimension vector space V into a vector space W. Then the sum of the dimensions of the range and the kernel is equal to the dimension of the domain. That is,rank(T) + nullity(T) = nordim(range) + dim(kernel) = dim(domain)

Section 6-2


Proof of Theorem 6.5Proof of Theorem 6.5 The linear transformation from an n-dimension vector

space into an m-dimension vector space can be represented by a matrix, i.e., T(x) = Ax where A is an m n matrix. Assume that the matrix A has a rank of r. Then,

rank(T) = dim(range of T) = dim(column space) = rank(A) = r

From Thm 4.7, we have

nullity(T) = dim(kernel of T) = dim(solution space) = n rThus,

rank(T) + nullity(T) = n + (n r) = n

Section 6-2


Ex 8: Finding the rank & nullityEx 8: Finding the rank & nullity

Find the rank and nullity of T: R3 R3 defined by

the matrix

Sol: Because rank(A) = 2, the rank of T is 2.

The nullity is dim(domain) – rank = 3 – 2 = 1.

000

110

201

A

Section 6-2


Ex 8: Finding the rank & nullityEx 8: Finding the rank & nullity

Let T: R5 R7 be a linear transformation Find the dimension of the kernel of T if the

dimension of the range is 2.dim(kernel) = n – dim(range) = 5 – 2 = 3

Find the rank of T if the nullity of T is 4rank(T) = n – nullity(T) = 5 – 4 = 1

Find the rank of T if ker(T) = {0}rank(T) = n – nullity(T) = 5 – 0 = 5

Section 6-2


One-to-One MappingOne-to-One Mapping A linear transformation T :VW is said to be

one-to-one if and only if for all u and v in V,T(u) = T(v) implies that u = v.

One-to-One & Onto Linear Transformation

Section 6-2

V

W

T

One-to-one

V

W

T

Not one-to-one


Thm 6.6: One-to-one LinearThm 6.6: One-to-one Linear transformation transformationLet T :VW be a linear transformation. Then T is one-to-one

if and only if ker(T) = {0}.

pf: 」 Suppose T is one-to-one. Then T(v) = 0 can have only one solution: v = 0. In this case, ker(T) = {0}. 」 Suppose ker(T) = {0} and T(u) = T(v). Because T is a linear transformation, it follows that T(u – v) = T(u) – T(v) = 0 This implies that u – v lies in the kernel of T and must therefore equal 0. Hence u – v = 0 and u = v, and we can conclude that T is one-to-one.

Section 6-2


Example 10Example 10 The linear transformation T: Mm,n Mn,m given by

is one-to-one because its kernel consists of only the m n zero matrix.

The zero transformation T: R3 R3 is not one-to-one because its kernel is all of R3.

TAAT )(

Section 6-2


Onto Linear TransformationOnto Linear Transformation A linear transformation T :VW is said to be onto

if every element in W has a preiamge in V. T is onto W when W is equal to the range of T. [Thm 6.7] Let T :VW be a linear transformation,

where W is finite dimensional. Then T is onto if and only if the rank of T is equal to the dimension of W, i.e., rank(T) = dim(W).

One-to-one: ker(T) = {0} or nullity(T) = 0

Section 6-2


Thm 6.8: One-to-one and ontoThm 6.8: One-to-one and onto linear transformation linear transformation Let T :VW be a linear transformation with vector

spaces V and W both of dimension n.Then T is one-to-one if and only if it is onto.

pf: 」 If T is one-to-one, then ker(T) = {0} and dim(ker(T)) = 0. In this case, dim(range of T) = n – dim(ker(T)) = n = dim(W). By Theorem 6.7, T is onto. 」 If T is onto, then dim(range of T) = dim(W) = n. Which by Theorem 6.5 implies that dim(ker(T)) = 0 By Theorem 6.6, T is onto-to-one.

Section 6-2


Example 11Example 11The linear transformation T:RnRm is given by T(x) = Ax.

Find the nullity and rank of T and determine whether T is one-to-one, onto, or either.

000

110

021

)(,110

021)(,

00

10

21

)(,

100

110

021

)( AdAcAbAa

T:RnRm dim(domain) rank(T) nullity(T) one-to-one onto

(a) T:R3 R3 3 3 0 Yes Yes

(b) T:R2 R3 2 2 0 Yes No

(c) T:R3 R2 3 2 1 No Yes

(d) T:R3 R3 3 2 1 No No

Section 6-2


IsomorphismIsomorphism

Def: A linear transformation T :VW that is one-to-one and onto is called isomorphism. Moreover, if V and W are vector spaces such that there exists an isomorphism from V to W, then V and W are said to be isomorphic to each other.

Theorem 6.9: Isomorphism Spaces & DimensionTwo finite-dimensional vector spaces V and W are isomorphic if and only if they are of the same dimension.

Section 6-2

Isomorphisms of Vector Spaces


Ex 12: Isomorphic Vector SpacesEx 12: Isomorphic Vector Spaces

The following vector spaces are isomorphic to each other.

R4 = 4-space M4,1 = space of all 4 1 matrices

M2,2 = space of all 2 2 matrices

P3 = space of all polynomials of degree 3 or less

V = {(x1, x2, x3, x4, 0): xi is a real number} (subspace of R5)

Section 6-2


6.3 6.3 Matrices forMatrices for Linear Transformation Linear Transformation Which one is better?

The key to representing a linear transformation T:VW by a matrix is to determine how it acts on a basis of V.

Once you know the image of every vector in the basis, you can use the properties of linear transformations to determine T(v) for any v in V.

)43,23,2(),,( 32321321321 xxxxxxxxxxxT

3

2

1

430

231

112

)(

x

x

x

AT xxSimpler to write. Simpler to read, and more adapted for computer use.


Thm 6.10: Standard matrix forThm 6.10: Standard matrix for a linear transformation a linear transformationLet T: RnRm be a linear transformation such that

Then the m n matrix whose n columns corresponds to

is such that T(v) = Av for every

v in Rn. A is called the standard

matrix for T.

Section 6-3

mn

n

n

n

mm a

a

a

TT

a

a

a

TT

a

a

a

TT

2

1

2

22

12

2

1

21

11

1 )

1

0

0

()(,,)

0

1

0

()(,)

0

0

1

()( eee

),( iT e

mnmm

n

n

aaa

aaa

aaa

A

21

22221

11211


Proof of Theorem 6.10Proof of Theorem 6.10Let

Because T is a linear transformation, we have

On the other hand,

.221121n

nnT

n Rvvvvvv eeev

)()()()()()(

)()(

2211

2211

2211

nn

nn

nn

TvTvTvvTvTvT

vvvTT

eeeeee

eeev

)()()()( 2211

2

1

2

22

12

2

1

21

11

12

1

21

22221

11211

veee

v

TTvTvTva

a

a

v

a

a

a

v

a

a

a

v

v

v

v

aaa

aaa

aaa

A

nn

mn

n

n

n

mmnmnmm

n

n

Section 6-3


Example 1 Example 1 Find the standard matrix for the linear transformation

T: R3R2 defined by T(x, y, z) = ( x – 2y, 2x + y)

Sol:

0

0

1

0

0

)()0,0()1,0,0()(

1

2

0

1

0

)()1,2()0,1,0()(

2

1

0

0

1

)()2,1()0,0,1()(

33

22

11

TTTT

TTTT

TTTT

ee

ee

ee

Section 6-3


Example 1 (cont.)Example 1 (cont.)

Note that

which is equivalent to T(x, y, z) = ( x – 2y, 2x + y).

012

021)()()( 321 eee TTTA

yx

yx

z

y

x

z

y

x

A2

2

012

021

Section 6-3


Example 2Example 2The linear transformation T: R2R2 is given by projecting

each point in R2 onto to the x-axis. Find the standard matrix for T.

Sol: This linear transformation is given by T(x, y) = (x, 0). Therefore, the standard matrix for T is

00

01)1,0()0,1( TTA

x

y

),( yx

)0,(x

Section 6-3


Composition of linearComposition of linear transformation transformation The composition T, of T1: RnRm with T2: RmRp is

defined by T(v) = T2( T1(v) ) = where v is a vector in Rn.

The domain of T is defined to the domain of T1.

The composition is not defined unless the range of T1 lies within the domain of T2.

Section 6-3

Composition of Linear Transformation

12 TT

nR vwmR pR u

1T 2T

T


Theorem 6.11Theorem 6.11

Let T1: RnRm and T2: RmRp be linear transformation with standard matrix A1 and A2.

The composition T: RnRp, defined by T(v) = T2( T1(v) ), is linear transformation.

Moreover, the standard matrix of A for T is given by the matrix product A = A2A1.

Section 6-3


Proof of Theorem 6.11Proof of Theorem 6.111. Let u and v be vectors in Rn and let c be any scalar.

Because T1 and T2 are linear transformation,T(u + v) = T2(T1(u + v)) = T2(T1(u) + T1(v)) = T2(T1(u)) + T2(T1(v)) = T(u) + T(v). T(cv) = T2(T1(cv)) = T2(cT1(v)) = cT2(T1(v)) = cT(v).

Thus, T is a linear transformation.2. T(v) = T2(T1(v)) = T2(A1v) = A2(A1v) = A2A1v

In general, the composition is not the same as

Section 6-3

12 TT .21 TT


Example 3Example 3Let T1 and T2 be linear transformation R3 from R3 such that andFind the standard matrices for the compositionsand .Sol: The standard matrices for T1 and T2 are

Section 6-3

),0,2(),,(1 zxyxzyxT ),,(),,(2 yzyxzyxT 12 TTT

21 TTT

010

100

011

,

101

000

012

21 AA

001

000

122

,

000

101

012

1221 AAATAAAT


Inverse Linear TransformationInverse Linear Transformation One benefit of matrix representation is that it can

represent the inverse of a linear transformation.

[Definition] If T1:RnRn and T2:RnRn are linear transformations such that T2(T1(v)) = v and T1(T2(v)) = v, then T2 is called the inverse of T1 and T1 is said to be invertible.

Section 6-3


Theorem 6.12Theorem 6.12 Let T:RnRn be linear transformation with

standard matrix A. Then the following conditions are equivalent.1. T is invertible.2. T is an isomorphism.3. A is invertible.And, if T is invertible with standard matrix A, then the standard matrix for is .

Section 6-3

1T 1A


Example 4Example 4The linear transformation T: R3R3 is defined by

Show that T is invertible, and find its inverse.

Sol:

A is invertible. Its inverse is

Therefore T is invertible and its standard matrix is

Section 6-3

)42,33,32(),,( zyxzyxzyxzyxT

0)det(

142

133

132

AA

326

101

0111A

.1A


Example 4 (cont.)Example 4 (cont.)Using the standard matrix for the inverse, we can find the rule

for by computing the image of an arbitrary vector

Section 6-3

1T).,,( zyxv

).326,,(),,(1 zyxzxyxzyxT

yyx

zx

yx

z

y

x

T

326326

101

011

)(1 v


Nonstandard BasesNonstandard Bases Finding a matrix for a linear transformation T:VW, where

B and are ordered bases for V and W, respectively. The coordinate matrix of v relative to B is [v]B. To represent the linear transformation T, A must be

multiplied by a coordinate matrix relative to B. The result of the multiplication will be a coordinate

matrix relative to . A is called the matrix of T relative to the

bases B and .

Section 6-3

Nonstandard Bases and General Vector Spaces

B

.][)]([ BB AT vv

B

B


Transformation MatrixTransformation MatrixLet V and W be finite-dimensional vector spaces with bases

B and , respectively, whereIf T:VW is a linear transformation such that

then the m n matrix whose n columns correspond to

is s.t. for every v in V.

Section 6-3

B }...,,,{ 21 nB vvv

,)]([,,)]([,)]([ 2

1

2

22

12

2

1

21

11

1

mn

n

n

Bn

m

B

m

B

a

a

a

T

a

a

a

T

a

a

a

T

vvv

,)]([ BT v

,

21

22221

11211

mnmm

n

n

aaa

aaa

aaa

A

BB AT ][)]([ vv


Example 5Example 5Let T: R2R2 be a linear transformation defined by

. Find the matrix of T relative to the

bases and

Sol:

Therefore the coordinate matrices of T(v1) and T(v2) relative

to are

The matrix for T relative to B and is

Section 6-3

)2,(),( yxyxyxT )}1,1(),2,1{( B )}1,0(),0,1{(B

v1 v2 w1 w2

212

211

30)3,0()1,1()(

03)0,3()2,1()(

wwv

wwv

TT

TT

B .30)]([,03)]([ T2

T1 BB TT vv

B

30

03A


Example 6Example 6For the linear transformation T: R2R2 given in Example 5,

use the matrix A to find T(v), where v = (2, 1).

Sol:

Section 6-3

)}1,1(),2,1{( B

1

1][)1,1(1)2,1(1)1,2( Bvv

3

3

1

1

33

03][)]([ BB AT vv

).3,3()1,0(3)0,1(3)()}1,0(),0,1{( vTB

)3,3()1,2()2,(),(:5 TyxyxyxTEx


6.4 6.4 Transition Matrix andTransition Matrix and Similarity Similarity The matrix of a linear transformation T:VV

depends on the basis of V. The matrix of T relative to a basis B is different

from the matrix of T relative to another basis Is is possible to find a basis B such that the matrix

of T relative to B is diagonal?

B


Transition MatricesTransition Matrices A linear transformation is defined by T:VV Matrix of T relative to B: A Matrix of T relative to :Transition matrix from to B: PTransition matrix from B to :

BB

B

A

1P

AB][vBT )]([ v

AB][v BT )]([ v

1PP

APPA

APPT

AT

BB

BB

1

1)(

)(

vv

vv

BBBB

BB

BB

PP

AT

AT

][][;][][

)(

)(

1 vvvv

vv

vv

Section 6-4


Example 1Example 1Find the matrix for T: R2R2,

relative to the basis

Sol: The standard matrix for T is

The transition matrix from to the standard basis

Therefore the matrix for T relative to is

A )3,22(),( yxyxyxT )}.1,1(),0,1{(B

31

22A

B

10

11

10

11 is )}1,0(),0,1{( 1PPB

B

21

231APPA

Section 6-4


Example 2Example 2Let and

be bases for R2, and let be the matrix for

T: R2R2 relative to B. Find

Sol: In Example 5 in Section 4.7,

The matrix of T relative to is given by

)}.2,2(),2,1{( B)}2,4(),2,3{( B

73

72A

.A

32

21,

12

23 1PP

B

31

121APPA

Section 6-4


Example 3Example 3For the linear transformation T: R2R2 given in Example 2,

find , , and for the vector v whose coordinate matrix

Sol:

B][v BT )]([ v BT )]([ v T13][ Bv

B

BB

BB

BB

A

TPT

AT

P

][

0

7

14

21

32

21)]([)]([

14

21

5

7

73

72][)]([

5

7

1

3

12

23][][

1

v

vv

vv

vv

Section 6-4


Similar MatricesSimilar Matrices[Definition] For square matrices A and of order n,

is said to be similar to A if there exists an invertible matrix

P such that

[Theorem 6.13] Let A, B, and C be square matrices of order n, Then the following properties are true.1. A is similar to A.2. If A is similar to B, then B is similar to A.3. If A is similar to B and B is similar to C, then A is similar to C.

Section 6-4

Similar Matrices

AA

.1APPA


Example 5Example 5

Suppose is the matrix for T: R3R3 relative

the standard basis. Find the matrix for T relative to the basis

Sol: The transition matrix from to the standard matrix is

Section 6-4

200

013

031

A

)}1,0,0(),0,1,1(),0,1,1{( B

B

200

020

004

100

0

0

100

011

0111

21

21

21

21

1 APPAPP


6.5 6.5 Applications of LinearApplications of Linear Transformation Transformation

The geometry of linear transformations in the planeReflection in the y-axis Reflection in the x-axis

y

x

y

x

yxyxT

10

01

),(),(

y

x

y

x

yxyxT

10

01

),(),(

x

y),( yx),( yx

x

y),( yx

),( yx


Reflection in the PlaneReflection in the PlaneReflection in the line y = x

x

y

y

x

xyyxT

01

10

),(),(

x

y

),( yx

),( xy

xy

Section 6-5


Expansions & ContractionsExpansions & Contractions in the Plane -- Horizontal in the Plane -- Horizontal

y

kx

y

xk

ykxyxT

10

0

),(),(

x

y

),( yx

Contraction: 0 < k <1

),( ykx

x

y

),( yx

Expansion: k >1

),( ykx

Section 6-5


Expansions & ContractionsExpansions & Contractions in the Plane -- Vertical in the Plane -- Vertical

ky

x

y

x

k

kyxyxT

0

01

),(),(

x

y

),( yx

Contraction: 0 < k <1

),( ykx

Expansion: k >1

x

y

),( yx

),( ykx

Section 6-5


Shears in the PlaneShears in the Plane

y

kyx

y

xk

ykyxyxT

10

1

),(),(Horizontal shear:

x

y),( yx

),( ykyx

ykx

x

y

x

k

kxyxyxT

1

01

),(),(Vertical shear:

x

y

),( yx

),( ykxx

Section 6-5

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Chap. 6 Linear Transformations Linear Algebra Ming-Feng Yeh Department of Electrical Engineering...

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