Linear Algebra - National Chiao Tung Universityyi/Courses...Linear Algebra Linear Transformations...

Linear Algebra

Linear Algebra

Chih-Wei Yi

Dept. of Computer ScienceNational Chiao Tung University

November 11, 2009

Linear Algebra

Linear Transformations

Section 1 De�nition and Examples


Linear Algebra



De�nition

V and W are vector spaces. A function L : V!W is called alinear transformation if L (au+ bv) = aL (u) + bL (v) for anyscalars a, b and vectors u, v 2 V.

Examples

Let L��

xy

��= [x ]. We have

L�a�u1u2

�+ b

�v1v2

��= L

��au1 + bv1au2 + bv2

��= [au1 + bv1]

and

aL��

u1u2

��+ bL

��v1v2

��= a [u1] + b [v1] = [au1 + bv1] .

Linear Algebra



Example

Let L ([x ]) =�x2x

�. We have

L (a [u] + b [v ]) = L ([au + bv ]) =�au + bv2au + 2bv

�and

aL ([u]) + bL ([v ]) = a�u2u

�+ b

�v2v

�=

�au + bv2au + 2bv

�.

Problem

Is L([x ]) = [x + 1] a linear transformation?

Linear Algebra



Problem

Is L��

xy

��=

�cos θ � sin θsin θ cos θ

� �xy

�=

�x cos θ � y sin θx sin θ + y cos θ

�a linear transformation?

Problem

Prove that the di¤erential operation is a linear transformation inthe vector space C1 [a, b].

Problem

Prove that if L : V!W is a linear transformation, thenL (0) = 0. (Note: the �rst 0 is the zero vector in V and thesecond 0 is the zero vector in W. If necessary, we use 0V to denotethe zero vector in V.)

Linear Algebra



Kernel and Range

De�nition (Kernel and Range)

If L : V!W is a linear transformation, then

kernel (L) = fv 2 V : L (v) = 0wg is called the kernel of L.range (L) = fw 2W : w = L (v) for some v 2 Vg is calledthe range of L.

Note: For any S � V,L (S) = fw 2W : w = L (v) for some v 2 Sg is called the imageof S.

Linear Algebra



Theorem

If L : V!W is a linear transformation, then

kernel (L) is a subspace of V.range (L) is a subspace of W.

Linear Algebra



Example

Let A is a 2� 3 matrix and v 2 R3.

1 Prove that L (v) = Av is a linear transformation form R3 toR2.

2 Find the kernel and range of L for A =�1 2 3�2 5 4

�.

Solution (1)

We have

L (au+ bv) = A (au+ bv) = A (au) +A (bv)= a (Au) + b (Av)= aL (u) + bL (v) .

Linear Algebra



Solution (How to �nd the kernel of A)

Solve the homogeneous linear system

�1 2 3�2 5 4

� 24 x1x2x3

35 = � 00

�.

By Gauss-Jordan elimination, we have�1 2 3�2 5 4

�!�1 2 30 9 10

�!�1 2 30 1 10

9

�!�1 0 7

90 1 10

9

�.

Let x3 = s. Then, x1 = � 79 s and x1 = �

109 s. Therefore,24 x1

x2x3

35 =24 � 7

9 s� 10

9 ss

35 = s24 � 7

9� 10

91

35 .

Linear Algebra



Solution (How to �nd the range of A)

The range of A is span��

1�2

�,

�25

�,

�34

��. Applying

elementary column operations,�1 2 3�2 5 4

�!

�1 0 0�2 9 10

�!�1 0 0�2 1 10

�!

�1 0 00 1 0

�.

So,

span��

1�2

�,

�25

�,

�34

��= span

��10

�,

�01

��= R2.

Linear Algebra



Exercise

Problem

Prove that if A is an m� n matrix, then1 L (v) = Av is a linear transformation form Rn to Rm .

2 dim (kernel (A)) + dim (range (A)) = n.

Linear Algebra


Section 2 Matrix Representations of Linear Transformations

Section 2 Matrix Representations of LinearTransformations

Linear Algebra



Matrix Representations of Linear Transformations

Theorem

Assume V and W are respectively n and m dimension vectorspaces, and B1 = fv1, v2, � � � , vng and B2 = fw1,w2, � � � ,wmgare respectively bases of V and W. If L : V!W is a lineartransformation, then [L (v)]B2 = A [v]B1 with aj = [L (vj )]B2 forj = 1, 2, � � � , n, and A is called a matrix representation of L w.r.t.ordered bases B1 and B2.

Linear Algebra



Proof.

Assuem [v]B1 =

26664c1c2...cn

37775. In other words,v = c1v1 + c2v2 + � � �+ cnvn.

[L (v)]B2 = [L (c1v1 + c2v2 + � � �+ cnvn)]B2= [c1L (v1) + c2L (v2) + � � �+ cnL (vn)]B2= c1 [L (v1)]B2+ c2 [L (v2)]B2 + � � �+ cn [L (vn)]B2= c1a1 + c2a2 + � � �+ cnan= A [v]B1 .

Linear Algebra



Example

Consider the linear transformation L��

xy

��=

�2x + 3y4x � 5y

�.

We have

L (e1) = L��

10

��=

�24

�= 2e1 + 4e2,

L (e2) = L��

01

��=

�3�5

�= 3e1 � 5e2.

Hence,

L��

xy

��=

�2 34 �5

� �xy

�.

Linear Algebra



Example

Let D : a2x2 + a1x + a0 ! 2a2x + a1 denote the di¤erentialoperation from P2 to P1, and B1 =

�x2, x , 1

and B2 = fx , 1g.

Then,

D

0B@24 100

35B1

1CA = D�x2�= 2x =

�20

�B2

;

D

0B@24 010

35B1

1CA = D (x) = 1 =�01

�B2

;

D

0B@24 001

35B1

1CA = D (1) = 0 =�00

�B2

.

Linear Algebra



Example (Cont.)

So,

A =�2 0 00 1 0

�= DB1 :B2 .

Linear Algebra



Example

Consider the linear transformation L��

xy

��=

24 2x � 2y3x + 4y�7x

35.We have

L (e1) = L��

10

��=

24 23�7

35 = 2e1 + 3e2 + (�7) e3,L (e2) = L

��01

��=

24 �240

35 = (�2) e1 + 4e2 + 0e3.Hence,

L��

xy

��=

24 2 �23 4�7 0

35 � xy

�.

Linear Algebra



Matrix Representation on Non-Standard Bases

Let L��

xy

��=

24 2x � 2y3x + 4y�7x

35, B1 = fe1, e2g, andB2 =

8<:24 1�10

35 ,24 010

35 ,24 1

0�1

359=;. Then,

L (e1) = L��

10

��=

24 23�7

35= �5

24 1�10

35� 224 010

35+ 724 1

0�1

35 =24 �5�2

7

35B2

,

and

Linear Algebra




(Cont.)

L (e2) = L��

01

��=

24 �240

35= �2

24 1�10

35+ 224 010

35+ 024 1

0�1

35 =24 �22

0

35B2

.

Hence, �L��

xy

��B2

=

24 �5 �2�2 27 0

35B1 :B2

�xy

�B1

.

Linear Algebra




Let L��

xy

��=

24 2x � 2y3x + 4y�7x

35, B3 = �� 11�,

��11

��, and

B4 = fe1, e2, e3g. Then,

L��

11

��=

24 07�7

35 = 0e1 + 7e2 � 7e3;L��

�11

��=

24 �417

35 = �4e1 + e2 + 7e3.Hence, �

L��

xy

��B4

=

24 0 �47 1�7 7

35B3 :B4

�xy

�B3

.

Linear Algebra




Let L��

xy

��=

24 2x � 2y3x + 4y�7x

35, B3 = �� 11�,

��11

��, and

B2 =

8<:24 1�10

35 ,24 010

35 ,24 1

0�1

359=;. Then,

L��

11

��=

24 07�7

35 = �724 1�10

35+ 024 010

35+ 724 1

0�1

35 ;L��

�11

��=

24 �417

35 = 324 1�10

35+ 424 010

35� 724 1

0�1

35 .

Linear Algebra




(Cont.)Hence,

�L��

xy

��B2

=

24 �7 30 47 �7

35B3 :B2

�xy

�B3

.

Linear Algebra




Verify the following equality for the linear transformation L.24 �7 30 47 �7

35B3 :B2

=

0B@24 1 0 1�1 1 00 0 �1

35B2!B4

1CA�1 24 2 �2

3 4�7 0

35B1 :B4

�1 �11 1

�B3!B1

.

Here

0B@24 1 0 1�1 1 00 0 �1

35B2!B4

1CA�1

is the transition matrix

B4 ! B2.

Linear Algebra



Flow Chart

TB3!B1TB1!B3 = I; TB1!B3TB3!B1 = I; TB2!B4TB4!B2 = I;TB4!B2TB2!B4 = I.LB1 :B2 = TB4!B2LB3 :B4TB1!B3 ;LB3 :B4 = TB2!B4LB1 :B2TB3!B1 .LB1 :B4 = LB3 :B4TB1!B3 ; LB3 :B2 = LB1 :B2TB3!B1 .

LB1:B2

LB3:B4

TB3→B1TB1→B3

[v]B3

[v]B1 [L(v)]B2

[L(v)]B4

TB4→B2TB2→B4

LB3:B2

LB1:B4

V WLB1:B2

LB3:B4

TB3→B1TB1→B3

[v]B3

[v]B1 [L(v)]B2

[L(v)]B4

TB4→B2TB2→B4

LB3:B2

LB1:B4

V W

Linear Algebra



Composition of Linear Transformations

Theorem

U, V, and W are vector spaces. B1, B2, and B3 are bases of U, V,and W, respectively. L1 : U! V and L2 : V!W are lineartransformations. Let A be the matrix representation of L1 w.r.t.B1 and B2, and B be the matrix representation of L2 w.r.t B2 andB3. Then,

L2 � L1 is a linear transformation.The matrix representation of L2 � L1 w.r.t. B1 and B3 is BA.

Linear Algebra



2D Image Processing

Many image processing techniques can be implemented bylinear transformations, e.g.

Dilations and contractionsRe�ectionsRotationsTranslations 1

1O¢ cially, translations are not linear transformations.

Linear Algebra



Dilations and Contractions

The transitionmatrices of dilationsand contractions:

T =�c1 00 c2

�

Linear Algebra



Re�ections

The transitionmatrices ofre�ections:

T =��1 00 �1

�

Linear Algebra



Rotations and Translations

The transition matrices ofrotations and translations:

TT =

�c1 00 c2

�+

�x0y0

�TR =

�cos θ � sin θsin θ cos θ

�.

Linear Algebra


Section 3 Similarity


Linear Algebra



De�nition (Similarity)

A and B are n� n matrices. We say B is similar to A if thereexists a nonsingular matrix S such that B = S�1AS.

Example

Let A =�2 01 1

�, B =

�2 �10 1

�, and U =

�1 �11 1

�. We

have U�1 =� 1

212

� 12

12

�and

B = U�1AU.

Linear Algebra



Linear Transformations and Transition Matrices

Theorem

V is a n dimension vector space, and L : V! V is a lineartransformation. B1 and B2 are bases of V, and S is a transitionmatrix from B1 to B2. Let A be the matrix representation of Lw.r.t. B2, and B be the matrix representation of L w.r.t. B1.Then, we have B = S�1AS.

Linear Algebra



Example

Let D be the di¤erential operation on P3. Let A and Brespectively denote the matrices representation of D w.r.t. basesB1 =

�x3, x2, x , 1

and B2 =

�(2x � 1)3, (2x � 1)2, 2x � 1, 1

.

Find A, B, and transition matrices for B1 ! B2 and B2 ! B1, andverify the similarity of A and B.

Solution

We have

D

0BBB@26641000

3775B1

1CCCA = D�x3�= 3x2 =

26640300

3775B1

,

Linear Algebra



Solution (Cont.)

D

0BBB@26640100

3775B1

1CCCA = D�x2�= 2x =

26640020

3775B1

,

D

0BBB@26640010

3775B1

1CCCA = D (x) = 1 =

26640001

3775B1

,

D

0BBB@26640001

3775B1

1CCCA = D (1) = 0 =

26640000

3775B1

.

Linear Algebra



Solution (Cont.)

So,

DB1 :B1 =

26640 0 0 03 0 0 00 2 0 00 0 1 0

3775 .

Linear Algebra



Solution (Cont.)

D

0BBB@26641000

3775B2

1CCCA = D�(2x � 1)3

�= 6(2x � 1)2 =

26640600

3775B2

,

D

0BBB@26640100

3775B2

1CCCA = D�(2x � 1)2

�= 4(2x � 1) =

26640040

3775B2

,

D

0BBB@26640010

3775B2

1CCCA = D (2x � 1) = 2 =

26640002

3775B2

.

Linear Algebra



Solution (Cont.)

D

0BBB@26640001

3775B2

1CCCA = D (1) = 0 =

26640000

3775B2

.

So,

DB2 :B2 =

26640 0 0 06 0 0 00 4 0 00 0 2 0

3775 .

Linear Algebra



Solution (Cont. (transition matrix TB2!B1))

26641000

3775B2

= (2x � 1)3 = 8x3 � 12x2 + 6x � 1 =

26648�126�1

3775B1

;

26640100

3775B2

= (2x � 1)2 = 4x2 � 4x + 1 =

266404�41

3775B1

;

26640010

3775B2

= 2x � 1 =

2664002�1

3775B1

;

26640001

3775B2

= 1 =

26640001

3775B1

.

Linear Algebra




TB2!B1 =

26648 0 0 0�12 4 0 06 �4 2 0�1 1 �1 1

3775 .

Linear Algebra




26641000

3775B1

= x3 =18(2x � 1)3 + 3

8(2x � 1)2 + 3

8(2x � 1) + 1

8

=

266418383818

3775B2

;

26640100

3775B1

= x2 =14(2x � 1)2 + 1

2(2x � 1)� 1

4=

266401412� 14

3775B2

;

Linear Algebra




26640010

3775B1

= x =12(2x � 1) + 1

2=

2664001212

3775B2

;

26640001

3775B1

= 1 =

26640001

3775B2

.

Linear Algebra




So,

TB1!B2 =

266418 0 0 038

14 0 0

38

12

12 0

18 � 1

412 1

3775 . (Note: TB2!B1TB1!B2 = I;TB1!B2TB2!B1 = I.

Linear Algebra



Flow Chart

TB2!B1TB1!B2 = I; TB1!B2TB2!B1 = I.LB1 :B1 = TB2!B1LB2 :B2TB1!B2 ;LB2 :B2 = TB1!B2LB1 :B1TB2!B1 .

LB2 :B1 = LB1 :B1TB2!B1 ; LB1 :B2 = LB2 :B2TB1!B2 .

LB1:B1

LB2:B2

TB2→B1TB1→B2

[v]B2

[v]B1 [L(v)]B1

[L(v)]B2

TB2→B1TB1→B2

LB2:B1

LB1:B2

LB1:B1

LB2:B2

TB2→B1TB1→B2

[v]B2

[v]B1 [L(v)]B1

[L(v)]B2

TB2→B1TB1→B2

LB2:B1

LB1:B2

Linear Algebra - National Chiao Tung Universityyi/Courses...Linear Algebra Linear Transformations...

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