5.1 Eigenvalues and Eigenvectors

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5.1 Eigenvalues and Eigenvectors 5.2 Diagonalization 5.3 Symmetric Matrices and Orthogonal Diagonalizatio

n

Chapter 5 Eigenvalues and EigenvectorsChapter 5 Eigenvalues and Eigenvectors

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5.1 Eigenvalues and Eigenvectors Eigenvalue problem:

If A is an nn matrix, do there exist nonzero vectors x in Rn such that Ax is a scalar multiple of x？

Eigenvalue and eigenvector:

A ： an nn matrix ： a scalarx： a nonzero vector in Rn

xAx

Eigenvalue

Eigenvector

Geometrical Interpretation

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Ex 1: (Verifying eigenvalues and eigenvectors)

10

02A

01

1x

11 20120

201

1002 xAx

Eigenvalue

22 )1(1011

010

1002 xAx

Eigenvalue

Eigenvector

Eigenvector

10

2x

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Thm 5.1: (The eigenspace of A corresponding to )

If A is an nn matrix with an eigenvalue , then the set of all eigenvectors of together with the zero vector is a subspace of Rn. This subspace is called the eigenspace of .

Pf:x1 and x2 are eigenvectors corresponding to

) , ..( 2211 xAxxAxei

) toingcorrespondr eigenvectoan is ..( )()( )1(

21

21212121

λxxeixxxxAxAxxxA

) toingcorrespondr eigenvectoan is ..( )()()()( )2(

1

1111

cxeicxxcAxccxA

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Ex 3: (An example of eigenspaces in the plane)

Find the eigenvalues and corresponding eigenspaces of

1001

A

yx

yx

A 1001

v

If ) ,(v yx

0

100

1001 xxx

For a vector on the x-axis Eigenvalue 11

Sol:

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For a vector on the y-axis

yyy0

100

1001

Eigenvalue 12

Geometrically, multiplying a vector (x, y) in R2 by the matrix A corresponds to a reflection in the y-axis.

The eigenspace corresponding to is the x-axis.

The eigenspace corresponding to is the y-axis.

11

12

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Thm 5.2: (Finding eigenvalues and eigenvectors of a matrix AMnn )

0)Idet( A(1) An eigenvalue of A is a scalar such that .

(2) The eigenvectors of A corresponding to are the nonzero

solutions of .

Characteristic polynomial of AMnn:

11 1 0det( I ) ( I ) n n

nA A c c c L

Characteristic equation of A:

0)Idet( A

0)I( xA

Let A is an nn matrix.

If has nonzero solutions iff . 0)I( xA 0)Idet( A0)I( xAxAx

Note:(homogeneous system)

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Ex 4: (Finding eigenvalues and eigenvectors)

51

122A

Sol: Characteristic equation:

0)2)(1(23

51122

)I(

2

A

Eigenvalue: 2 ,1 21

2 ,1

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2)2( 2

0 ,133

00

31124

)I(

2

1

2

12

tttt

xx

xx

xA

1)1( 1

0 ,144

00

41123

)I(

2

1

2

11

tttt

xx

xx

xA

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200020012

A


0)2(200

020012

I 3

A

Eigenvalue: 2

Ex 5: (Finding eigenvalues and eigenvectors)

Find the eigenvalues and corresponding eigenvectors for the matrix A. What is the dimension of the eigenspace of each eigenvalue?

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The eigenspace of A corresponding to :2

000

000000010

)I(3

2

1

xxx

xA

0, ,100

001

03

2

1

tsts

t

s

xxx

2 toingcorrespondA of eigenspace the:,100

001

Rtsts

Thus, the dimension of its eigenspace is 2.

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Notes:

(1) If an eigenvalue 1 occurs as a multiple root (k times) for

the characteristic polynominal, then 1 has multiplicity k.

(2) The multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace.

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Ex 6 ： Find the eigenvalues of the matrix A and find a basis

for each of the corresponding eigenspaces.

30010201105100001

A


0)3)(2()1(

30010201

105100001

I

2

A

Eigenvalue: 3 ,2 ,1 321

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1)1( 1

0000

20010101

105000000

)I(

4

3

2

1

1

xxxx

xA

0, ,

1202

0010

2

2

4

3

2

1

tsts

tt

st

xxxx

1202

,

0010

is a basis for the eigenspace of A corresponding to 1

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2)2( 2

0000

10010001

105100001

)I(

4

3

2

1

2

xxxx

xA

0 ,

0150

0

50

4

3

2

1

tttt

xxxx

0150


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3)3( 3

0000

00010101

105200002

)I(

4

3

2

1

3

xxxx

xA

0 ,

105

0

050

4

3

2

1

tt

t

t

xxxx

105

0


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Thm 5.3: (Eigenvalues for triangular matrices)If A is an nn triangular matrix, then its eigenvalues are the entries on its main diagonal.

Ex 7: (Finding eigenvalues for diagonal and triangular matrices)

335011002

)( Aa

3000004000000000002000001

)( Ab

Sol:)3)(1)(2(

335011002

I )(

Aa

3 ,1 ,2 321

3 ,4 ,0 ,2 ,1 )( 54321 b

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Eigenvalues and eigenvectors of linear transformations:

. of eigenspace thecalled is vector)zero (with the of rseigenvecto all setof theand

, toingcorrespond ofr eigenvectoan called is vector The.)(such that vector nonzero a is thereif :n nsformatiolinear tra a of eigenvaluean called is number A

TTVVT

xxxx

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Ex 8: (Finding eigenvalues and eigenspaces)

.200

013031

seigenspace ingcorrespond and seigenvalue theFind

A

Sol:

200013031

AI )4()2( 2

2 ,4 seigenvalue 21

2for Basis )}1 ,0 ,0(),0 ,1 ,1{(4for Basis )}0 ,1 ,1{(follows. as are seigenvalue twofor these seigenspace The

22

11

BB

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Notes:

)}1 ,0 ,0(),0 ,1 ,1(),0 ,1 ,1{('

diagonal. is,basis the torelative ofmatrix the,Then 8. Ex.in found rseigenvectot independen

linear threeof up made of basis thebelet and 8, Ex.in A is

matrix standard n whosensformatiolinear tra thebe Let (1)3

33

B

B'TA'

RB'

RT:R

200020004

'A

A of rsEigenvecto A of sEigenvalue

. of seigenvalue theare matrix theof entries diagonalmain The (2) AA'

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Keywords in Section 5.1:

eigenvalue problem: 特徵值問題 eigenvalue: 特徵值 eigenvector: 特徵向量 characteristic polynomial: 特徵多項式 characteristic equation: 特徵方程式 eigenspace: 特徵空間 multiplicity: 重數

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5.2 Diagonalization Diagonalization problem:

For a square matrix A, does there exist an invertible matrix P such that P-1AP is diagonal?

Diagonalizable matrix:

A square matrix A is called diagonalizable if there exists an invertible matrix P such that P-1AP is a diagonal matrix.

(P diagonalizes A) Notes:

(1) If there exists an invertible matrix P such that , then two square matrices A and B are called similar.

(2) The eigenvalue problem is related closely to the diagonalization problem.

APPB 1

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Thm 5.4: (Similar matrices have the same eigenvalues)

If A and B are similar nn matrices, then they have the same eigenvalues.

Pf:

APPBBA 1similar are and

A

APPAPPPAP

PAPAPPPPAPPB

I

III

)I(III111

1111

Thus A and B have the same eigenvalues.

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Ex 1: (A diagonalizable matrix)

200013031

A


0)2)(4(200

013031

I 2

A

2 ,2 ,4s：eigenvalue The 321

reigenvecto the4)1(

011

1p

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r eigenvecto the2)2(

100

,01

132 pp

200020004

such that

， 100011011

][

1

321

APP

pppP

200040002

100011011

][

1

312

APP

pppP Note: If

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Thm 5.5: (Condition for diagonalization)

An nn matrix A is diagonalizable if and only if it has n linearly independent eigenvectors.

Pf:ablediagonaliz is )( A

there exists an invertible such that is diagonalLet and

1

1 2 1 2[ ] ( , , , )n n

P D P APP p p p D diag

L L

1

21 2

1 1 2 2

0 00 0

[ ]

0 0[ ]

n

n

n n

PD p p p

p p p

LLL M M O ML

L

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1 2[ ]nAP Ap Ap ApAP PD

Q L

are eigenvectors of , 1, 2, ,

( . . the column vector of )i i i

i

Ap p i ni e p P A

K

is invertible are linearly independent.1 2 , , , nP p p pQ Lrs.eigenvectot independenlinearly has nA

has linearly independent eigenvectors 1 2

1 2

( ) , , with corresponding eigenvalues , ,

n

n

A n p p p

LL

, 1, 2, , i i iAp p i n K

Let 1 2[ ]nP p p p L

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1 2 1 2

1 1 2 2

1

21 2

[ ] [ ][ ]

0 00 0

[ ]

0 0

n n

n n

n

n

AP A p p p Ap Ap App p p

p p p PD

L LL

LLL M M O ML

are linearly independent is invertible

is diagonalizable

1 1

1

, , , np p p P

P AP DA

Q L

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Ex 4: (A matrix that is not diagonalizable)

1021

able.diagonaliznot ismatrix following that theShow

A


0)1(1021I 2

A

1：eigenvalue The 1

1

0 2 0 1 1I ~ eigenvector

0 0 0 0 0A I A p

A does not have two linearly independent eigenvectors, so A is not diagonalizable.

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Steps for diagonalizing an nn square matrix:

Step 2: Let1 2[ ]nP p p p L

Step 1: Find n linearly independent eigenvectors

for A with corresponding eigenvalues.1 2, , np p pL

Step 3: 1

21

0 00 0

0 0 n

P AP D

LL

M M O ML

where, , 1, 2, , i i iAp p i n K

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Ex 5: (Diagonalizing a matrix)

diagonal. is such that matrix a Find

113131111

1APPP

A


0)3)(2)(2(113

131111

I

A

3 ,2 ,2s：eigenvalue The 321

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21

000010101

~313111

111I1 A

101

r eigenvecto 0 1

3

2

1

pt

t

xxx

22

0001001

~113151

113I 4

141

2 A

41

1r eigenvecto 24

141

3

2

1

pt

tt

xxx

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33

000110

101~

413101

112I3 A

111

r eigenvecto 3

3

2

1

pttt

xxx

300020002

s.t.

， 141110111

][

1

321

APP

pppP

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Notes:

1 1

2 2

0 0 0 00 0 0 0

(1)

0 0 0 0

k

kk

kn n

d dd d

D D

d d

L LL L

M M O M M M O ML L

1

11

)2(

PPDA

PAPDAPPDkk

kk

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Thm 5.6: (Sufficient conditions for diagonalization)

If an nn matrix A has n distinct eigenvalues, then the corresponding eigenvectors are linearly independent and A is diagonalizable.

Ex 7: (Determining whether a matrix is diagonalizable)

300100121

A

Sol: Because A is a triangular matrix, its eigenvalues are

3 ,0 ,1 321

These three values are distinct, so A is diagonalizable.

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Ex 8: (Finding a diagonalizing matrix for a linear transformation)

diagonal. is torelative for matrix thesuch that for basis a Find

)33()( bygiven n nsformatiolinear tra thebe Let

3

321321321321

33

BTRB

xxx, xx, xxxx,x,xxTRT:R

Sol:

113131111

bygiven is for matrix standard The

A

T

From Ex. 5 you know that A is diagonalizable. Thus, the three linearly independent eigenvectors found in Ex. 5 can be used to form the basis B. That is

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

300020002

][ ][ ][][ ][ ][

321

321

BBB

BBB

AvAvAvvTvTvTD

The matrix for T relative to this basis is

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diagonalization problem: 對角化問題 diagonalization: 對角化 diagonalizable matrix: 可對角化矩陣

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5.3 Symmetric Matrices and Orthogonal Diagonalization

Symmetric matrix:A square matrix A is symmetric if it is equal to its transpose:

TAA

Ex 1: (Symmetric matrices and nonsymetric matrices)

502031210

A

13

34B

501041123

C

(symmetric)

(symmetric)

(nonsymmetric)

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Thm 5.7: (Eigenvalues of symmetric matrices)

If A is an nn symmetric matrix, then the following properties are true.

(1) A is diagonalizable.

(2) All eigenvalues of A are real.

(3) If is an eigenvalue of A with multiplicity k, then has k linearly independent eigenvectors. That is, the eigenspace of has dimension k.

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Ex 2: Prove that a symmetric matrix is diagonalizable.

bccaA

Pf: Characteristic equation:

0)( 22

cabbabccaAI

022

222

22222

4)(

42

442)(4)(

cba

cbaba

cabbabacabba

As a quadratic in , this polynomial has a discriminant of

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04)( (1) 22 cba

0 , cba

diagonal. ofmatrix a is 0

0

aa

A

04)( )2( 22 cba

The characteristic polynomial of A has two distinct real roots, which implies that A has two distinct real eigenvalues. Thus, A is diagonalizable.

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A square matrix P is called orthogonal if it is invertible and

Orthogonal matrix:

TPP 1

Thm 5.9: (Properties of symmetric matrices)

Let A be an nn symmetric matrix. If 1 and 2 are distinct eig

envalues of A, then their corresponding eigenvectors x1 and x2 a

re orthogonal.

Thm 5.8: (Properties of orthogonal matrices)

An nn matrix P is orthogonal if and only if its column vectors form an orthogonal set.

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Ex 5: (An orthogonal matrix)

535

534

532

51

52

32

32

31

0P

Sol: If P is a orthogonal matrix, then I 1 TT PPPP

I100010001

00

535

32

534

51

32

532

52

31

535

534

532

51

52

32

32

31

TPP

535

32

3

5345

132

2

53252

31

1 0 , ,Let ppp

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1

0 roducesp

321

323121

ppp

pppppp

set. lorthonormaan is } , ,{ 321 ppp

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Thm 5.10: (Fundamental theorem of symmetric matrices)

Let A be an nn matrix. Then A is orthogonally diagonalizable and has real eigenvalue if and only if A is symmetric.

Orthogonal diagonalization of a symmetric matrix:Let A be an nn symmetric matrix.(1) Find all eigenvalues of A and determine the multiplicity of each.(2) For each eigenvalue of multiplicity 1, choose a unit eigenvector. (3) For each eigenvalue of multiplicity k2, find a set of k linearly indepen

dent eigenvectors. If this set is not orthonormal, apply Gram-Schmidt orthonormalization process.

(4) The composite of steps 2 and 3 produces an orthonormal set of n eigenvectors. Use these eigenvectors to form the columns of P. The matrix will be diagonal.DAPPAPP T 1

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Ex 7: (Determining whether a matrix is orthogonally diagonalizable)

111101111

1A

081812125

2A

102

0233A

2000

4A

Orthogonally diagonalizable

Symmetricmatrix

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Ex 9: (Orthogonal diagonalization)

142412222

A. esdiagonaliz that matrix orthogonalan Find

A

P

Sol:0)6()3( )1( 2 AI

)2 ofty multiplici a has( 3 ,6 21

) , ,( )2 ,2 ,1( ,6 )2( 32

32

31

1

1111

vvuv

)1 ,0 ,2( ),0 ,1 ,2( ,3 )3( 322 vv

Linear Independent

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Gram-Schmidt Process:

)1 , ,( ),0 ,1 ,2( 54

52

222

233322

wwwwvvwvw

) , ,( ),0 , ,(53

553

453

2

3

335

15

2

2

22

wwu

wwu

535

32

534

51

32

532

52

31

0P

300030006

1APP

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symmetric matrix: 對稱矩陣 orthogonal matrix: 正交矩陣 orthonormal set: 單範正交集 orthogonal diagonalization: 正交對角化

5.1 Eigenvalues and Eigenvectors

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