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Chapter 7

Eigenvalues and Eigenvectors

7.1 Eigenvalues and Eigenvectors

7.2 Diagonalization

7.3 Symmetric Matrices and Orthogonal Diagonalization

Elementary Linear Algebra 投影片設計編製者R. Larsen et al. (6 Edition) 淡江大學電機系翁慶昌教授

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7.1 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 scalar

x： a nonzero vector in Rn

xAx

Eigenvalue

Eigenvector

Geometrical Interpretation

Elementary Linear Algebra: Section 7.1, pp.421-422

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Ex 1: (Verifying eigenvalues and eigenvectors)

10

02A

01

1x

11 201

202

01

1002

xAx

Eigenvalue

22 )1(10

11

010

1002

xAx

Eigenvalue

Eigenvector

Eigenvector

10

2x

Elementary Linear Algebra: Section 7.1, p.423

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Thm 7.1: (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

λxxei

xxxxAxAxxxA

) toingcorrespondr eigenvectoan is ..(

)()()()( )2(

1

1111

cxei

cxxcAxccxA


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

Find the eigenvalues and corresponding eigenspaces of

10

01A

y

x

y

xA

10

01v

If ) ,(v yx

0

100

10

01 xxx

For a vector on the x-axis Eigenvalue 11

Sol:


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For a vector on the y-axis

yyy

01

00

10

01

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 7.2: (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:

011

1)I()Idet( cccAA nn

n

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

51

122)Idet(

2

A

2 ,1 :sEigenvalue 21

2 ,1


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2)2( 2

0 ,1

33

00

31~

31

124

0

0

31

124)I(

2

1

2

12

ttt

t

x

x

x

xxA

1)1( 1

0 ,1

44

00

41~

41

123

0

0

41

123)I(

2

1

2

11

ttt

t

x

x

x

xxA


:Check xAxi

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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

0

0

0

000

000

010

)I(

3

2

1

x

x

x

xA

0, ,

1

0

0

0

0

1

0

000

000

010

~

000

000

010

3

2

1

tsts

t

s

x

x

x

2 toingcorrespondA of eigenspace the:,

1

0

0

0

0

1

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

3 ,2 ,1 :sEigenvalue 321 Elementary Linear Algebra: Section 7.1, p.429

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1)1( 1

0

0

0

0

2001

0101

10500

0000

)I(

4

3

2

1

1

x

x

x

x

xA

0, ,

1

2

0

2

0

0

1

0

2

2

0000

0000

2100

2001

~

2001

0101

10500

0000

4

3

2

1

tsts

t

t

s

t

x

x

x

x

1

2

0

2

,

0

0

1

0

is a basis for the eigenspace of A corresponding to 1


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2)2( 2

0

0

0

0

1001

0001

10510

0001

)I(

4

3

2

1

2

x

x

x

x

xA

0 ,

0

1

5

0

0

5

0

0000

1000

0510

0001

~

1001

0001

10510

0001

4

3

2

1

tt

t

t

x

x

x

x

0

1

5

0



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3)3( 3

0

0

0

0

0001

0101

10520

0002

)I(

4

3

2

1

3

x

x

x

x

xA

0 ,

1

0

5

0

0

5

0

0000

0100

5010

0001

~

0001

0101

10520

0002

4

3

2

1

tt

t

t

x

x

x

x

1

0

5

0



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Thm 7.3: (Eigenvalues of 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)

335

011

002

)( Aa

3000004000000000002000001

)( Ab

Sol:

)3)(1)(2(

335

011

002

I )(

Aa

3 ,1 ,2 321

3 ,4 ,0 ,2 ,1 )( 54321 bElementary Linear Algebra: Section 7.1, p.430

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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

T

TVVT

x

xxx


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Ex 8: (Finding eigenvalues and eigenspaces)

.

200

013

031

seigenspace ingcorrespond and seigenvalue theFind

A

Sol:

)4()2(

200

013

0312

AI

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

B

B


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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 7.1:

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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7.2 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 7.4: (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(III

111

1111

A and B have the same characteristic polynomial. 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 ,4 :sEigenvalue 321

Ex.5) p.403 (See

0

1

1

:rEigenvecto 4)1( 1

p


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Ex.5) p.403 (See

1

0

0

,

0

1

1

:igenvectorE 2)2( 32

pp

200

020

004

100

011

011

][ 1321 APPpppP

400

020

002

010

101

101

][ (2)

200

040

002

100

011

011

][ (1)

1132

1312

APPpppP

APPpppP

Notes:


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Thm 7.5: (Condition for diagonalization)

An nn matrix A is diagonalizable if and only if

it has n linearly independent eigenvectors. Pf:

ablediagonaliz is )( A

),,,( and ][Let

diagonal is s.t. invertiblean exists there

2121

1

nn diagDpppP

APPDP

][

00

0000

][

2211

2

1

21

nn

n

n

ppp

pppPD


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][][2121 nn

ApApAppppAAP

) of rseigenvecto are of tor column vec the..(

,,2 ,1 ,

APpei

nipAp

PDAP

i

iii

t.independenlinearly are ,,, invertible is 21 npppP

rs.eigenvectot independenlinearly has nA

n

npppnA

,, seigenvalue ingcorrespondh wit

,, rseigenvectot independenlinearly has )(

21

21

nipAp iii ,,2 ,1 , i.e.

][Let 21 n

pppP


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PDppp

ppp

ApApAp

pppAAP

n

n

nn

n

n

00

00

00

][

][

][

][

2

1

21

2211

21

21

ablediagonaliz is

invertible is tindependenlinearly are ,,,1

11

A

DAPP

Pppp n


Note: If n linearly independent vectors do not exist,

then an nn matrix A is not diagonalizable.video.edhole.com

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Ex 4: (A matrix that is not diagonalizable)

10

21

able.diagonaliznot ismatrix following that theShow

A


0)1(10

21I 2

A

1 :Eigenvalue 1

0

1 :rEigenvecto

00

10~

00

20I 1pAIA

A does not have two (n=2) linearly independent eigenvectors,

so A is not diagonalizable.Elementary Linear Algebra: Section 7.2, p.439

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Steps for diagonalizing an nn square matrix:

Step 2: Let ][ 21 npppP

Step 1: Find n linearly independent eigenvectors

for A with corresponding eigenvalues

nppp ,,, 21

Step 3:

nipApDAPP iii

n

,,2 ,1 , where ,

00

00

00

2

1

1


n ,,, 21

Note:

The order of the eigenvalues used to form P will determine

the order in which the eigenvalues appear on the main diagonal of D.video.edhole.com

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Ex 5: (Diagonalizing a matrix)

diagonal. is such that matrix a Find

113

131

111

1APPP

A


0)3)(2)(2(113

131111

I

A

3 ,2 ,2 :sEigenvalue 321


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21

000010101

~313111

111I1 A

1

0

1

:rEigenvecto

1

0

1

0 1

3

2

1

pt

t

t

x

x

x

22

0001001

~113151

113I 4

141

2 A

4

1

1

:rEigenvecto

4

1

1

241

41

41

3

2

1

pt

t

t

t

x

x

x


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33

000110

101~

413101

112I3 A

1

1

1

:rEigenvecto

1

1

1

3

3

2

1

pt

t

t

t

x

x

x

300

020

002

141

110

111

][Let

1

321

APP

pppP


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Notes: k is a positive integer

kn

k

k

k

n d

dd

D

d

dd

D

00

0000

00

0000

)1( 2

1

2

1

1

1

1

1111

111

1

1

)()()(

)())((

)(

)2(

PPDA

PAP

APAAP

APPPPPAPPAP

APPAPPAPP

APPD

APPD

kk

k

kk


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Thm 7.6: (Sufficient conditions for diagonalization)

If an nn matrix A has n distinct eigenvalues, then the

corresponding eigenvectors are linearly independent and

A is diagonalizable.


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Ex 7: (Determining whether a matrix is diagonalizable)

300

100

121

A

Sol: Because A is a triangular matrix,

its eigenvalues are the main diagonal entries.

3 ,0 ,1 321

These three values are distinct, so A is diagonalizable. (Thm.7.6)


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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

B

TRB

xxx, xx, xxxx,x,xxT

RT:R

Sol:

113

131

111

)()()(

bygiven is for matrix standard The

321 eTeTeTA

T


From Ex. 5, there are three distinct eigenvalues

so A is diagonalizable. (Thm. 7.6)

3,2 ,2 321

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300

020

002

][ ][ ][

][ ][ ][

])([ )]([ ])([

332211

321

321

BBB

BBB

BBB

ppp

ApApAp

pTpTpTD

The matrix for T relative to this basis is


)}1 ,1 ,1(),4 ,1 ,1(),1 ,0 ,1{(},,{ 321 pppB

Thus, the three linearly independent eigenvectors found in Ex.

5

can be used to form the basis B. That is

)1 ,1 ,1(),4 ,1 ,1(),1 ,0 ,1( 321 ppp

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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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7.3 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

1334

B

501041123

C

(symmetric)

(symmetric)

(nonsymmetric)


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Thm 7.7: (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.

bcca

A

Pf: Characteristic equation:

0)( 22

cabbabc

caAI

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

a

aA

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


. 01

10 because orthogonal is

01

10 (a) 1

TPPP

Ex 4: (Orthogonal matrices)

.

5

30

5

40105

40

5

3

because orthogonal is

5

30

5

40105

40

5

3

(b) 1

TPPP

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Thm 7.8: (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

0

0

535

32

534

51

32

532

52

31

535

534

532

51

52

32

32

31

TPP

535

32

3

534

51

32

2

532

52

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 7.9: (Properties of symmetric matrices)

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

eigenvalues of A, then their corresponding eigenvectors x1

and x2 are orthogonal.


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50/53Elementary Linear Algebra: Section 7.3, p.452

.orthogonal are seigenvaluedistinct toingcorrespond

01

10 of rseigenvecto any two that Show

A

Ex 6: (Eigenvectors of a symmetric matrix)

Sol: Characteristic function

0)4)(2(8631

13I 2

A

4 ,2 :sEigenvalue21

0 ,1

1 x

00

11~

11

11I2 (1)

111

ssA

0 ,1

1 x

00

11~

11

11I4 (2)

222

ttA

.orthogonal are xand x 0xx2121

stst

t

t

s

s

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Thm 7.10: (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 independent 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

102023

3A

2000

4A

Orthogonally diagonalizable

Symmetricmatrix


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Ex 9: (Orthogonal diagonalization)

142

412

222

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

v

vuv

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

Linear IndependentElementary Linear Algebra: Section 7.3, p.455

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Gram-Schmidt Process:

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

52

222

233322

www

wvvwvw

) , ,( ),0 , ,(53

553

453

2

3

335

15

2

2

22

w

wu

w

wu

535

32

534

51

32

532

52

31

321

0

] [ (4) pppP

300

030

006

1 APPAPP T


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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

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