Eigen,Diag,Sumetrix

8/3/2019 Eigen,Diag,Sumetrix

1/50

Eigenvalues and Eigenvectors


2/50

16- 2



Diagonalization

Symmetric Matrices and Orthogonal Diagonalization


3/50

7.1 Eigenvalues and Eigenvectors

Eigenvalue problem:

IfA is an nvn matrix, do there exist nonzero vectors x in Rn

such thatAx is a scalar multiple ofx

Eigenvalue and eigenvector:Aan nvn matrix

Pa scalar

x a nonzero vector in Rn

xAx P!

Eigenvalue

Eigenvector

Geometrical Interpretation

16- 3


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Ex 1: (Verifying eigenvalues and eigenvectors)

-

!

10

02A

-

!

0

11x

11 2012

02

01

1002 xAx !

-!

-!

--

!

Eigenvalue

22 )1(1

01

1

0

1

0

10

02xAx !

-

!

-

!

-

-

!

Eigenvalue

Eigenvector

Eigenvector

-

!

1

02

x

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Ex 2: (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!

-

!

-

!

-

-

01

00

10

01 xxx

For a vector on thex-axis Eigenvalue 11!P

Sol:

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For a vector on they-axis

-

!

-

!

-

-

yyy

01

00

10

01

Eigenvalue 12!P

Geometrically, multiplying a vector (x, y)in R2by the matrixA corresponds to a

reflection in they-axis.

The eigenspace corresponding to is thex-axis.

The eigenspace corresponding to is they-axis.

11 !P

12 !P

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Thm 7.2: (Finding eigenvalues and eigenvectors of a matrixAMnvn )

0)Idet( !AP(1) An eigenvalue ofA is a scalarP such that .

(2) The eigenvectors ofA corresponding to P are the nonzero

solutions of .

Characteristic polynomial ofAMnvn:

01

1

1)I()Idet( cccAAn

n

n!! PPPPP .

Characteristic equation ofA:

0)Idet( !AP

0)I( ! xAP

LetA is an nvn matrix.

If has nonzero solutions iff .0)I( ! xAP 0)Idet( !AP

0)I( !! xAxAx PP Note:

(homogeneous system)

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Ex 4: (Finding eigenvalues and eigenvectors)

-

!

51

122A

Sol: Characteristic equation:

0)2)(1(23

51

122)I(

2 !!!

!

PPPP

P

PP A

Eigenvalue: 2,1 21 !! PP

2,1 ! P

16- 8


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2)2( 2 !P

0,1

33

0

0

31

124)I(

2

1

2

1

2

{

-

!

-

!

-

-

!

-

-

!

ttt

t

x

x

x

xxAP

1)1( 1 !P

0,1

44

0

0

41

123

)I(

2

1

2

1

1

{

-

!

-

!

-

-

!

-

-

!

ttt

t

x

x

x

x

xAP

16- 9


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-

!200

020012

A


0)2(

200

020012

I 3 !!

! PP

PP

P A

Eigenvalue: 2!P

Ex 5: (Finding eigenvalues and eigenvectors)

Find the eigenvalues and corresponding eigenvectors for

the matrixA. What is the dimension of the eigenspace of

each eigenvalue?

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The eigenspace of A corresponding to :2!P

-

!

-

-

!

0

0

0

000

000

010

)I(

3

2

1

x

x

x

xAP

0,,

1

00

0

01

0

3

2

1

{

-

-

!

-

!

-

tsts

t

s

x

xx

2toingcorrespondAofeigenspacethe:,

1

0

0

0

0

1

!

-

-

PRtsts

Thus, the dimension of its eigenspace is 2.

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Notes:

(1) If an eigenvalue P1occurs as a multiple root (ktimes)for

the characteristic polynominal, then P1has multiplicity k.

(2) The multiplicity of an eigenvalue is greater than or equal

to the dimension of its eigenspace.

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Ex 6Find the eigenvalues of the matrixA and find a basis

for each of the corresponding eigenspaces.

-

!

3001

0201

10510

0001

A


0)3)(2()1(

3001

0201

10510

0001

I

2 !!

!

PPP

PP

PP

P A

Eigenvalue:3,2,1

321

!!! PPP16- 13


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1)1( 1 !P

-

!

-

-

!

0

0

0

0

2001

0101

10500

0000

)I(

4

3

2

1

1

x

x

x

x

xAP

0,,

1

2

02

0

0

10

2

2

4

3

2

1

{

-

-

!

-

!

-

tsts

t

t

st

x

x

xx

1

2

0

2

,

0

0

1

0

-

-

is a basis for the eigenspaceof A corresponding to 1!P

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2)2( 2 !P

-

!

-

-

!

0

0

0

0

1001

0001

10510

0001

)I(

4

3

2

1

2

x

x

x

x

xAP

0,

0

1

5

0

0

5

0

4

3

2

1

{

-

!

-

!

-

ttt

t

x

x

x

x

0

1

5

0

-

is a basis for the eigenspaceof A corresponding to 2!P

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3)3( 3 !P

-

!

-

-

!

0

0

0

0

0001

0101

10520

0002

)I(

4

3

2

1

3

x

x

x

x

xAP

0,

1

0

5

0

0

5

0

4

3

2

1

{

-

!

-

!

-

tt

t

t

x

x

x

x

1

0

5

0

-

is a basis for the eigenspace

of A corresponding to 3!P

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Thm 7.3: (Eigenvalues for triangular matrices)

IfA is an nvn triangular matrix, then its eigenvalues arethe entries on its main diagonal.

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

-

!

335

011002

)( Aa

-

!

30000040000000000020

00001

)( Ab

Sol:

)3)(1)(2(

335

011

002

I)( !

! PPPP

P

P

P Aa

3,1,2 321 !!! PPP

3,4,0,2,1)( 54321 !!!!! PPPPPb

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Eigenvalues and eigenvectors of linear transformations:

.ofeigenspacethecalled

isvector)zero(with theofrseigenvectoallsetoftheand

,toingcorrespondofreigenvectoancalledisvectorThe

.)(such thatvectornonzeroaisthereif:

nnsformatiolinear traaofeigenvalueancalledisnumberA

P

P

P

P

P

T

TVVT

x

xxx !p

16- 18


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Keywords in Section 7.1:

eigenvalue problem:

eigenvalue:

eigenvector:

characteristic polynomial:

characteristic equation:

eigenspace:

multiplicity:

16- 19


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16- 20

Lecture 16: Eigenvalues and Eigenvectors

Today


Diagonalization



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7.2 Diagonalization

Diagonalization problem:

For a square matrixA, does there exist an invertible matrixP

such thatP-1APis diagonal?

Diagonalizable matrix:

A square matrixA is called diagonalizable if there exists an

invertible matrixPsuch thatP-1APis a diagonal matrix.

(PdiagonalizesA) Notes:

(1) If there exists an invertible matrixPsuch that ,then two square matricesA andB 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)

IfA andB are similarnvn matrices, then they have the

same eigenvalues.

Pf:

APPBBA1

similarareand

!

A

APPAPPPAP

PAPAPPPPAPPB

!

!!!

!!!

I

III

)I(III

111

1111

P

PPP

PPPP

ThusA andB have the same eigenvalues.

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Ex 1: (A diagonalizable matrix)

-

!

200

013

031

A


0)2)(4(

200

013

031

I 2 !!

! PPP

PP

P A

2,2,4seigenvalueThe 321 !!! PPP

reigenvectothe4)1( !P

-

!

0

1

1

1p

16- 23


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reigenvectothe2)2( !P

-

!

-

!10

0

,01

1

32 pp

-

!

-

!!

200

020

004

such that

100

011

011

][

1

321

APP

pppP

-

!

-

!

!

200

040

002

100

011

011

][

1

312

APP

pppP Note: If

16- 24


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Thm 7.5: (Condition for diagonalization)

An nvn matrixA is diagonalizable if and only if it has n

linearly independent eigenvectors.

Pf:

ablediagonalizis)( A

),,,(and][Let

diagonaliss.t.invertibleanexiststhere

2121

1

nn diagDpppP

APPDP

PPP .. !!

!

][

00

00

00

][

2211

2

1

21

nn

n

n

ppp

pppPD

PPP

P

P

P

.

.

/1//

.

.

.

!

-

!

16- 25


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PDAP

ApApApAP n

!@

! ][ 21 .3

)ofrseigenvectoareoftorcolumn vecthe..(

,,2,1,

APpei

nipAp

i

iii -!! P

t.independenlinearlyare,,,invertibleis21 n

pppP .3

rs.eigenvectotindependenlinearlyhas nA@

n

npppnA

PPP .

.

,,seigenvalueingcorrespondhwit

,,rseigenvectotindependenlinearlyhas)(

21

21

nipAp iii ,,2,1, -!! P

][Let 21 npppP .!

16- 26


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PDppp

pppApApAppppAAP

n

n

nn

nn

!

-

!

!!!

P

PP

PPP

./1//

.

.

.

.

..

00

00

00

][

][][][

2

1

21

2211

2121

ablediagonalizis

invertibleistindependenlinearlyare,,,

1

11

A

DAPP

Pppp n

!@

.3

16- 27


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Ex 4: (A matrix that is not diagonalizable)

-

!

10

21

able.diagonaliznotismatrixfollowingthat theShow

A


0)1(10

21I 2 !!

! PP

PP A

1eigenvalueThe 1 !P

-

!

-

-

!!0

1reigenvecto

00

10~

00

20I 1pAIAP

A does not have two linearly independent eigenvectors,

soA is not diagonalizable.

16- 28


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Steps for diagonalizing an nvn square matrix:

Step 2: Let ][ 21 npppP .!

Step 1: Find n linearly independent eigenvectors

for A with corresponding eigenvalues.nppp .,, 21

Step 3:

-

!!

n

DAPP

P

PP

.

/1//

.

.

00

00

00

2

1

1

nipAp iii ,,2,1,where, -!! P

16- 29


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Ex 5: (Diagonalizing a matrix)

diagonal.issuch thatmatrixaFind

113

131

111

1APPP

A

-

!


0)3)(2)(2(

113

131

111

I !!

! PPP

P

PP

P A

3,2,2seigenvalueThe 321 !!! PPP

16- 30


31/50

21 !P

-

-

!

000

010

101

~

313

111

111

I1 AP

-

!

-

!

-

1

0

1

reigenvecto0 1

3

2

1

p

t

t

x

x

x

22 !P

-

-

!

000

10

01

~

113

151

113

I41

41

2 AP

-

!

-

!

-

4

1

1

reigenvecto 241

41

3

2

1

p

t

t

t

x

x

x

16- 31


32/50

33 !P

-

-

!

000

110

101

~

413

101

112

I3 AP

-

!

-

!

-

1

1

1

reigenvecto 3

3

2

1

p

t

t

t

x

x

x

-

!

-

!!

300

020

002

s.t.

141

110

111

][

1

321

APP

pppP

16- 32


33/50

Notes:

-

!

-

!

k

n

k

k

k

n d

d

d

D

d

d

d

D

.

/1//

.

.

.

/1//

.

.

00

00

00

00

00

00

)1( 2

1

2

1

1

11

)2(

!

!!

PPDA

PAPDAPPD

kk

kk

16- 33


34/50

Thm 7.6: (Sufficient conditions for diagonalization)

If an nvn matrixA has n distinct eigenvalues, then the

corresponding eigenvectors are linearly independent and

A is diagonalizable.

Ex 7: (Determining whether a matrix is diagonalizable)

-

!

300

100

121

A

Sol: BecauseA is a triangular matrix, its eigenvalues are

3,0,1 321 !!! PPP

These three values are distinct, so A is diagonalizable.

16- 34


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

diagonal.istorelative

formatrixthesuch thatforbasisaFind

)33()(

bygivennnsformatiolinear trathebeLet

3

321321321321

33

B

TRB

xxx,xx, xxxx,x,xxT

RT:R

!

p

Sol:

-

!

113

131

111

bygivenisformatrixstandardThe

A

T

From Ex.5 you know thatA is diagonalizable.Thus, the three

linearly independent eigenvectors found in Ex.5 can be used

to form the basisB.That is16- 35


36/50

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

? A

? A

-

!

!

!

300

020

002][][][

][][][

321

321

BBB

BBB

AvAvAv

vTvTvTD

The matrix forTrelative to this basis is

16- 36


37/50


diagonalization problem:

diagonalization:

diagonalizable matrix:

16- 37


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Lecture 16: Eigenvalues and Eigenvectors

Today


Diagonalization


16- 38


39/50

7.3 Symmetric Matrices and Orthogonal Diagonalization

Symmetric matrix:

A square matrixA is symmetric if it is equal to its transpose:

TAA !

Ex 1: (Symmetric matrices and nonsymetric

matrices)

-

!

502

031210

A

-

!

13

34B

-

!

501

041

123

C

(symmetric)

(symmetric)

(nonsymmetric)

16- 39


40/50

Thm 7.7: (Eigenvalues of symmetric matrices)

IfA is an nvn symmetric matrix, then the following properties

are true.

(1)A is diagonalizable.

(2) All eigenvalues ofA are real.

(3) IfP is an eigenvalue ofA with multiplicity k, then P has k

linearly independent eigenvectors.That is, the eigenspace

ofP has dimension k.

16- 40


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Ex 2:

Prove that a symmetric matrix is diagonalizable.

-

!

bc

caA

Pf: Characteristic equation:

0)( 22 !!

! cabba

bc

caAI PP

PP

P

0u22

222

22222

4)(

42

442)(4)(

cba

cbaba

cabbabacabba

!

!

!

As a quadratic in P, this polynomial has a discriminant of

16- 41


42/50

04)((1) 22 ! cba

0, !! cba

diagonal.ofmatrixais0

0

-

!

a

aA

04)()2( 22 " cba

The characteristic polynomial ofA has two distinct real roots,

which implies thatA 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 7.9: (Properties of symmetric matrices)

LetAbe an nvn symmetric matrix. IfP1and P2 are distinct

eigenvalues ofA, then their corresponding eigenvectorsx1

andx2 are orthogonal.

Thm 7.8: (Properties of orthogonal matrices)

An nvn matrixPis orthogonal if and only if its column vectors

form an orthogonal set.

16- 43


44/50

Ex 5: (An orthogonal matrix)

-

!

53

5

53

4

53

2

5

1

5

2

32

32

31

0P

Sol: IfPis a orthogonal matrix, then I1 !!TT PPPP

I

100

010

001

0

0

53

5

3

253

4

5

132

53

2

5

231

53

5

53

4

53

25

1

5

232

32

31

!

-

!

-

-

!

TPP

-

!

-

!

-

!

535

32

3

534

51

32

2

532

52

31

1 0,,Let ppp

16- 44


45/50

1

0

roducesp

321

323121

!!!

!!!

ppp

pppppp

set.lorthonormaanis},,{ 321 ppp

16- 45


46/50

Thm 7.10: (Fundamental theorem of symmetric matrices)

LetAbe an nvn matrix.ThenA is orthogonally diagonalizable

and has real eigenvalue if and only ifA is symmetric.

Orthogonal diagonalization of a symmetric matrix:

LetAbe an nvn symmetric matrix.(1) Find all eigenvalues ofA and determine the multiplicity of each.

(2) For each eigenvalue of multiplicity 1, choose a unit eigenvector.

(3) For each eigenvalue of multiplicity ku2, find a set ofk 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 ofn

eigenvectors. Use these eigenvectors to form the columns ofP.The

matrix will be diagonal.DAPPAPPT

!!1

16- 46


47/50

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

-

!

111

101

111

1A

-

!

081

812

125

2A

-! 1020233A

-

!20

004A

Orthogonally

diagonalizable

Symmetric

matrix

16- 47


48/50

Ex 9: (Orthogonal diagonalization)

-

!

142

412

222

A.esdiagonalizthatmatrixorthogonalanFind

A

P

Sol:

0)6()3()1( 2 !! PPP AI

)2oftymultipliciahas(3,6 21 !! PP

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

3

2

3

1

1

1

111

!!!! v

v

uvP

)1,0,2(),0,1,2(,3)3( 322 !!! vvP

Linear Independent

16- 48


49/50

Gram-Schmidt Process:

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

52

2

22

233322

!

!!! w

ww

wvvwvw

),,(),0,,(53

5

53

4

53

2

3

335

1

5

2

2

22

!!!!w

wu

w

wu

-

!

53

532

53

4

5

132

53

2

5

231

0

P

-

!

300

030

006

1APP

16- 49


50/50


symmetric matrix:

orthogonal matrix:

orthonormal set:

orthogonal diagonalization:

16 50

Eigen,Diag,Sumetrix

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