Solve:

, , , repeated real eigenvalues ' X A X

:Example 1, 1,5 Solve:

1 2 2

' 2 1 2

2 2 1

X X

3

1

1 for =5

1

K

Sec7.5: Multiple Eigenvalue Solution

Lin. indep eigenvectors

nvv v ,,, 21 One single eigenvector

v

nt

ntt veXveX veX ,,, 2211

, , , repeated real eigenvalues ' X A X

Repeated real EigenvaluesRepeated real Eigenvalues

One single eigenvector 1v

1 0 vIA 21 vvIA 32 vvIA

kk vvIA 1

tevX 11

tevtvX 212

tt evtvvX

322

2

13

tkkk

ktk evtvvX

1)!1(

1

1


:Example Solve:

3 18'

2 9X X

3, 3


kk vvIA 1

tevX 11

tevtvX 212

tt evtvvX

322

2

13

tkkk

ktk evtvvX

1)!1(

1

1


:Example Solve:

2,2,2 2 1 6

' 0 2 5

0 0 2

X X


kk vvIA 1

tevX 11

tevtvX 212

tt evtvvX

322

2

13

tkkk

ktk evtvvX

1)!1(

1

1


:Example Solve:


kk vvIA 1

tevX 11

tevtvX 212

tt evtvvX

322

2

13

tkkk

ktk evtvvX

1)!1(

1

1

XX

010

122

001

' 1 ,1 ,1

Homog Linear System

2X2 system ' X AX

2 complex1,2 = i

2 real distinct

1 2, 2 real repeated ,

2 lin indep eig-vec Chain G-eigvec

21 , vv 21 , vv

3X3 system ' X AX

1 real + 2 complex 3 real distinct 2 real repeated + 1 real

3 lin indep eig-vec 2 lin indep eig-vec

321 , , vvv 121 , , wvv

321 ,, i 3,21,321 ,

3 real repeated

3 li

n i

nd

ep e

ig-v

ec

2 li

n i

nd

ep e

ig-v

ec

3

2

1

,

,

v

v

v

1

1,

w

v

321

1 li

n i

nd

ep e

ig-v

ec

1v


:Example 2,2,2 2 1 6

' 0 2 5

0 0 2

X X

DEF

completenot isit if defective called is 1kty multiplici of eigenvalueAn

rseigenvecto missing ofnumber ofdefect

0

0

1

1v

2 defect

:Example

3 18'

2 9X X

3, 3

1

31v

1 defect


:Example 2,2,2 2 1 6

' 0 2 5

0 0 2

X X

0

0

1

1v

2 defect

TvIA ]0,0,1[ 02 1

TvvIA ]0,1,0[ 2 21

TvvIA ],,0[ 25

1

5

632

rank 2 generalized eigenvector

rank 3 generalized eigenvector

02

BUT

02

21

22

vIA

vIA

02

BUT

02

22

23

vIA

vIA

DEF: A rank r generalized eigenvctor associated with is a vector v such that

02 BUT 02 1 vIAvIA rr


:Example 2,2,2 2 1 6

' 0 2 5

0 0 2

X X

0

0

1

1v

2 defect

TvIA ]0,0,1[ 02 1

TvvIA ]0,1,0[ 2 21

TvvIA ],,0[ 25

1

5

632

, , 321 vvv

1r veigenvecto on the based rseigenvecto dgeneralize ofchain

1r veigenvecto on the based rseigenvecto dgeneralize ofchain 3length A


DEF A length k chain of generalized eigenvectors based on the eigenvector is a set of of k generalized eigenvectors such that

, , , 21 kvvv 1v

1 kk vvIA 21 kk vvIA

12 vvIA

0 kk vIA

Example

kxk

300

030

003

1) Find charc. Equ.2) Find all eigenvalues3) How many free variables4) How many lin. Indep eigvct5) defect

Example

kxk

30000

13000

01000

00300

00130

00013

1) Find charc. Equ.2) Find all eigenvalues3) How many free variables4) How many lin. Indep eigvct5) defect

kxk

J

000

100

010

001

Definition:

Jordan block with eigenvalue

, , , 21 kvvv Chain of generalized eigenvectors Examples

70

17

50

15

200

120

012

9000

1900

0190

0019

Jordan Block

sJ

J

J

J

00

00

00

2

1Definition:

Where each submatix is a jordan block of the form

iJ

Jordan Normal Form

kxki

i

i

i

iJ

000

100

010

001

Exmples:

90000

19000

00200

00120

00012

1A

1000

1100

0050

0015

2A

200

040

014

3A

100

020

004

4A

1) Find eigenvalues2) multiplicity3) How maany lin. Indep eigenvectors4) How many chain and length

Theorem 1: Any nxn matrix A is similar to a Jordan normal form matrix

Jordan Normal Form

Theorem 1:

Let A be nxn matrix there exits an invertable Q such that:

JQAQ 1

where J is in Jordan normal form

:Example3, 3

1

31v

92

183A

30

13J

:Example

123

031

115

A

300

030

013

J

Find the Jordan form Find the Jordan form

3,3,3

Jordan Normal FormTheorem 1:

Let A be nxn matrix there exits an invertable Q such that:

JQAQ 1

where J is in Jordan normal form

sJ

J

J

00

00

00

2

1

If all generalized eigenvectors are arranged as column vectors in proper order corresponding to the appearance of the Jordan blocks in (*), the results is the matrix Q

30000

03000

01300

00020

00012

J

2,2,3,3,3

:Example

Let A be 5x5 matrix3,3,3

121 ,, wvv

2,2

21,uu],,,,[ 12121 wvvuuQ

, , , repeated real eigenvalues

Another method to compute: Generalized eigenvectorsAnother method to compute: Generalized eigenvectors


kk vvIA 1

kvvv ,,, 21

Chain of generalized eigenvectors

Compute:

)( IA 2)( IA kIA )(

kk vIA 0)(

Solve:

Compute:

kk vIAv )(1

32 )( vIAv

21 )( vIAv

kvvv ,,, 21

Chain of generalized eigenvectors

:Example Find all generalized eigenvectors:

1 ,1 1,

12 )( kk vIAv

010

122

001

A

Another method to compute: Generalized eigenvectorsAnother method to compute: Generalized eigenvectors

Compute:)( IA 2)( IA kIA )(

kk vIA 0)(

Solve:

Compute:

kk vIAv )(1

32 )( vIAv

21 )( vIAv

kvvv ,,, 21 Chain of generalized eigenvectors

:Example Find all generalized eigenvectors:

1 ,1 1,

12 )( kk vIAv

010

122

001

A

:Solution

0110

0112

0000

0IA

0110

0001

000023 RR

22

1 R

1 lin indepeigenvector

Length of chain =3

321 ,, vvv

110

112

000

)( IA

002

002

000

)( 2IA

000

000

000

)( 3IA

0

0

1

3v

0

2

0

0

0

1

110

112

000

2v

2

2

0

0

2

0

110

112

000

1v

Documents

Solve: