1

Lecture 2: Eigenvalue-Eigenvector and Orthogonal Matrix (Part B Chapter 7 in Advanced Eng. Math. By E. Kreyszig)

Definition For a given square matrix A, if equation AX = X, with is a nonzero number and X is a nonzero vector exists then is the eigenvalue and X the eigenvector of A. Both and X are derived from A, they represent certain characteristics of A. The definition shows that, square matrix A multiplying its eigenvector only changes the eigenvectors magnitude with a proportional factor of (eigenvalue). Calculation

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)3)(2(2)1)(6(2))(1)(2(

0-02-1-6--123-2-2-

0100010001

02-1-6-123-22-

02-1-6-123-22-

0 |I| from Find (1)

steps Two 0 |I-| requests RuleCramer the,for solution nontrivial has LES shomogeneou theIf

)(

32123

3

2

1

3

2

1

111

=====++

=

++

=

=

=

=

=

====

xxx

xxx

A-

AX

OXIAOIXAXXIAXXXA nnnnnnn

Roots of a polynomial: the number of roots equals the polynomial order and the product of all roots equals the polynomial constant. (2) Find eigenvector X from homogeneous LES, 0)( = XIA with the calculated using GE a) For = 5

=

=

=

=

=

=

=+

=++

121

20002052

000210521

16-8-016-805-2-1-

32-16-016-805-2-1-

3-27-6-425-2-1-

5-2-1-6-423-27-

5-02-1-6-5-123-25-2-

1

13

12

11

132

321

ccxcxcx

Xxxxxx

Notice: there are 2 Equations including 3 unknown components of the eigenvector (x1, x2, x3) so 1 independent component x1 can take any value, c1, and the other components, x2 and x3, have the values corresponding to c1. In general, for a LES with n unknowns, after Gauss Elimination, we have r = rank (A) being the number of independent equations in the LES then, (n r) equals the number of independent unknowns. The eigenvector needs to be represented as a linear combination of the (n - r) independent vectors. b) For = -3 (the double roots)

+

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103

012

103

and 012

10

01

take Usually,23

r,eigrnvecto in the componentst independen as ; takeIf

0322

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)3(-02-1-6-)3(-123-2)3(-2-

32323

2

33

22

23

3322

321

kkXXXkkxx

cxcxxx

X

cxcx

xxxrn

rn

Note: you can use any two independent vectors as

3

2

xx then calculate x1 and obtain the eigenvector X.

2

Rules 1. Values of equal the roots of characteristic polynomial derived from |A| = 0. If the order of the polynomial is k, you have k real and/or complex roots, they may be distinct or multiple. A square matrix, Ann can have at maximum n distinct roots (eigenvalues). 2. All eigenvectors corresponding to distinct eigenvalues are independent.

21 2

1

11

2

2

12 1 2

2

5 3 5 30 (5 ) 9 0 8, 2.

3 5 3 58,

5 8 3 -3 3 1

3 5 8 0 0 12,

5 2 3 3 3 1 and independent.

3 5 2 0 0 1

A

x X

x

xX X X

x

= = = = =

=

= =

=

= =

3. AT has the same eigenvalues as A. Example eigenvalues and eigenvectors for 2d square matrices: Applications

1. Principal Directions in vector transformation (by multiplying a matrix) A circle elastic membrane in the x1-x2 plane with boundary equation: x12 + x22 = 1 is stretched so that any point P (x1 x2) on the boundary goes to point Q (y1 y2) by = !! = = 5 33 5 !!

In the Principal Direction, points P moves to Q in the same direction: 5 33 5 !! = !! 5 33 5 = 5 ! 9 = 0 ! = 8, ! = 2 eigenvectors ! : 5 8 33 5 8 !! = 0 ! ! = 0 1 independent ! = ! = c! !! = c! 11 ! : 5 2 33 5 2 !! = 0 ! + ! = 0 1 independent ! = ! = d! !! = d! 11

Illustration

Matrix

1011

Eq. 2 2+ 1 = 0 2 2k + k2 = 0 ( k1)( k2) = 0 2 2 cos + 1 = 0 1, 2 1= 2=1 1=2= k 1= k1 2= k2 1,2= cos isin

X1, 2 ( )01 ( ) 1) (0, 0 1, ( ) 1) (0, 0 1, (1, -i) (1, i)

3

2. Discrete Time Markov Chain A random process of state vector Xn1 characterized as memoryless: the next state Xt+1 depends only on the current state Xt but not on the sequence of events that preceded it: !!!!! = !!!!! , !!!!!!!!!!!! = 0.8 0.1 0.10.1 0.7 0.20 0.1 0.9

!! = 1!! = 0!! = 1 = !!!! = 0.9!!!! = 0.3!!!! = 0.9 !" = probability of transforming from !! to !!!! !!!! = 0.9 includes 0.8 from !! ; 0.1 from !! and 0.1 from !! .

3. Vibration analysis

Two masses, m1 and m2, hang on two springs (k1 and k2) as shown. With a small disturb, the system will vibrate around it force-balanced positions y1 = 0 and y2 = 0. Calculate the displacements of m1 and m2: y1(t) and y2(t).

From Newtons 2nd law and Hooks law.

221211112221

2

1 )( )( ykykkykyykdtydm ++==

221212222

2

2 )( ykykyykdtydm ==

=

+=

2

1

22

2

21

2

2

1

22

221

22

2

2

21

2

1

2225)(

yy

dtyddtyd

yy

kkkkk

dtydm

dtydm

and If

22

11

=

=t

t

exyexy

=

t

t

exex

dtyd

dtyd

2

2

21

22

2

21

2

Xxx

xx

xx

xx

exex

exex

t

t

t

t

r Eigenvecto ; Eigenvalue 22

2522

2522

25 22

1

2

12

2

12

2

21

2

12

2

21

=

=

=

y1

y2

m1 = 1 k2 = 2 m2 = 1

0

0

k1 = 3

4

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=

==

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ti

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ceycey

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i

ceexyceexyxcX

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62

61

22

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1

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1

2

12

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1111

2

1

2

1

2

11

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2

0021

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622

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0012

12

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The system can vibrate in different modes! 2. Diagonalization of a square matrix Ann with n independent eigenvectors (distinct eigenvalues) Diagonalization means transforming a non-diagonal matrix into an equivalent diagonal matrix, hence it becomes simple for operations. If matrix Ann with distinct eigenvalues, it must have independent eigenvectors. Form a new matrix P whose column vectors are these distinct eigenvectors (P is called the modal matrix of A). It can be shown that P-1 exists (|A| 0) and the matrix product P-1AP = D. For this diagonal matrix D the diagonal elements are the distinct eigenvalues. The order or the eigenvalues in D is the same as the order of column vectors in P:

==

n

DAPP

...000.000...00...0

2

1

1

=

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1005

1111

2332

1111

21

1111

21 and

1111

in order r eigenvecto theChange5001

1111

2332

1111

21

1111

21 and

1111

11

5 ;1

1 1

2332

Given

11

1

1222111

APPDPP

P

APPD

PPcXcXA

=

==

=

=

==

==

=

1006

1114

2145

and 11

141-1

1 ;14

62145

Given

54

51

51

51

1

54

51

51

51

12211

AXXD

XXXXA -

We can retrieve A from D and P: AIAIPPAPPPAPPPPDP ==== )()()( 11111 3. Matrix Power calculation using 1= PDPA

. 112112 === PPDAPPDPDPPDPA kk

=

==

=

=

=

5001

1111

21

5001

1111

then 1111

21 and

1111

2332

Given 23123231 PPDAPPA

5

4. Similar Matrix Matrix C is formed by two given matrices A and B with ABBC 1= then C and A are similar matrices and the process is called a Similarity Transformation of A. Features of Similar Matrix C and A share the same eigenvalues {i} and their eigenvectors have the relation AC XBX

1= 1 25 52 15 5

1 25 51 12 15 5

5 41) Given matrix and test the similarity transformation.

1 2

Let and

Eigenvalues and eigenvectors of 5 4 5- 4

1 2 1 2-

T

X A

P X P X X

A

A

= =

= = = =

=

21 2

1

2

0 (5- )(2- ) 4 0

7 6 0 ( 6)( 1) 0 6, 1.Eigenvectors

65-6 4 -1 4 -1 4 41 2-6 1 -4 0 0 1

15-1 4 4 4 1 1 11 2-1 1 1 0 0 1

Similarity Matrix o

X

X

= =

+ = = = =

=

=

=

=

1 2 1 2 23 65 5 5 5 5 51 1 12 1 2 1 21 125 5

5 55 5 5 5

23 65 5

21 125 5

f

5 4 1 2 5 4 1 21 2 2 1 1 2 2 1

Eigenvalues and eigenvectors of

23 5 60 0

21 12 5

-

A

A P AP

A

= =

= =

%1

2

123 65 5

21 125 5

1 25 51 1 12 1 5 55 5

6 as

1

6

6 23 5 6 6 7 6 7 6 6121 12 5 6 21 18 0 0 756

4 1 2 4 6 Similarity!

1 2 1 1 7

s A

P X

=

=

=

=

= = =

5. Orthogonal Matrix Orthogonal Vectors Two vectors A1n and B1n have the scalar product (dot) AB = 0. A1n (B1n)T = 0 Orthogonal vectors must be independent, but independent vectors may not be orthogonal. Perpendicular is limited to 3D, Orthogonal for nay Dimensions. Orthogonal Matrix A square matrix has all column/row vectors {ai} orthogonal. aiak= 0 j k Orthonormal Matrix (Normalize the column/row vectors in the orthogonal matrix) A real square matrix is orthogonal matrix, iff its column/row vectors: (a1,,an) have:

6

aiak=1length have vectorscolumn/row all othereach vectorscolumn/rowdifferent

if 1 if 0

=

=

kjkj

Features of an orthonormal matrix: (1) AT = A1; (2) |A| = 1, (3) Eigenvalue || =1, and (4) forming a unit perpendicular coordinate system. Check agiven orthnormal matrix A

IAAAA TT =

=

=

=

32

31

32

32

32

31

31

32

32

32

32

31

31

32

32

32

31

32

32

31

32

32

32

31

31

32

32

;

32

32

31

31

32

32

32

31

32

Rule A symmetric matrix A can construct an orthonormal matrix X using its normalized eigenvectors. Two steps: 1) calculate eigenvectors (2) normalization the eigenvectors.

=+

=+

=

=====

=

51

52

21

525

1

21

2

12

12-

020021

4221

042201

21

020012

1224

542251

05

0504)4)(1(042

214221

xx

xx

A

1|| 5212)51(

04)51(0512

251 0

of s

1

matrix? orthogonalan 1001

0) product dot (vector lorthonorma0)(:Tests

2

51

52

52

51

51

52

52

51

51

52

52

51

51

52

52

51

1

51

52

52

51

51

52

52

51

=

==

=+=

=

=

=

=

==

==+

=

ii

X

X

XXXX

X

T

Specific matrices which can be transformed into an orthogonal matrix: 1) A square matrix An n has n distinct eigenvalues can construct an orthogonal matrix from its independent eigenvectors. The column or row vectors in this orthogonal matrix construct a basis for Rn (a Cartesian coordinate system). 2) An symmetric matrix A can derive an orthonormal system from its eigenvectors and form an orthogonal matrix (by normalization) whose column/row vectors is a basis of Rn (a Cartesian coordinate system)