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8/18/2019 06. Eigenvalue Problem (1)
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The Eigenvalue Problem
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Homogeneous Systems
• A homogeneous linear system
A x = 0• If det( A)!0, the unique solution is the trivial
solution x = 0.• If det( A)=0, there exist nontrivial solutions.
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Example 1
Find the nontrivial solutions to the
homogeneous system
045
02
02
321
321
321
=++
=++
=!
+
x x x
x x x
x x x
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Solution 1
The system reduces to:
Letting x3=t , the solution can be expressed as
x 1+ x
2 = 0
!
x 2 + x
3 = 0
x 1+ 2 x
2 ! x
3 = 0
2 x 1
+ x 2
+ x 3
= 0
5 x 1+ 4 x
2 + x
3 = 0
t x
t x
t x
=
=
!
=
3
2
1
!
"
!#
$
!
%
!&
'(=
1
1
1
t X
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Applications
•
Quantum mechanics, geology, mathematics,image processing, etc.
•
MechanicsPrincipal stresses and the orientation of theprincipal planes
•
Structural Dynamics
Natural frequencies and mode shapes ofstructures
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The Eigenvalue Problem
Let A be a square matrix of dimension n x n, and
let x be a vector of dimension n. The problem is
to find scalars ! for which there exists a
nonzero vector x such that
Eigenvalue
(characteristic value)
Eigenvector
(characteristic vector)
x x A ! =
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Solution Methods
• Analytical Solution
– Solution of the characteristic equation
• Numerical Solution
– Power method
– Inverse power method
– Jacobi’s method, QR method
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Analytical Method
•
The eigenvalues of A are the solutions of the characteristic equation
• When the determinant is expanded, it becomesa polynomial of degree n, called the
characteristic polynomial .
( ) 0 x I A =! "
( ) 0det
21
22221
11211
=
!
!
!
=!
"
"
"
"
nnnn
n
n
aaa
aaa
aaa
!
"#""
!
!
I A
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Eigenvalues
If A is an n x n real matrix, then its n eigenvalues! 1, ! 2, !, ! n. are the real and complex roots of
the characteristic polynomial
Properties:
( ) ( ) I A p ! ! "= det
( ) !!==
==
n
k
k
n
i
iiatrace
11
" A ( ) !=
=
n
k
k
1
det " A
( ) ( ) ( )( ) ( )nn p ! ! ! ! ! ! !
""""
=
!211
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Eigenvector
If ! is an eigenvalue of A and the nonzero vector v
has the property that
Then v is called the eigenvector of A
corresponding to the eigenvalue ! . v is obtainedby solving the homogeneous system ( A - ! I ) v =
0.
vv A ! =
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Analytical Method
Hand computations used to solve the eigenvalue
problem when the dimension n is small.
1.
Find the coefficients of the characteristicpolynomial p(! )= det( A - ! I )
2.
Find its roots (eigenvalues).
3. Find the nonzero solutions (eigenvectors) of
the homogeneous linear system ( A - ! I ) v = 0.
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Example 2
Find the eigenpairs ! j , v j for the matrix
!
!!
"
#
$
$$
%
&
'
'''
=
310
121
013
A
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Solution 2
Eigenvalues. The characteristic equation det( A - ! I )=0 is
Therefore, the three eigenvalues are
! 1 = 1, ! 2 = 3, and ! 3 = 4.
012198
310
121
013
23 =+!
+!
=
!!
!!!
!!
" " "
"
"
"
( )( )( ) 0431 =!!!! " " "
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Solution 2
Eigenvectors.
For ! 1 = 1,
For ! 2 = 3,
!"
!#
$
!%
!&
'
=
!"
!#
$
!%
!&
'
(((
)
*
+++
,
-
...
.
0
0
0
210
111
012
3
2
1
x
x
x
!"
!#
$
!%
!&
'
=
1
2
1
1 av
02
02
32
21
=+!
=!
x x
x x
a x
a x
a x
=
=
=
3
1
2 2
!"
!#
$
!%
!&
'
=
!"
!#
$
!%
!&
'
(((
)
*
+++
,
-
.
...
.
0
0
0
010
111
010
3
2
1
x
x
x
!"
!#
$
!%
!&
'
(
=
1
0
1
2 bv
0
0
2
31
=
=+
x
x x
b x
b x
x
!=
=
=
3
1
2 0
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Solution 2
Eigenvectors.
For ! 3 = 4,
!"
!#
$
!%
!&
'
=
!"
!#
$
!%
!&
'
(((
)
*
+++
,
-
..
...
..
0
0
0
110
121
011
3
2
1
x
x
x
!"
!#
$
!%
!&
'
(=
1
1
1
2 cv
0
0
32
21
=+
=+
x x
x x
c x
c x
c x
=
!=
=
1
2
3
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Numerical Methods
1. Power Method
2.
Shifted-inverse Power Method
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Power Method
Assume that the n x n matrix A has n distinct
eigenvalues ! 1, ! 2, !, ! n and that they are
ordered in decreasing magnitude; that is,
|! 1| > |! 2| " |! 3| " ! " |! n|
eigenvector v1 corresponding to ! 1 is called adominant eigenvector .
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Power Method
If x 0 is chosen appropriately, then the sequences{ x k = [ x1
(k ) x2(k ) ! xn
(k ) ]’} and {ck } generated
recursively by
yk = A x kand
x k +1 = (1/ck +1) yk ,
where
ck +1 = max{| xi
(k )|
will converge to the dominant eigenvector v1 andeigenvalue ! 1, respectively.
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Example 3
Use the power method to find the dominant
eigenpair for the matrix
!!!
"
#
$$$
%
&
''
'''
=
10264
7172
5110
A
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Solution 3
Start with x 0 = {1 1 1}T and use to generate the
sequence of vectors { x k } and constants {c
k }.
The first iteration produces
113
2
2
1
1
12
12
8
6
1
1
1
10264
7172
5110
x c=
!"
!#
$
!%
!&
'
=
!"
!#
$
!%
!&
'
=
!"
!#
$
!%
!&
'
(
((
)
*
+
++
,
-
..
..
.
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Solution 3
The second iteration produces
Iteration generates the sequence
The sequence converges to v = {2/5 3/5 1}T, ! =4.
228
5
16
7
316
3
10
3
7
3
2
2
1
13
16
110264
7172
5110
x c=
!"
!#
$
!%
!&
'
=
!"
!#
$
!%
!&
'
=
!"
!#
$
!%
!&
'
(((
)
*
+++
,
-
..
..
.
!,
119
78,
19
38,
12
9,
13
16,
1
1278
47
52
21
38
23
76
31
18
11
12
5
8
5
16
7
3
2
2
1
!"
!#
$
!%
!&
'
!"
!#
$
!%
!&
'
!"
!#
$
!%
!&
'
!"
!#
$
!%
!&
'
!"
!#
$
!%
!&
'
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C-Implementation
Vector PowerMethod(
Matrix A, /* nxn matrix A */
Vector x0, /* nx1 starting vector */
double e, /* tolerance */
int N, /* maximum iterations */
double * ev /* eigenvalue */
);
Returns: The dominant eigenvector
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Shifted-inverse Power Method
Shifting Eigenvalues. Suppose that ! , v is an
eigenpair of A. If " is any constant, then ! - #, v
is an eigenpair of the matrix A - " I.
Inverse Eigenvalues. Suppose that ! , v is an
eigenpair of A. If ! !0, then 1 / ! , v is an
eigenpair of the matrix A-1.
Suppose that ! , v is an eigenpair of A. If " !! , then
1 / (! - #), v is an eigenpair of the matrix ( A-# I )-1.
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Shifted-inverse Power Method
A constant " can be chosen so that µ 1 = 1/(! i - #)is the dominant eigenvalue of ( A - # I )-1.
Furthermore, if x 0 is chosen appropriately, then
the sequences { x k = [ x
1
(k ) x2
(k ) ! xn
(k ) ]’} and {ck }
generated recursively by
yk = ( A - " I )-1 x
k
and
x k +1 = (1/ck +1) yk ,
where
ck +1
= max{| xi(k )|}
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Shifted-inverse Power Method
will converge to the dominant eigenpair of the
matrix ( A - # I )-1. Finally, the corresponding
eigenvalue for the matrix A is given by the
calculation
! µ
" +=
1
1
j
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Example 4
Employ the shifted-inverse power method to find theeigenpairs of the matrix
Note: The eigenvalues of A are ! 1 = 4,! 2 = 2 and ! 3 = 1,
and select and appropriate " and starting vector for
each case
!!!
"
#
$$$
%
&
''
''
'
=
10264
7172
5110
A
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Solution 4
Case (1): for the eigenvalue ! 1 = 4, we select" = 4.2 and the starting vector x
0 = {1 1 1}T. First
form the matrix ( A – 4.2 I )-1, iteration produces
0
1
1
1
2.14264
78.122
5112.4
0 x y ==
!!
!!!!
"#$
%&'
()
*+,
-
11
1
6078431373.0
4117647059.0
18181818.23
18181818.23
09090909.14
545454545.9
0 x y c=!=
!
!
!
=
"#$
%&'
"#$
%&'
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Solution 4
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Solution 4
the sequence converges to µ1 = -5 which is
the dominant eigenvalue of ( A – 4.2 I )-1 ,
and xk converges to v1 = [2/5, 3/5,1]
T.
the eigenvalue ! 1 of A is given by
! 1 = (1/ µ
1 ) + " = 1/(-5) + 4.2 = -0.2+4.2 = 4
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Solution 4
Case (2): for the eigenvalue ! 2 = 2, we select " =
2.1 and the starting vector x 0 = {1 1 1}T then form
the matrix (A – 2.1I)-1.
the sequence converges to µ1 = -10 and the
eigenpair of the matrix is
! 2 = (1/ µ2 ) + " = (1/-10) + 2.1 = -0.1+2.1 =2
v1 = [1/4, 1/2,1]T.
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Solution 4
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Solution 4
Case (3): for the eigenvalue ! 3 = 1, we select " =
0.875 and the starting vector x 0 = {0 1 1}T then
form the matrix (A – 0.875I)-1.
the sequence converges to µ1 = 8 and the
eigenpair of the matrix is
! 2 = (1/ µ1 ) + " = (1/8) + 0.875 = 0.125+0.875 = 1
v3 = [1/2, 1/2,1]T.
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Solution 4