Eigenvalues and eigenvectors
4
Eigenvalues and Eigenvectors Consider multiplying nonzero vectors by a given square matrix, such as We want to see what influence the multiplication of the given matrix has on the vectors. In the first case, we get a totally new vector with a different direction and different length when compared to the original vector. This is what usually happens and is of no interest here. In the second case something interesting happens. The multiplication produces a vector which means the new vector has the same direction as the original vector. The scale constant, which we denote by is 10. The problem of systematically finding such ’s and nonzero vectors for a given square matrix will be the theme of this chapter. It is called the matrix eigenvalue problem or, more commonly, the eigenvalue problem. We formalize our observation. Let be a given nonzero square matrix of dimension Consider the following vector equation: ......(1) .Ax lx n n. A [a jk ] l l [30 40] T 10 [3 4] T , c 6 3 4 7 dc 5 1 d c 33 27 d , c 6 3 4 7 dc 3 4 d c 30 40 d . (2) ) X 0 ( 0 A I AX X To have a non-zero solution of this set of homogeneous linear equation (2) | A-λI | must .be equal to zero i.e (3) A I 0 .The following procedure can find the eigen values & eigen vector of n order matrix A 1. to find the characteristic polynomial P(λ) = det [A−λI 2. to find the roots of the characteristic equations the roots are eigen values that we required P( ) 0 3. To solve the homogenous system ] .To find n- eigen vectors [Α−λΙ]Χ=0 1 wallaa alebady [email protected]
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Transcript of Eigenvalues and eigenvectors
Eigenvalues and EigenvectorsConsider multiplying nonzero vectors by a given square matrix, such as
We want to see what influence the multiplication of the given matrix has on the vectors.In the first case, we get a totally new vector with a different direction and different lengthwhen compared to the original vector. This is what usually happens and is of no interesthere. In the second case something interesting happens. The multiplication produces avector which means the new vector has the same direction asthe original vector. The scale constant, which we denote by is 10. The problem ofsystematically finding such ’s and nonzero vectors for a given square matrix will be thetheme of this chapter. It is called the matrix eigenvalue problem or, more commonly, theeigenvalue problem.
We formalize our observation. Let be a given nonzero square matrix ofdimension Consider the following vector equation:
......(1).Ax � lxn � n.
A � [ajk]
l
l
[30 40]T � 10 [3 4]T,
c6 3
4 7d c5
1d � c33
27d , c6 3
4 7d c3
4d � c30
40d .
c08.qxd 11/9/10 3:07 PM Page 323
(2))X 0(
0
A I
AX X
To have a non-zero solution of this set of homogeneous linear equation (2) | A-λI | must .be equal to zero i.e
( 3 )A I 0
.The following procedure can find the eigen values & eigen vector of n order matrix A
1. to find the characteristic polynomial P(λ) = det [A−λI
2. to find the roots of the characteristic equations the roots are eigen
values that we requiredP() 0
3. To solve the homogenous system
]
.To find n- eigen vectors
[Α−λΙ]Χ=0
1
wallaa alebady [email protected]
Example: Determine the eigen value and corresponding eigen vector at
the matrix
23
14A
Solution:
X
1
a
x
4
x
x
X
x x x x x
(4
0
1/ 2
1/ 2
3/ 10
1/ 10
the eigenvectors 1/ 2
1/ 22
1
5the eigen vector corresponding to
,0523
05
523
145
3/ 10
1/ 10
3
1
1031 for X
X is defind by
eigenvector may be normalized to unit length the normalized eigenvector3
The eigen vector corresponding to 1
3 where a is arbitrary constant
0323
034123
14
To find the eigen vector for 1
51 ,&
05)(05
)(2) 3 0023
14
00
0
23
14
22
2
2121121
21121
2
1
2
1
21
1
21
21221
211212
1
2
1
2
length aaa
X
is
a x ax x x x x
x x x x x
xxAX Xfor
a
alength a
length
XX
a
aX
is
alet x a , x
x xx x x
x
x
x
xAX X
AI
1)( 6
2
wallaa alebady [email protected]
Example: Determine the eigen values and corresponding eigen vectors of
the matrix
A
P
P
X
3c
2 / 3
X
9 6
X
c
A
2 / 31 / 3
2 / 3 1 / 32 / 3
2 / 31 / 32 / 3
0(:
0(:
1 / 3
2 / 3
2 / 3
1
2
2
3144,
1
2
2
2
2
2
042
022
0223
0
402
022
223
0)(3
)(96 ,3 ,
9 18 ) 09)((01629918( )
0
702
052
226
0( )
702
052
226
333
222
1
1
213
31
21
321
3
2
1
11
321
223
I ) XA
case
I ) XAcase
c
clength clength
X
c
c
c
c
the eigenvecto rs is X
cx xlet x
xx
xx
xxx
x
x
x
I XAcase
distinct
Simplifyin g we have
I
3
wallaa alebady [email protected]
.Find the eigenvalues. Find the corresponding eigenvectors
1. 2. 3. 4. D13 5 2
2 7 �8
5 4 7
TD3 5 3
0 4 6
0 0 1
Tc5 �2
9 �6dc3.0 0
0 �0.6d
5. E�3 0 4 2
0 1 �2 4
2 4 �1 �2
0 2 �2 3
U
c08.qxd 10/30/10 10:56 AM Page 329
H.W
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wallaa alebady [email protected]
