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Eigen Values and Eigen Vectors
Eigenvalues and Eigenvectors• Eigenvalue problem:
If A is an nn matrix, do there exist nonzero vectors x in Rn
such that Ax is a scalar multiple of x？
Eigenvalue and eigenvector:
A：an nn matrix
：a scalar
x： a nonzero vector in Rn
xAx
Eigenvalue
Eigenvector
Geometrical Interpretation
Eigen value and Eigen vector
A matrix eigenvalue problem considers the vectorequation:
Ax = λx (1)
Here A is a given square matrix, λ an unknown scalar, and xan unknown vector. In a matrix eigenvalue problem, thetask is to determine λ’s and x’s
Thm 5.2: (Finding eigenvalues and eigenvectors of a matrix AMnn )
Let A is an nn matrix.
(1) An eigenvalue of A is a scalar such that .
(2) The eigenvectors of A corresponding to are the nonzero
solutions of .
If has nonzero solutions iff .0)I( xA 0)Idet( A
0)I( xAxAx Note:
(homogeneous system)
Characteristic polynomial of AMnn:
1
1 1 0det( I ) ( I )
n n
nA A c c c
Characteristic equation of A:
0)Idet( A
How to find the eigenvalues and eigenvectors of:
Ax = λx (1)
11 1 1 1
21 1 2 2
1 1
.
n n
n n
n nn n n
a x a x x
a x a x x
a x a x x
11 1 12 2 1
21 1 22 2 2
1 1 2 2
( ) 0
( ) 0
( ) 0.
n n
n n
n n nn n
a x a x a x
a x a x a x
a x a x a x
( ) . A I x 0
11 12 1
21 22 2
1 2
( ) det( ) 0.
n
n
n n nn
a a a
a a aD
a a a
A I
7 - 7
• Example: Finding eigenvalues and eigenvectors of the matrix
51
122A
Sol: Characteristic equation:
0)2)(1(23
51
122)I(
2
A
Eigenvalue: 2 ,1 21
2 ,1
7 - 8
2)2( 2
0 ,1
33
0
0
31
124)I(
2
1
2
1
2
ttt
t
x
x
x
xxA
1)1( 1
0 ,1
44
0
0
41
123)I(
2
1
2
1
1
ttt
t
x
x
x
xxA
1015
745
1427
A
Find the eigenvalues and eigenvectors
next
Example 2:
1015
745
1427
A
next
1015
745
1427
AI
1015
745
1427
A
next
1015
745
1427
AI
1015
745
1427
det)det(
AI
1015
745
1427
A
next
1015
745
1427
AI
1015
745
1427
det)det(
AI
])[5(])[5(])[7(
1015
745
1427
A
next
1015
745
1427
AI
1015
745
1427
det)det(
AI
])[5(])[5(]3314)[7( 2
7 - 14
• Ex 4: (Finding eigenvalues and eigenvectors)
51
122A
Sol: Characteristic equation:
0)2)(1(23
51
122)I(
2
A
Eigenvalue: 2 ,1 21
2 ,1
1015
745
1427
A
next
1015
745
1427
AI
1015
745
1427
det)det(
AI
])[5(]62)[5(]3314)[7( 2
1015
745
1427
A
next
1015
745
1427
AI
1015
745
1427
det)det(
AI
]4214)[5(]62)[5(]3314)[7( 2
1015
745
1427
A
next
1015
745
1427
AI
1015
745
1427
det)det(
AI
]4214)[5(]62)[5(]3314)[7( 2
9157
21070
3010
231987
3314
23
2
23
1015
745
1427
A
next
1015
745
1427
AI
1015
745
1427
det)det(
AI
]4214)[5(]62)[5(]3314)[7( 2
9157
21070
3010
231987
3314
23
2
23
characteristic polynomial
1015
745
1427
A
next
09157 23
The roots of the characteristic equations are
potential rational roots:1,-1,3,-3,9,-9
0)1)(3)(3(
)34)(3(9157 223
1015
745
1427
A
next
09157 23
The eigenvalues are: -3, -3, -1
To find an eigenvector belonging to the repeated root –3,
consider the null space of the matrix –3I - A
1015
745
1427
A
next
09157 23
The eigenvalues are: -3, -3, -1
To find an eigenvector belonging to the repeated root –3,
consider the null space of the matrix –3I - A
000
000
4.12.1
715
715
14210
3 toreducesAI
The 2 dimensional null space of this
matrix has basis =
5
0
7
,
0
5
1
1015
745
1427
A
next
09157 23
The eigenvalues are: -3, -3, -1
To find an eigenvector belonging to the repeated root –1,
consider the null space of the matrix –1I - A
000
110
201
915
735
1428
1 toreducesAI
The null space of this
matrix has basis =
1
1
2
1015
745
1427
A
09157 23
The eigenvalues are: -3, -3, -1
1
1
2
5
0
7
,
0
5
1The eigenvectors are:
