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A nonzero vector x is an eigenvector (or characteristic vector) of a square matrix A if there exists a scalar λ such that Ax = λx. Then λ is an eigenvalue (or characteristic value) of A.
Note: The zero vector can not be an eigenvector even though A0 = λ0. But λ = 0 can be an eigenvalue.
Example:
Show x 2
1
isaneigenvector for A
2 4
3 6
Solution : Ax 2 4
3 6
2
1
0
0
But for 0, x 02
1
0
0
Thus,x isaneigenvector of A,and 0 isaneigenvalue.
An n×n matrix A multiplied by n×1 vector x results in another n×1 vector y=Ax. Thus A can be considered as a transformation matrix.
In general, a matrix acts on a vector by changing both its magnitude and its direction. However, a matrix may act on certain vectors by changing only their magnitude, and leaving their direction unchanged (or possibly reversing it). These vectors are the eigenvectors of the matrix.
A matrix acts on an eigenvector by multiplying its magnitude by a factor, which is positive if its direction is unchanged and negative if its direction is reversed. This factor is the eigenvalue associated with that eigenvector.
Example 1: Find the eigenvalues of
two eigenvalues: 1, 2 Note: The roots of the characteristic equation can be repeated. That
is, λ1 = λ2 =…= λk. If that happens, the eigenvalue is said to be of multiplicity k.
Example 2: Find the eigenvalues of
λ = 2 is an eigenvector of multiplicity 3.
51
122A
)2)(1(23
12)5)(2(51
122
2
AI
200
020
012
A
0)2(
200
020
0123
AI
5
6
7
Example 1 (cont.):
00
41
41
123)1(:1 AI
0,1
4
,404
2
11
2121
ttx
x
txtxxx
x
00
31
31
124)2(:2 AI
0,1
3
2
12
ss
x
xx
To each distinct eigenvalue of a matrix A there will correspond at least one eigenvector which can be found by solving the appropriate set of homogenous equations. If λi is an eigenvalue then the corresponding eigenvector xi is the
solution of (A – λiI)xi = 0
Example 2 (cont.): Find the eigenvectors of
Recall that λ = 2 is an eigenvector of multiplicity 3.
Solve the homogeneous linear system represented by
Let . The eigenvectors of = 2 are of the form
s and t not both zero.
0
0
0
000
000
010
)2(
3
2
1
x
x
x
AI x
txsx 31 ,
,
1
0
0
0
0
1
0
3
2
1
ts
t
s
x
x
x
x
200
020
012
A
10
11
Definition: The trace of a matrix A, designated by tr(A), is the sum of the elements on the main diagonal.
Property 1: The sum of the eigenvalues of a matrix equals the trace of the matrix.
Property 2: A matrix is singular if and only if it has a zero eigenvalue.
Property 3: The eigenvalues of an upper (or lower) triangular matrix are the elements on the main diagonal.
Property 4: If λ is an eigenvalue of A and A is invertible, then 1/λ is an eigenvalue of matrix A-1.
Property 5: If λ is an eigenvalue of A then kλ is an eigenvalue of kA where k is any arbitrary scalar.
Property 6: If λ is an eigenvalue of A then λk is an eigenvalue of Ak for any positive integer k.
Property 8: If λ is an eigenvalue of A then λ is an eigenvalue of AT.
Property 9: The product of the eigenvalues (counting multiplicity) of a matrix equals the determinant of the matrix.
Algebraic Multiplicity of an eigenvalue λ is defined as the order of the eigenvalue as a root of the characteristics equation and is denoted by multa(λ) (or M λ)
Geometric multiplicity of λ is defined as the number of linearly independent eigenvectors corresponding and is denoted by multa(λ) (or M λ)
Theorem :- If A is a square matrix then for every eigenvalue of A, the algebraic multiplicity is greater than geometric multiplicity.
This theorem states that Every square matrix A satisfies its own characteristics equation; that is, if
λn+kn-1λn-1+kn-2λn-2+…..+k1λ+k0=0Is the characteristic equation of an
n×n matrix, thenAn+kn-1An-1+kn-2An-2+…..+k1A+k0I=0
16
If P = [ p1 p2 ], then AP = PD even if A is not diagonalizable.
Since AP 1 21 5
3 2 5 0
A
rp1
rp2
A
rp1 A
rp2
5 4 5 10
5(1) 2( 2)
5(1) 2(5)
5 1
1
2 2
5
1
rp1 2
rp2
rp1
rp2
1 0
0 2
11
25
5 00 2
1 2
1 5
5 00 2
PD.
17
So, our two distinct eigenvalues both have algebraic multiplicity 1 and geometric multiplicity 1. This ensures that p1 and p2 are not scalar multiples of each other; thus, p1 and p2 are linearly independent eigenvectors of A.
Since A is 2 x 2 and there are two linearly independent eigenvectors from the solution of the eigenvalue problem, A is diagonalizable and P-1AP = D.
We can now construct P, P-1 and D. Let
Then, P 1 5 / 7 2 / 7 1 / 7 1 / 7
and D
1 0
0 2
5 0
0 2
.
P rp1
rp2
1
1
25
1 2
1 5
.
18
Note that, if we multiply both sides on the left by P 1, then
AP Arp1
rp2
A
rp1 A
rp2
5 4
5 10
5(1) 2( 2)
5(1) 2(5)
5 11
2 2
5
1
rp1 2
rp2
rp1
rp2
1 0
0 2
11
25
5 00 2
1 2
1 5
5 00 2
PD becomes
P 1AP 5 / 7 2 / 7 1 / 7 1 / 7
1 21 5
3 2 5 0
5 / 7 2 / 7 1 / 7 1 / 7
5 4 5 10
35 / 7 0 / 7
0 / 7 14 / 7
5 0
0 2
D.
19
Example 2: Find the eigenvalues and eigenvectors for A.
Step 1: Find the eigenvalues for A.
Recall: The determinant of a triangular matrix is the product of the elements at the diagonal. Thus, the characteristic equation of A is
λ1 = 1 has algebraic multiplicity 1 and λ2 = 3 has algebraic multiplicity 2.
.
A 3 4 00 3 00 0 1
p() det(I A) I A 0
3 4 0
0 3 00 0 1
( 3)2 ( 1)1 0.
20
Step 2: Use Gaussian elimination with back-substitution to solve (λI - A) x = 0. For λ1 = 1, the augmented matrix for the system is
Column 3 is not a leading column and x3 = t is a free variable. The geometric multiplicity of λ1 = 1 is one, since there is only one free variable. x2 = 0 and x1 = 2x2 = 0.
I A |r0
2 4 00 2 00 0 0
000
:
12 r1 r1
r2r3
1 2 00 2 00 0 0
000
:r1
12 r2 r2
r3
1 2 00 1 00 0 0
000
.
21
The eigenvector corresponding to λ1 = 1 is
The dimension of the eigenspace is 1 because the eigenvalue has only one linearly independent eigenvector. Thus, the geometric multiplicity is 1 and the algebraic multiplicity is 1 for λ1 = 1 .
x
x1
x2
x3
00t
t
001
. If we choose t 1, then
rp1
001
is
our choice for the eigenvector.
B1 {rp1} is a basis for the eigenspace, E
1, with dim(E
1) 1.
22
The augmented matrix for the system with λ2 = 3 is
Column 1 is not a leading column and x1 = t is a free variable. Since there is only one free variable, the geometric multiplicity of λ2 is one.
3I A |r0
0 4 00 0 00 0 2
000
:
14 r1 r1
r2r3
0 1 00 0 00 0 2
000
:r1
r3 r2r2 r3
0 1 00 0 20 0 0
000
:r1
12 r2 r2
r3
0 1 00 0 10 0 0
000
.
23
x2 = x3 = 0 and the eigenvector corresponding to λ2 = 3 is
The dimension of the eigenspace is 1 because the eigenvalue has only one linearly independent eigenvector. Thus, the geometric multiplicity is 1 while the algebraic multiplicity is 2 for λ2 = 3 . This means there will not be enough linearly independent eigenvectors for A to be diagonalizable. Thus, A is not diagonalizable whenever the geometric multiplicity is less than the algebraic multiplicity for any eigenvalue.
x
x1
x2
x3
t00
t
100
,we choose t 1, and
rp2
100
is
our choice for the eigenvector.
B2 {rp2} is a basis for the eigenspace, E
2, with dim(E
2) 1.
24
This time, AP Arp1
rp2
rp2
A
rp1 A
rp2 A
rp2
1
rp1 2
rp2 2
rp2
rp1
rp2
rp2
1 0 0
0 2 0
0 0 2
PD.
P 1 does not exist since the columns of P are not linearly independent.
It is not possible to solve for D P 1AP, so A is not diagonalizable.
25
AP Arp1
rp2 ...
rpn
A
rp1 A
rp2 ... A
rpn
1
rp1 2
rp2 ... n
rpn
PD rp1
rp2 ...
rpn
1 0O
0 n
For A, an n x n matrix, with characteristic polynomial roots
for eigenvalues λi of A with corresponding eigenvectors pi.
P is invertible iff the eigenvectors that form its columns are linearly independent iff
1,2 ,...,n , then
a lgbraic multiplicity for each distinct i .dim(Ei
) geometric multiplicity
26
P 1AP P 1PD D 1 0
O
0 n
.
This gives us n linearly independent eigenvectors for P, so
P-1 exists. Therefore, A is diagonalizable since
The square matrices S and T are similar iff there exists a nonsingular P such that S = P-1TP or PSP-1 = T.
Since A is similar to a diagonal matrix, A is diagonalizable.
27
Example 4: Solve the eigenvalue problem Ax = λx and find the eigenspace, algebraic multiplicity, and geometric multiplicity for each eigenvalue.
Step 1: Write down the characteristic equation of A and solve for its eigenvalues.
A 4 3 60 1 0 3 3 5
p() I A 4 3 6
0 1 03 3 5
( 1)( 1) 4 6
3 5
28
Since the factor (λ - 2) is first power, λ1 = 2 is not a repeated root. λ1 = 2 has an algebraic multiplicity of 1. On the other hand, the factor (λ +1) is squared, λ2 = -1 is a repeated root, and it has an algebraic multiplicity of 2.
p()
( 1)( 4)( 5) 18( 1) 0
(2 5 4)( 5) 18( 1) 0
( 3 52 4 52 25 20) 18 18 0
3 3 2 ( 1)(2 2) ( 1)( 2)( 1) 0
( 2)( 1)2 0.
So the eigenvalues are 1 2,2 1.
29
Step 2: Use Gaussian elimination with back-substitution to solve (λI - A) x = 0 for λ1 and λ2.
For λ1 = 2 , the augmented matrix for the system is
2I A |r0
603
333
60 3
000
~
16 r1 r1
13 r2 r2
r3
103
1 / 213
10 3
000
~r1r2
3r1 r3 r3
100
1 / 21
3 / 2
100
000
:r1r2
32 r2 r3 r3
100
1 / 210
100
000
.
In this case,
x3 = r, x2 = 0, and
x1 = -1/2(0) + r
= 0 + r = r.
30
Thus, the eigenvector corresponding to λ1 = 2 is
x
x1
x2
x3
r0r
r
101
,r 0. If we choose
rp1
101
,
then B1 101
is a basis for the eigenspace of 1 2.
E1span({
rp1}) and dim(E1
) 1, so the geometric multiplicity is 1.
Arx 2
rx or (2I A)
rx
r0.
4 3 60 1 0 3 3 5
101
4 60
3 5
202
2
101
.
31
For λ2 = -1, the augmented matrix for the system is
( 1)I A |r0
303
303
60 6
000
~
13 r1 r1
r2r3
103
103
20 6
000
~r1r2
3r1 r3 r3
100
100
200
000
x3 = t, x2 = s, and x1 = -s + 2t. Thus, the solution has two linearly independent eigenvectors for λ2 = -1 with
x
x1
x2
x3
s 2tst
s
110
t
201
, s 0, t 0.
32
If we chooserp2
110
, and
rp3
201
, then B2
110
,201
is a basis for E2span({
rp2 ,
rp3}) and dim(E2
) 2,
so the geometric multiplicity is 2.
33
Thus, we have AP PD as follows :
4 3 60 1 0 3 3 5
1 1 20 1 01 0 1
1 1 20 1 01 0 1
2 0 00 1 00 0 1
2 1 20 1 02 0 1
2 1 20 1 02 0 1
.
Since the geometric multiplicity is equal to the algebraic multiplicity for each distinct eigenvalue, we found three linearly independent eigenvectors. The matrix A is diagonalizable since P = [p1 p2 p3] is nonsingular.
34
P | I 1 1 20 1 01 0 1
1 0 00 1 00 0 1
:1 1 20 1 00 1 1
1 0 00 1 0 1 0 1
:
1 0 20 1 00 0 1
1 1 00 1 01 1 1
:1 0 00 1 00 0 1
1 1 20 1 01 1 1
. So, P 1
1 1 20 1 01 1 1
.
We can find P-1 as follows:
35
AP PD gives us A APP 1 PDP 1.
Thus, PDP 1 1 1 20 1 01 0 1
2 0 00 1 00 0 1
1 1 20 1 01 1 1
2 1 20 1 02 0 1
1 1 20 1 01 1 1
4 3 60 1 0 3 3 5
A
Note that A and D are similar matrices.
36
D 1 1 20 1 01 1 1
4 3 60 1 0 3 3 5
1 1 20 1 01 0 1
2 2 40 1 0 1 1 1
1 1 20 1 01 0 1
2 0 00 1 00 0 1
.
Also, D = P-1 AP =
So, A and D are similar with D = P-1 AP and A = PD P-1 .
37
If P is an orthogonal matrix, its inverse is its transpose, P-1 =
PT . Since
PTP
rp1
T
rp2
T
rpT
3
rp1
rp2
rp3
rp1
T rp1
rp1
T rp2
rp1
T rp3
rpT
2
rp1
rp2
T rp2
rp2
T rp3
rp3
T rp1
rp3
T rp2
rpT
3
rp3
rp1
rp1
rp1
rp2
rp1
rp3
rp2
rp1
rp2
rp2
rp2
rp3
rp3
rp1
rp3
rp2
rp3
rp3
rpi
rp j I because
rpi
rp j 0
for i j andrpi
rp j 1 for i j 1,2, 3. So, P 1 PT .
38
A is a symmetric matrix if A = AT. Let A be diagonalizable so that A = PDP-1. But A = AT and
AT (PDP 1)T (PDPT )T (PT )T DTPT PDPT A.
This shows that for a symmetric matrix A to be diagonalizable, P must be orthogonal.
If P-1 ≠ PT, then A ≠AT. The eigenvectors of A are mutually orthogonal but not orthonormal. This means that the eigenvectors must be scaled to unit vectors so that P is orthogonal and composed of orthonormal columns.
39
Example 5: Determine if the symmetric matrix A is diagonalizable; if it is, then find the orthogonal matrix P that orthogonally diagonalizes the symmetric matrix A.
Let A 5 1 0 1 5 00 0 2
, then det(I A)
5 1 01 5 00 0 2
( 1)31( 2) 5 1
1 5( 2)( 5)2 ( 2)
3 82 4 48 ( 4)(2 4 12) ( 4)( 2)( 6) 0
Thus, 1 4, 2 2, 3 6.
Since we have three distinct eigenvalues, we will see that we are guaranteed to have three linearly independent eigenvectors.
40
Since λ1 = 4, λ2 = -2, and λ3 = 6, are distinct eigenvalues, each of the eigenvalues has algebraic multiplicity 1.
An eigenvalue must have geometric multiplicity of at least one. Otherwise, we will have the trivial solution. Thus, we have three linearly independent eigenvectors.
We will use Gaussian elimination with back-substitution as follows:
41
For 1 4 ,
1I Ar0
1 1 01 1 00 0 6
000
:1 1 00 0 00 0 1
000
:
1 1 00 0 10 0 0
000
x2 s, x3 0, x1 s .
rx
x1
x2
x3
s
110
orrp1
1 / 2
1 / 20
.
42
For 2 2 ,
2I Ar0
7 1 01 7 00 0 0
000
:1 7 00 1 00 0 0
000
x3 s, x2 0, x1 0 .
rx
x1
x2
x3
s
001
orrp2
001
.
43
For 3 6,
3I Ar0
1 1 01 1 00 0 8
000
:
1 1 00 0 10 0 0
000
x2 s, x3 0, x1 s.
rx
x1
x2
x3
s
110
orrp3
1 / 2
1 / 20
.
44
As we can see the eigenvectors of A are distinct, so {p1, p2, p3} is linearly independent, P-1 exists for P =[p1 p2 p3] and
Thus A is diagonalizable.
Since A = AT (A is a symmetric matrix) and P is orthogonal with approximate scaling of p1, p2, p3, P-1 = PT.
AP PD PDP 1.
PP 1 PPT 1 / 2 0 1 / 2
1 / 2 0 1 / 20 1 0
1 / 2 1 / 2 00 0 1
1 / 2 1 / 2 0
1 0 00 1 00 0 1
I .
45
As we can see the eigenvectors of A are distinct, so {p1, p2, p3} is linearly independent, P-1 exists for P =[p1 p2 p3] and
Thus A is diagonalizable.
Since A = AT (A is a symmetric matrix) and P is orthogonal with approximate scaling of p1, p2, p3, P-1 = PT.
AP PD PDP 1.
PP 1 PPT 1 / 2 0 1 / 2
1 / 2 0 1 / 20 1 0
1 / 2 1 / 2 00 0 1
1 / 2 1 / 2 0
1 0 00 1 00 0 1
I .
46
PDPT 1 / 2 0 1 / 2
1 / 2 0 1 / 20 1 0
4 0 00 2 00 0 6
1 / 2 1 / 2 00 0 1
1 / 2 1 / 2 0
4 / 2 0 6 / 2
4 / 2 0 6 / 20 2 0
1 / 2 1 / 2 00 0 1
1 / 2 1 / 2 0
5 1 0 1 5 00 0 2
A.
Note that A and D are similar matrices. PD P-1 =
47
1 / 2 1 / 2 0
0 0 1
1 / 2 1 / 2 0
5 1 0 1 5 00 0 2
1 / 2 0 1 / 2
1 / 2 0 1 / 20 1 0
4 / 2 4 / 2 0
0 0 2
6 / 2 6 / 2 0
1 / 2 0 1 / 2
1 / 2 0 1 / 20 1 0
4 0 00 2 00 0 6
.
Also, D = P-1 AP = PTAP
So, A and D are similar with D = PTAP and A = PDPT .
48
1 / 2 0 1 / 2
1 / 2 0 1 / 20 1 0
4 0 00 2 00 0 6
T1 / 2 0 1 / 2
1 / 2 0 1 / 20 1 0
T
1 / 2 0 1 / 2
1 / 2 0 1 / 20 1 0
4 0 00 2 00 0 6
1 / 2 1 / 2 00 0 1
1 / 2 1 / 2 0
4 / 2 0 6 / 2
4 / 2 0 6 / 20 2 0
1 / 2 1 / 2 00 0 1
1 / 2 1 / 2 0
5 1 0 1 5 00 0 2
AT = (PD P-1)T = (PD PT )T = (PT)T DT PT = P DT PT
= A. This shows that if A is a symmetric matrix, P must be orthogonal with P-1 = PT.
49
From Example 1, the diagonal matrix for matrix A is :
D 1 0
0 2
5 0
0 2
P 1AP D A PDP 1,
and A3 PDP 1PDP 1PDP 1 PD3P 1
1 21 5
( 5)3 0
0 (2)3
5 / 7 2 / 7 1 / 7 1 / 7
125 16
125 40
5 / 7 2 / 7 1 / 7 1 / 7
641 / 7 234 / 7585 / 7 290 / 7
. For A3, the eigenvalues, are 1
3 125 aand 23 8.
In general, the power of a matrix, Ak PDkP 1. and the eigenvalues are ik ,
where i is on the main diagonal of D.
50