Example
3.
Eigenvalues and Eigenvectors
3.1
Introduction
Let A be an n x n square matrix and consider the equation
A value of
for which A x
A x
x .
x has a non-trivial solution x
0 is called an eigenvalue (or characteristic value) of A and the corresponding solution x
0 is an eigenvector (or characteristic vector). The set of all eigenvalues of A is known as the spectrum of A .
Many physical situations lead to eigenvalue problems.
Example – Two Port Network
Consider a two-port network with transmission matrix T . it is of interest to know what impedance may be put across the output terminals such that at the input terminals that same impedance may be seen.
V
1 i
1
T i
2
V
2 z
We require v
2 i
2
z and v
1 i
1
z
i v
1 a c
1
i zi
1
1
a c
a c b d
1 z
or A x
x b d
i v
2
2
b d
i zi
2
2
i i
2
1
z
1
an eigenvalue problem
This impedance is known as the characteristic or iterative impedance of the network. The input impedance of any number of identical two port networks connected in cascade and terminated by their characteristic impedance, will be that characteristic impedance. i
1 T T T
V
1 z v
1 z i
1
Example – Electric Circuit
Consider the electrical circuit
1
The currents flowing in the loops satisfy
L di
2 dt
1
C
L di
1 i
2
1
dt
i
3
C dt
i
1
i
2
dt
1
C
0
i
2
i
1
dt
L di
3 dt
1
C
i
3
i
2
dt
0
LCi
1
' '
i
1
0
LCi
3
' '
i
3
i
2 i
2
LCi
2
' '
2 i
2
0
0
i
3
Or
i
1
0
1
1
0
1
2
1
0
1
1
i i i
1
2
3
LC
i i
i
1
"
"
2
"
3
If we look for solutions which are sinusoidal and all have the same frequency i.e. i
1 i
3
I
1
I
3 sin( sin( wt wt
1
), i
2
3
)
I
2 sin( wt
2
),
Then i
1
'' w
2 i
1
, i
"
2
w
2 i
2 and i
3
" w
2 i
3
and the problem becomes
1
1
0
1
2
1
0
1
1
i i i
1
2
3
w
2
LC
i i i
1
2
3
This is an eigenvalue problem with
Example – Mechanical System
Consider the mass-spring system
w
2
LC
2
k
1
, k
2
- spring stiffness, m
1
, m
2
- mass
From Newton’s Law the equations of motion are m
1 m
2 x
1
" x
"
2
k
1
k
2 x
1
x
2 k
2
x
1 x
2
x
1
or
( k
1
m
1 k
2 m
2 k
2
) k
2
m k
1
2 m
2
x x
2
1
x x
2
"
1
"
If we look for “normal mode” solutions i.e. both particles oscillating with the same frequency, we need x
1
l
1 sin( wt
1
), x
2
l
2 sin( wt
2
)
.
Thus gives x
1
" w
2 x
1
, x
2
"
( k
1
m
1 k
2 m
2 k
2
)
w
2 x
2 k
2
m k
1
2 m
2
x x
2
1
w
2
x x
2
1
This is an eigenvalue problem where
w
2 gives the natural frequencies of vibration and the eigenvector gives the relative motion of the two particles.
3.2
Determination of Eigenvalues and Eigenvectors
A x
x iff ( A
I ) x
0
For this equation to have non-trivial solutions ( x
for otherwise x
( A
I )
1
0
0 is the unique solution.
0 ) we must have ( A
I )
1
not existing,
Hence
is an eigenvalue iff A
I
0 .
A
I is a polynomial of degree n in
, known as the characteristic polynomial of A .
A
I
0 is the characteristic equation of A .
3
Clearly a matrix has at most n (real or complex) eigenvalues and corresponding eigenvectors.
Example
A
1
2
4
3
From
A
I
1
2
3
4
1
3
8
3
4
2
8
2
4
5
5
1
Hence the eigenvalues are
5 and
1
To find the corresponding eigenvectors we solve:
When
5
4
2
4
2 x x
2
1
0
0
0
2
4
4
2
0
0
~
1
0
1
0
0
0
x
1
x
2
x x
2
1
k
1
1
, k
0
One of the eigenvectors is for k
1 ,
1
1
When
2
2
1
4
4 x x
2
1
0
0
0
2
2
x
1
For k
2 x
2
x x
2
1
k
1
2
, k
1 ,
1
2
4
4
0
0
~
1
0
0
2
0
0
0
Example
A
1
0
1
1
2
1
0
1
1
4
1
1 0
A
I
1 2
1
0
1 1
1
1
1
3
2
1
1
1
2
3
2
2
When
0
1
1
0
x
2
2
1
1 x
3
, x
1
0
0
1
x
1 x
2 x
3
x
2
x
3
0
x
1
x
2 x
3
1
0
1
2
1
1
k
1
1
1
, k
0
0
1
1
0
0
0
~
1
0
0
1
1
1
0
1
1
0
0
0
~
1
0
0
1
For k
1 , the eigenvectors is
1
1
When
1
1
0
1
0
1
1
0
0
1
x
1 x
2 x
3
0
0
0
1
1
1
1
0
1
0
0
0
0
~
1
0
0
1
1
1
1
0
0
0
0
0
1
~ 0
0
x
2
0 , x
1
x
2
x
3
x
3 x
1
x
2 x
3
k
When
3
1
0
2
1
1
1
0
1
2
x
1 x x
2
3
0
1
1
0
0
1
1
, k
0
1
0
0
0
0
0
1
1
0
0
1
0
0
0
0
5
0
2
1
x
2
1
1
1
2 x
3
, x
1
0
1
2
x
0
0
0
~
0
1
2
2
x
3
x
3
1
1
1
0
1
2
0
0
0
1
~ 0
0
1
1
1 x
1
x
2 x
3
k
1
1
2
, k
1
2
2
0
0
0
~
1
0
0
0
1
1
0
1
2
0
0
0
0
Example
Consider the 3-mass, 2-spring system, where the masses are all the same ( m ) and the springs have equal stiffness ( k )
-displacements from equilibrium
The equations of motion are: mx
1
" mx
"
2
k
k x
2
x
2
x
1
x
1
F
1 x
3
x
2
F
1
F
2
mx
3
"
1
1
0
1 k
2
1
x
3
0
1
1
x x
1 x
2 x
3
2
F
2
m k
x
1
" x
2
" x
3
"
w
2 m k
x
1 x x
3
2
In the case of
“normal mode” solutions
From the previous example we have that the normal mode frequencies and the corresponding particle displacements are: w
2 w
3 w
1
k
0
3 k m m x x
2 x
3
1
T
T
1 , 0 ,
1 ,
1
2 , 1
T
In the above examples the eigenvalues are real and distinct and the eigenvectors are linearly independent. There are other possibilities.
Example
Let A
1
4
1
1
.
Then
A
I
1
4
1
1
2
2
5
1
2 i
6
For
1
2 i , an eigenvector x
1
For
1
2 i , an eigenvector x
1
i
2
1
i
1
2
.
.
Example
A
5
3
3
1
Then
A
3
3
I
5
3
3
3
x x
2
1
0
0
3
1
x
i.e.
2 is the only eigenvalue
2
2 k
1
1
It is also possible to have a repeated eigenvalue which does lead to more than one linearly independent eigenvector.
3.3
Eigenvalue Results
Two vectors x , y are called orthogonal if x
T y
0
Theorem
Eigenvectors corresponding to distinct eigenvalues are linearly independent.
Proof: Omitted.
Corollary
If an n
n matrix A has n distinct eigenvalues then its eigenvectors form a basis for C n
.
Theorem
If A is a real symmetric matrix, then
(a) the eigenvalues are all real (but not necessarily distinct).
(b) eigenvectors corresponding to distinct eigenvalues are orthogonal.
Proof:
(a)
Let A x
x ( 1 )
Transposing this gives x
T
A
x
T ( 2 ) (since A
A
T
)
Taking the complex conjugate of (2) gives x
T
A
x
T
( 3 )
7
x
T x
T
A x
x
T x
x
x
T
A x
x
T x
Subtracting
(
) x
T x
0
Now x
T x
x
1
2 x
2
2
...
x n
2
is not zero
and hence
and
is real.
(b)
Let A x
1
1 x
1
( 1 ) and A x
2
2 x
2
(2) and
1
2
Transpose (2)
x
T
2
x
A
1
2 x
T
2
( A
x
2
T
A x
1
2 x
T
2 x
1
A
T
) (3) x
T
2
x
T
2
A
Subtracting
0 x
1
1
(
1
x
T
2
2 x
1
) x
2
T x
1
1
2
hence x
T
2 x
1
0 and the eigenvectors are orthogonal.
Theorem
If A is an n
n real symmetric matrix, then it is possible to find n linearly independent eigenvectors (even though the eigenvalues may not be distinct) which form a basis for n
R .
Proof: Omitted.
An n x n matrix is said to be diagonalizable if there is a non-singular n x n matrix P such that
D
P
1
AP , where D is a diagonal matrix.
Theorem
If n x n matrix A has n linearly independent eigenvectors then it is diagonalizable.
.
Remark:
Let
A
a a
a
11
21 p 1 a
12 a
22
a p 2
a a
1 m
2
m a pm
, B
b b b
11
21 m 1 b
12 b
22
b m 2
b
1 n b
2 n
b mn
, a
11 a
1 m a
1
a
21
, , a m
a
2 m
b
1 b
11
b
21
, , b n b
1 n
b
2 n
a p 1
We observe that a pm b m 1 b mn
8
AB
b
11 a
1
a a
a
11
21 p 1
b
21 a a
12 a
22 a
p 2
2
a a
2 m a
1 m pm
b b
11
21 b m 1 b m 1 a m b
12 a
1 b
12 b
22
b m 2
b
22 a
2 b b
1 n
2
b mn n
A b
1
b m 2 a m
A b
2
b
1 n a
1
A b n
b
2 n a
2
b mn a m
Proof:
First we note that A will have n linearly independent eigenvectors if it is symmetric or if it has n distinct eigenvalues i.e. there exist x
1
, , x n
,
1
, ,
n
such that A x
1
1 x
1
, , A x n
n x n
Let
AP
1
P
x
1
A
x
1 x
1
0 x x
A
2
2 x
2
x n
A
be the matrix whose columns are the n eigenvectors of A . Then x
0 n x n
0
x
1
1
x
1
2
2 x
2 x
2
0 x
n
3
x n
0 x n
0 x
1
0 x n
1
n x n
So we have AP = PD where
D
1
0
0
AS
x
1
,
0
2
x
2
,
, x n
0
n
diag
1
, ,
n
is linearly independent, P
1 exist. Thus we have D
P
1
AP
Example
Show that A
1
2
4
3
is diagonalizable.
Proof:
For A
1
2
4
3
, we have eigenvalues
1
5 ,
2
1 . For
1
5 one of the corresponding eigenvector is
1
1
. For
2
1
2
1
, one of the corresponding eigenvector is .
Take P
1
1 exists and P
1 that A
1
2
2
1
. As the eigenvectors are linearly independent, P
1
1
1 /
/ 3
3
2
1
/
/
3
3
. Also
4
3
is diagonalizable.
P
1
AP
1
1 /
/ 3
3
2
1
/
/
3
3
1
2
4
3
1
1
2
1
1
1
5
0
2
1
1
0
1
. It concludes
Example
Is A
5
3
3
1
diagonalizable?
9
Sol:
A
I
5
3
3
1
2
2
, i.e.
2 is the only eigenvalue.
3
3
3
3
x x
2
1
0
0
x
k
1
1
As there is only one linearly independent eigenvectors, A
5
3
3
1
is not diagonalizable.
Example
Diagonalize A
1
3
3
Proof:
A
I
1
3
3
5
3
3
5
3 3
3
1
3
, if possible..
3
3
1
3
3
2
4
(
1 )(
2 )
2
When
1
0
3
3
3
6
3
3
0
3
x
1 x x
2
3
0
3
0
3
3
6
3
0
3
3
0
0
0
1
~ 0
0
1
3
3
x
2
x
3
, x
1
x
2
x
3 x
1
x
2 x
3
k
1
1
1
,
k
When
2
3
3
3
3
0
0
3
3 x
1
x
2 x
3
3
3
0
0
3
0
0
3
3
3
x
1 x x
2
3
0
0
0
0
1
~ 0
0
1
0
0
1
0
0
0
0
0
x
3
t , x
2
s , x
1
s s
t t
t
0
1
1
s
1
1
0
, t
2 s
2
0
0
s
t
0
3
3
0
0
0
~
1
0
0
1
1
0
0
1
0
0
0
0
10
A
1
3
3
3
5
3
3
1
3
has three linearly independent eigenvectors, therefore diagonalizable and
1
0
1
1
1
0
1
1
1
1
1
3
3
3
5
3
3
1
3
1
0
1
1
1
0
1
1
1
2
0
0
A
1
3
3
0
2
0
0
0
1
Example
Show that A
5
1
0
0
5
1
0
0
5
is not diagonalizable.
Proof:
A
I
5
1
0
0
5
1
0
0
5
( 5
)
3
3
5
3
3
1
3
is
A
5
1
0
0
5
1
0
0
5
has only the eigenvalue
5 .
When
5
0
1
0
0
0
1
A
5
1
0
0
0
0
x
1 x
2 x
3
0
5
1
0
x
1
x
2
0 , x
3
t
x x x
1
2
3
t
0
0
1
, t
0 .
0
0
5
has only one linearly independent eigenvector, therefore A
5
1
0
0
5
1
0
0
5
is not diagonalizable.
An important application of diagonalisation occurs in the solution of systems of differential equations.
Example
Diagonalisation is useful if it is required to raise a matrix A to a high power i.e.
If P
1
AP
D , then A
PDP
1
and A m
( PDP
1
)( PDP
1
)......( PDP
1
)
A
PD m
P
1 m .
.
11
D
1
0
0
0
2
0
n
diag
1
, ,
n
D m
m
1
0
0
0
m
2
0
m n
diag
m
1
, ,
m n
Example (optional)
Find all 2 by 2 matrices A which satisfy A
2
4 A
4 I
3
3
2
4
.
Sol:
A
2
4 A
4 I
( A
2 I )
2
We find that the matrix B
3
3
3
3
1
1
,
2
3
1
respectively.
That is
3
3
2
4
1
1
1
1
1
,
3
3
2
4
(*)
2
4
has eigenvalues 1, 6 and their corresponding eigenvectors are
2
4
3
1
2
6
3
1
2
Therefore,
3
3
3
5
3
P
1
2
4
5
BP
1
1
3
1
2
1
1
2
3
1
1
0
2
3
5
3
3
5
2
4
1
1
2
3
1
0
6
1
0
1
1
2
3
1
1
3
3
0
6
D , where P
1
3
5
3
2
3
5
, P
1
1
2
3
1
,
2
4
D
5 5
1
2
1
3
1
1
0
1
0
0
6
Note : if P
3
1
2
1
1
, then D
6
0
0
1
Note that
1
1
2
3
1
1
3
5
3
2
3
5
5 5
1
1
3
1
2
1
both sides of (*)
1
2
1
3
1
. Then we have
0
6
12
1
1
2
3
1
1
( A
2 I )
2
1
2
1
3
1
1
1
2
3
1
1
3
3
2
4
1
1
3
1
2
Let
1
1
2
3
1
1
( A
Y
2 I )
1
2
1
3
1
1
1
2
3
1
1
( A
2 I )
1
2
1
3
1
1
1
2
3
1
1
( A
2 I )
1
1
2
3
1
1
0
, then we need to solve Y
2
0
6
1
0
We note that:
a c b d
2
p
0
0 q
a ac
2
bc cd ab bc
bd
d
2
p
0
0 q
0
6
We have
a ab
2
bc bd
ac bc
cd d
2
p
0
(
( 1
0 ( 2 )
3 q ( 4 )
(1) – (4), we have
)
) a
a b (
2 a
bc d )
c ( bc a
d d
2
)
2 d
2 p
q p
0
0
( 1 ) q ( 4 )
(
(
(
2
3 a
)
)
If p
q
0 , then a
d
Therefore, from (2), (3)
b
0 .
c
0 . d )( a
d )
Thus, the solutions to the matrix equation Y
2
p
0 p
q .
0 q
are of diagonal matrices if p
q
0 .
And there will be at most 4 distinct solutions.
Note: Y
2
a
0
Now, let’s solve
0 b
Y
2 a
0
0 b
1
0
a
0
2 b
0
2
p
0
0 q
a b
2
2
p q
0
6
. According to the above analysis, we have the solutions as
1
0
0
6
,
1
0
0
6
,
0
1 0
6
,
1
0
0
6
It follows that
1
0
0
6
1
1
3
1
2
1
( A
2 I )
1
1
2
3
1
,
1
0
0
6
0
1 0
6
1
1
2
3
1
1
Then, the solutions to
( A
A
2
2
4
I )
1
1
2
3
1
A
4 I
,
3
3
0
1
2
4
0
6
are:
1
1
2
3
1
1
( A
2 I )
1
2
1
3
1
,
1
1
2
3
1
1
( A
2 I )
1
2
1
3
1
13
A
1
1
1
2
3
1
1
0
A
3
1
1
2
3
1
0
1
0
6
1
2
1
3
1
1
0
6
1
1
2
3
1
1
2 I , A
2
2 I , A
4
1
1
2
3
1
1
0
1
1
2
3
1
1
0
0
6
1
1
2
3
1
1
0
6
1
1
2
3
1
1
2 I ,
2 I
-End-
14
