The document discusses eigenvalues and eigenvectors, including their definitions and physical applications. It provides an example of using eigenvalues and eigenvectors to model a spring-mass system involving two masses connected by springs. The document derives the characteristic equation for finding the eigenvalues of the system's matrix and solves it to obtain the two eigenvalues. It then finds the corresponding eigenvectors. Finally, it solves for the eigenvalues and eigenvectors of another 4x4 matrix as an example.

1. Autar Kaw
Humberto Isaza
http://nm.MathForCollege.com
Transforming Numerical Methods Education for STEM Undergraduates

2. http://nm.MathForCollege.com

3. 1. Define eigenvalues and eigenvectors of a square matrix
2. Find eigenvalues and eigenvectors of a square matrix
3. Relate eigenvalues to the singularity of a square matrix, and
4. Use the power method to numerically find the largest eigenvalue in magnitude of
a square matrix and the corresponding eigenvector.

4. What does eigenvalue mean?
The word eigenvalue comes from the German word Eigenwert where Eigen means
characteristic and Wert means value. However, what the word means is not on
your mind! You want to know why I need to learn about eigenvalues and
eigenvectors. Once I give you an example of an application of eigenvalues and
eigenvectors, you will want to know how to find these eigenvalues and
eigenvectors.

5. Can you give me a physical example application of eigenvalues and
eigenvectors?
Look at the spring-mass system as shown in the picture below.
Assume each of the two mass-displacements to be denoted by 𝑥1 and 𝑥2, and
let us assume each spring has the same spring constant 𝑘.
x1 x2
m1
m2
k k

6. )
( 1
2
1
2
1
2
1 x
x
k
kx
dt
x
d
m 



Then by applying Newton’s 2nd and 3rd law of motion to develop a force-balance
for each mass we have
)
( 1
2
2
2
2
2 x
x
k
dt
x
d
m 


Rewriting the equations, we have
0
)
2
( 2
1
2
1
2
1 


 x
x
k
dt
x
d
m 0
)
( 2
1
2
2
2
2 

 x
x
k
dt
x
d
m
Let 15
,
20
,
10 2
1 

 k
m
m
0
)
2
(
15
10 2
1
2
1
2



 x
x
dt
x
d 0
)
(
15
20 2
1
2
2
2


 x
x
dt
x
d
and
and
which gives,
and

7. 
From vibration theory, the solutions can be of the form

 
0
sin 

 t
A
x i
i 
Where
i
A

0

= amplitude of the vibration of mass ,
= frequency of vibration,
= phase shift.
Then
)
0
(
2
2
2



 t
Sin
w
A
dt
x
d
i
i


8. 

Substituting and in equations,
i
x 2
2
dt
x
d i
0
)
2
(
15
10 2
1
2
1 



 A
A
A  and 0
)
(
15
20 2
1
2
2 


 A
A
A 
gives
0
15
)
30
10
( 2
1
2



 A
A
 0
)
15
20
(
15 2
2
1 



 A
A 
and
or
0
5
.
1
)
3
( 2
1
2



 A
A
 0
)
75
.
0
(
75
.
0 2
2
1 



 A
A 
and

9. Substituting and in equations,
i
x
i
x 2
2
dt
x
d i
0
)
2
(
15
10 2
1
2
1 



 A
A
A  and 0
)
(
15
20 2
1
2
2 


 A
A
A 
gives
0
15
)
30
10
( 2
1
2



 A
A
 0
)
15
20
(
15 2
2
1 



 A
A 
and
or
0
5
.
1
)
3
( 2
1
2



 A
A
 0
)
75
.
0
(
75
.
0 2
2
1 



 A
A 
and
In matrix form, these equations can be rewritten as

























0
0
75
.
0
75
.
0
5
.
1
3
2
1
2
2
A
A






























0
0
75
.
0
75
.
0
5
.
1
3
2
1
2
2
1
A
A
A
A


10. Let 
 
2









75
.
0
75
.
0
5
.
1
3
]
[A







2
1
]
[
A
A
X
0
]
[
]
[
]
[ 
 X
X
A 
]
[
]
[
]
[ X
X
A 

In the above equation, is the eigenvalue and is the eigenvector corresponding
to . As you can see, if we know for the above example we can calculate the
natural frequency of the vibration

]
[X


 


11. Why are the natural frequencies of vibration important? Because you do not want
to have a forcing force on the spring-mass system close to this frequency as it
would make the amplitude very large and make the system unstable.
i
A

12. If is a matrix, then is an eigenvector of if
]
[A n
n 0
]
[


X
]
[
]
[
]
[ X
X
A 

]
[A
where is a scalar and . The scalar is called the eigenvalue of
and is called the eigenvector corresponding to the eigenvalue .


0
]
[ 
X  ]
[A
]
[A
What is the general definition of eigenvalues and eigenvectors of a square
matrix?

13.  ]
[A
To find the eigenvalues of a n n matrix , we have
]
[
]
[
]
[ X
X
A 

0
]
[
]
[
]
[ 
 X
X
A 
0
]
][
[
]
[
]
[ 
 X
I
X
A 
0
]
])[
][
[
]
([ 
 X
I
A 
Now for the above set of equations to have a nonzero solution,
0
])
[
]
det([ 
 I
A 

14. This left hand side can be expanded to give a polynomial in solving the above
equation would give us values of the eigenvalues. The above equation is called
the characteristic equation of .

]
[A
For a matrix, the characteristic polynomial of is of degree as
follows
n
n
]
[A A n
0
])
[
]
det([ 
 I
A 
giving
0
2
2
1
1 





 

n
n
n
n
c
c
c 


Hence. this polynomial has n roots

15. 

Find the eigenvalues of the physical problem discussed in the beginning of this
chapter, that is, find the eigenvalues of the matrix









75
.
0
75
.
0
5
.
1
3
]
[A
Solution















75
.
0
75
.
0
5
.
1
3
]
[
]
[ I
A
    0
)
5
.
1
)(
75
.
0
(
)
75
.
0
)(
3
(
)
det( 






 

 I
A
0
125
.
1
3
75
.
0
25
.
2 2




 



16. 0
125
.
1
75
.
3
2


 

)
1
(
2
)
125
.
1
)(
1
(
4
)
75
.
3
(
)
75
.
3
( 2







2
092
.
3
75
.
3 

3288
.
0
,
421
.
3

So the eigenvalues are 3.421 and 0.3288.

17. Find the eigenvectors of









75
.
0
75
.
0
5
.
1
3
A
Solution
3288
.
0
,
421
.
3 2
1 
 

Let







2
1
]
[
x
x
X
be the eigenvector corresponding to
421
.
3
1 

The eigenvalues have already been found in Example 1 as

18. Hence
0
]
])[
[
]
([ 1 
 X
I
A 
0
1
0
0
1
421
.
3
75
.
0
75
.
0
5
.
1
3
2
1




























x
x























0
0
671
.
2
75
.
0
5
.
1
421
.
0
2
1
x
x
If
s
x 
1
then
s
x
x
s
2808
.
0
0
5
.
1
421
.
0
2
2






19. The eigenvector corresponding to then is,
421
.
3
1 









s
s
X
2808
.
0
]
[








2808
.
0
1
s
The eigenvector corresponding to is
421
.
3
1 







 2808
.
0
1
Similarly, the eigenvector corresponding to is
3288
.
0
2 







781
.
1
1

20. Find the eigenvalues and eigenvectors of














0
0
5
.
0
5
.
0
5
.
0
5
.
0
1
0
5
.
1
]
[A
Solution
The characteristic equation is given by
0
])
[
]
det([ 
 I
A 
0
0
5
.
0
5
.
0
5
.
0
5
.
0
1
0
5
.
1
det 



















0
)]
5
.
0
)(
5
.
0
(
)
0
)(
5
.
0
)[(
1
(
)]
0
)(
5
.
0
(
)
)(
5
.
0
)[(
5
.
1
( 









 




21. 0
25
.
1
2 2
3



 


The roots of the above equation are
0
.
1
,
5
.
0
,
5
.
0


Note that there are eigenvalues that are repeated. Since there are only two distinct
eigenvalues, there are only two eigenspaces. But, corresponding to  = 0.5 there
should be two eigenvectors that form a basis for the eigenspace.
To find the eigenspaces, let











3
2
1
]
[
x
x
x
X

22. Given
0
]
[
)]
[( 
 X
I
A 
then





































0
0
0
0
5
.
0
5
.
0
5
.
0
5
.
0
1
0
5
.
1
3
2
1
x
x
x



For  = 0.5 ,



































0
0
0
5
.
0
0
5
.
0
5
.
0
0
5
.
0
1
0
1
3
2
1
x
x
x
Solving this system gives
a
x
b
x
a
x 


 3
2
1 ,

23. So





















a
b
a
x
x
x
3
2
1























0
0
0 b
a
a























0
1
0
1
0
1
b
a
So the vectors and form a basis for the eigenspace for the eigenvalue










1
0
1










0
1
0
5
.
0



24. 1


For




































0
0
0
1
0
5
.
0
5
.
0
5
.
0
5
.
0
1
0
5
.
0
3
2
1
x
x
x
Solving this system gives
a
x
a
x
a
x 5
.
0
,
5
.
0
, 3
2
1 




The eigenvector corresponding to is
1



























5
.
0
5
.
0
1
5
.
0
5
.
0 a
a
a
a
Hence the vector is a basis for the eigenspace for the eigenvalue of












5
.
0
5
.
0
1
1



25. Theorem 1: If is a triangular matrix – upper triangular, lower
triangular or diagonal, the eigenvalues of are the diagonal entries of .
Theorem 2: is an eigenvalue of if is a singular (noninvertible)
matrix.
Theorem 3: and have the same eigenvalues.
Theorem 4: Eigenvalues of a symmetric matrix are real.
Theorem 5: Eigenvectors of a symmetric matrix are orthogonal, but only for
distinct eigenvalues.
Theorem 6: is the product of the absolute values of the eigenvalues of
]
[A n
n
]
[A
]
[A ]
[A
0

 ]
[A
]
[A
T
]
[A
)
det(A ]
[A

26. What are the eigenvalues of














2
.
7
0
6
2
0
5
.
7
5
9
0
0
3
7
0
0
0
6
]
[A
Solution
Since the matrix is a lower triangular matrix, the eigenvalues of are
the diagonal elements of . The eigenvalues are
]
[A
]
[A
]
[A
2
.
7
,
5
.
7
,
3
,
6 4
3
2
1 



 




27. One of the eigenvalues of












7
1
2
9
5
3
2
6
5
]
[A
]
[A
is zero. Is invertible?
Solution
0

 ]
[A ]
[A
is an eigenvalue of , that implies is singular and is not
invertible.

28. Given the eigenvalues of









 

5
.
8
1
8
2
5
5
.
3
6
5
.
3
2
]
[A
are
711
.
4
,
33
.
12
,
547
.
1 3
2
1 


 


What are the eigenvalues of if
]
[B












5
.
8
2
6
1
5
5
.
3
8
5
.
3
2
]
[B

29. Solution
T
A
B ]
[
]
[ 
Since , the eigenvalues of and are the same. Hence eigenvalues
of also are
are the same. Hence eigenvalues of
also are
]
[A
]
[B
]
[B
711
.
4
,
33
.
12
,
547
.
1 3
2
1 


 



30. Given the eigenvalues of









 

5
.
8
1
8
2
5
5
.
3
6
5
.
3
2
]
[A
are
711
.
4
,
33
.
12
,
547
.
1 3
2
1 


 


Calculate the magnitude of the determinant of the matrix.

31. Solution
Since
3
2
1
]
det[ 



A
711
.
4
33
.
12
547
.
1


88
.
89


32. How does one find eigenvalues and eigenvectors numerically?
One of the most common methods used for finding eigenvalues and
eigenvectors is the power method. It is used to find the largest eigenvalue in an
absolute sense. Note that if this largest eigenvalues is repeated, this method will
not work. Also this eigenvalue needs to be distinct. The method is as follows:

33. ]
[ )
0
(
X
1. Assume a guess for the eigenvector in equation.
one of the entries of needs to be unity.
2. Find
3. Scale so that the chosen unity component remains unity
4. Repeat steps (2) and (3) with
to get
5. Repeat the steps 2 and 3 until the value of the eigenvalue converges.
]
[
]
[
]
[ X
X
A 

]
[ )
0
(
X
]
[
]
[
]
[ )
0
(
)
1
(
X
A
Y 
]
[ )
1
(
Y
]
[
]
[ )
1
(
)
1
(
)
1
(
X
Y 

]
[
]
[ )
1
(
X
X  ]
[ )
2
(
X

34. s
E
If is the pre-specified percentage relative error tolerance to which you would like
the answer to converge to, keep iterating until
s
i
i
i
E





100
)
1
(
)
(
)
1
(



where the left hand side of the above inequality is the definition of absolute
percentage relative approximate error, denoted generally by A pre-specified
percentage relative tolerance of Implies at least significant digits are
current in your answer. When the system converges, the value of is the largest (in
absolute value) eigenvalue of .
s
E
m

 2
10
5
.
0 m

]
[A

35. Using the power method, find the largest eigenvalue and the corresponding
eigenvector of














0
0
5
.
0
5
.
0
5
.
0
5
.
0
1
0
5
.
1
]
[A

36. Solution
Assume











1
1
1
]
[ )
0
(
X
























1
1
1
0
0
5
.
0
5
.
0
5
.
0
5
.
0
1
0
5
.
1
]
[
]
[ )
0
(
X
A













5
.
0
5
.
0
5
.
2













2
.
0
2
.
0
1
5
.
2
)
1
(
Y
5
.
2
)
1
(



37. ]
[ )
0
(
X
We will choose the first element of to be unity.













2
.
0
2
.
0
1
]
[ )
1
(
X


























2
.
0
2
.
0
1
0
0
5
.
0
5
.
0
5
.
0
5
.
0
1
0
5
.
1
]
[
]
[ )
1
(
X
A













5
.
0
5
.
0
3
.
1













3846
.
0
3846
.
0
1
3
.
1
]
[ )
2
(
X
3
.
1
)
2
(



38. 












3846
.
0
3846
.
0
1
]
[ )
2
(
X
The absolute relative approximate error in the eigenvalues is
100
)
2
(
)
1
(
)
2
(






a
100
5
.
1
5
.
1
3
.
1



%
307
.
92


39. )
(i

Conducting further iterations, the values of and the corresponding eigenvectors is
given in the table below

40. 1


The exact value of the eigenvalue is
and the corresponding eigenvector is













5
.
0
5
.
0
1
]
[X

41. Eigenvalue
Eigenvectors
Power method