AMOUD UNIVERSITY
FACULTY OF ENGINEERING
DEPARTMENT OF CIVIL ENGINEERING
Numerical Mathematics
LECTURE NOTES
CHAPTER FIVE
Eigenvalues and Eigenvectors
CLASS: Junior
FIRST SEMESTER
ACADEMIC YEAR: September, 2020 – July, 2021
LECTURER: Mr. Abdiqani M. Aden (Msc in Research & Statistics; Msc in Pure Mathematics)
Introduction
This is an important concept in mathematics because it has many applications in the areas of
physical sciences and engineering. This section is straightforward but it does rely on a number of
topics such as matrices, determinants, vectors etc. You need to thoroughly know how to evaluate
determinants to understand this chapter.
Definition of Eigenvalue and Eigenvector?
Before we define what is meant by an eigenvalue and eigenvector we do an example which
involves these.
Example (1)
 4 2 
 2
Let A  
 and u    then evaluate Au .
1 1 
1 
Solution
Multiplying the matrix and column vector we have
 4 2  2   6 
 2
Au  
      3  
 1 1 1   3 
1 
Note:
What do you notice about your result?
Well we have Au  3u . In geometric terms we have that the matrix A scalar multiplies the
vector u by 3:
y
3
2.5
2
 
 1
3
2
1.5
u=
1
 2
 
 1
0.5
1
2
3
4
5
6
x
Fig 1
In general terms this can be written as
Au  u
------------- Equation (1)
Where A is a square matrix, u is a column vector and  (lambda) is a scalar. Can you think of a
vector, u, which satisfies equation (1)?
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AU 2020
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The zero vector u  O . In this case we say we have the trivial solution. In this chapter we
consider the non-trivial solutions, u  O , and these solutions are powerful tools in linear algebra.
 For non-zero vector u the scalar  is called an eigenvalue of the matrix A and the vector u
is called an eigenvector belonging or corresponding to  which satisfies Au  u .
 In most linear algebra literature the Greek letter lambda,  , is used for eigenvalues. These
terms eigenvalue and eigenvector are derived from the German word ‘eigenwerte’ which
means ‘proper value.’ The word eigen is pronounced “i-gun”.
 Eigenvalues was initially developed in the field of differential equations by Jean D’
Alembert.
Example (2)
 1 1
Let A  
 . Verify the following:
 2 4 
 1
(a) u    is an eigenvector of the matrix A corresponding to the eigenvalue 1  2 .
 1
1 
(b) v    is an eigenvector of the matrix A corresponding to the eigenvalue 2  3 .
 2
Solution
(a) Multiplying the given matrix A and column vector u we have
 1 1 1  2 
 1
Au  
      2  
 2 4 1  2 
 1
 1
Thus u    is an eigenvector of the matrix A belonging to 1  2 because Au  2u .
 1
(b) Similarly we have
 1 1  1   3 
1 
Av  
      3  
 2 4  2   6 
 2
1 
Thus v    is an eigenvector of the matrix A belonging to 2  3 because Av  3v .
 2
Note:
What do you notice about your results?
A 2 by 2 matrix can have more than one eigenvalue and eigenvector.
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AU 2020
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 We can have eigenvalues and eigenvectors for any square matrices such as 3  3 , 4  4 , 5  5
, etc which satisfy Au  u .
Example (3)
 5 0 0
0


Let A   9 4 1 and u  1  . Determine an eigenvalue of this matrix A.
 6 2 1 
 2


 
Solution
Multiplying the matrix and column vector we have
 5 0 0 0 0
 0

   
 
Au   9 4 1 1    2   2  1 
 6 2 1   2   4 
 2

   
 
We have Au  u where   2 . Hence   2 is an eigenvalue of the matrix A with an
eigenvector u.
Characteristics Equation
From the above equation Au  u , we have Au = λI u by multiplying the identity keeps in the
same. We can rewrite as:
Au - λI u = 0
(A - λI )u = 0 -------- Equation (2)
We have none-zero vector u, If and only If
det(A - λI ) = 0
This is an important equation and is called Characteristic equation.
 We use this characteristic equation to find the eigenvalues and corresponding eigenvector,
and the procedure is as follows:
1. Solve the characteristic equation for the scla. Solve the characteristic equation for the scala  .
2. For the eigenvalue  determine the corresponding eigenvector u by solving the system
 A  I  u  O .
Note: Eigenvalues and Eigenvectors come in pairs, you can’t have one without the other.
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AU 2020
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Example (4)
 2 0
Determine the eigenvalues and corresponding eigenvectors of A  
.
 1 3
Solution
What do we find first, the eigenvalues or eigenvectors?
Eigenvalues because eigenvectors are born out of eigenvalues. Remember we carry out the
procedure outlined above:
Step 1. Solve the characteristic equation for the scalar  .
We need to find λ in det  A   I  . First we obtain A   I :
2
A  I  
1
2

1
0
1 0



3
0 1
0  0   2  
0 



3  0    1
3
Substituting this into det  A   I  gives
0 
2
det  A   I   det 
   2    3     0
3 
 1
For eigenvalues we equate this determinant to zero:
 2    3     0
1  2 or 2  3
Step 2. For the eigenvalue  determine the corresponding eigenvector u by solving the system
 A  I  u  O .
 2 0
Let u be the eigenvector corresponding to 1  2 . Substituting A  
 and 1    2 into
 1 3
 A  I  u  O
gives
 2 0   2 0  

 u  O
 1 3   0 2  
0 0

u  O
1 1
 A   I  u  
0
x
Remember O    and let u    and so we have
0
 y
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AU 2020
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 0 0  x   0 

    
 1 1  y   0 
Multiplying out gives
00  0
x y 0
Remember the eigenvector cannot be the zero vector, therefore at least one of the
values, x or y, must be non-zero. From the bottom equation we have x   y .
 2   3 
but we can have   ,   ,
 2   3 
Simplest solution is x  1, y  1
 
 ,
  
 5
  etc
 5 
We have an infinite number of solutions. We need to write down the general eigenvector u.
How?
Let x  s then y   s where s  0 and is a scalar. Thus the eigenvectors belonging to   2 are
 1

  is the simplest eigenvector 
 1 

 s
 1
u     s   where s  0
 s 
 1
Similarly we find the general eigenvector v corresponding to the other eigenvalue 2  3 . Putting
2    3 into  A   I  v  O gives
 2 0   3 0  
 1 0 

 v  O simplifies to 


v O
 1 0
 1 3   0 3  
 A   I  v  
x
0
By writing v    [different x and y from those above] and O    we obtain
 y
0
 1 0  x   0 

    
 1 0  y   0 
Multiplying out:
 x  0  0, x  0  0
From these equations we must have x  0 . What is y equal to?
We can choose y to be any real number apart from zero. Thus
y  s where s  0
The general eigenvector belonging to 2  3 is
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AU 2020
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 0 

  is the simplest eigenvector 
 1 

 x  0
0
v        s   where s  0
 y s
1 
Summarizing the above we have for the eigenvalue 1  2 the eigenvector
And for 2  3
the eigenvector
V=
u=
1
−1
0
1
Example (5)
Determine the Eigenvalues and corresponding Eigenvectors of matrix A.
1 0
4
A= 0 4
0
3 5 −3
Example (6)
Determine the Eigenvalues of matrix A.
1 −4 −2
A= 0
3
1
1
2
4
Reading Assignment!!!
Find out the practical applications of Eigenvalues and Eigenvectors in Engineering?
THE END
Numerical Methods Lecture Notes
AU 2020
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