Anale. Seria Informatică. Vol. XIX fasc.

1 – 2021
Annals. Computer Science Series. 19th Tome 1st Fasc. – 2021

SOME APPLICATION OF EIGENVALUE AND EIGENVECTOR PROBLEMS

Zahidullah Rehan

Sayed Jamaluddin Afghani University, Education Faculty,

Mathematics and physics Department, Kunar, Afghanistan

Corresponding author: Zahidullah Rehan rehanzahidullah@gmail.com

ABSTRACT: One of the most useful brand of mathematics is linear algebra, which have more application in science and
engineering, because it required the description of some measurable quantities. In this research paper we determine some
application of Eigen-value problems. So in this research work first we discuss how to obtain the Eigen-value and Eigen-
vector of square matrix and their characteristic equation and polynomial and then apply the solution of Eigen-value problem
to stretching of elastic membrane problems, eigenvalue problems arising from population model and vibrating system of two
masses on two springs problems.
KEYWORDS: Matrix, Eigenvalue, Eigenvector

1. INTRODUCTION finite element. The regular reducible matrix does not

have a unique Eigen-mode will all element are finite.
From the view point of engineering application [2]
eigenvalue and eigenvector problem are more useful According to our information and advance
in the connection with matrices. Linear equation technology the complexity of the social
𝐴𝑥 = 𝑏 come from steady state problem. [8] communication and biological networks surrounding
Eigenvalue have their great importance in dynamic us is increasing rapidly. Many important
𝑑𝑥
problems the solution of 𝑑𝑡 = 𝐴𝑥 is changing with combinatorial parameters of large networks are often
time growing or decaying or oscillating we cannot hard to calculate are approximate. Eigen-values
find if by elimination. Here we discuss linear algebra provide an effective and efficient tools for studying
properties of large graphs which arise in practice. [4]
based on 𝐴𝑥 = 𝜆𝑥. all the matrices in this paper or
square matrices. [1]
2. HOW TO FIND THE EIGEN-VALUE
Let A be 𝑛 × 𝑛 matrix then consider the following
AND EIGEN-VECTOR OF A MATRIX
vector equation. [4]
The problem of deterring the eigenvalue and
𝐴𝑥 = 𝜆𝑥 (1)
eigenvector of a matrix is called an Eigen-value
In the above equation 𝑥 and 𝜆 are uknown where 𝑥 is problem. To find the Eigen-value and Eigen-vector of
vector and 𝜆 is scalar, we must determine the value of a matrix first we write equation (1) in the component.
these unknown which satisfier equation (1). [6]
We know that 𝑥 = 0 is the solution of equation (1) for 𝑎11 𝑥1 + ⋯ ⋯ ⋯ + 𝑎1𝑛 = 𝜆𝑥1
the value of 𝜆, because 𝐴0 = 0. This is of no interest. 𝑎21 𝑥1 + ⋯ ⋯ ⋯ + 𝑎2𝑛 = 𝜆𝑥2
The solution of equation (1) for the value of 𝜆 is } (2)
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
known as an Eigen-value (characteristic value) of the 𝑎𝑛1 𝑥1 + ⋯ ⋯ ⋯ + 𝑎𝑛𝑛 = 𝜆𝑥𝑛
matrix A and the solution of equation (1) are known
as the Eigen-vector (characteristic vector) of A. Transform the term on the right side to the left side,
Corresponding to that Eigen value of A is called the we have
Spectrum of A. [6] (𝑎11 − 𝜆)𝑥1 + 𝑎12 𝑥2 + ⋯ ⋯ ⋯ + 𝑎1𝑛 = 0
On the base of eigenvalue and eigenvector and 𝑎21 𝑥1 + (𝑎22 − 𝜆)𝑥2 + ⋯ ⋯ ⋯ + 𝑎2𝑛 = 0
eigenmode characterization of reducible matrix does } (3)
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
not have necessary eigenvector. If the reducible 𝑎𝑛1 𝑥1 + 𝑎𝑛2 𝑥2 + ⋯ ⋯ ⋯ + (𝑎𝑛𝑛 − 𝜆) = 𝜆𝑥𝑛
matrix has Eigen-value the Eigen-value is not
necessarily unique and has a finite value. Eigen- In matrix notation equation (3) can be written as
vector of the corresponding to the Eigen-value of the
(𝐴 − 𝜆𝐼)𝑥 = 0 (4)
reducible matrix is not unique and contain at least a

120
 Anale. Seria Informatică. Vol. XIX fasc. 1 – 2021
Annals. Computer Science Series. 19th Tome 1st Fasc. – 2021

By Cramer theorem the above homogenous linear So that the required eigenvalues are 𝜆 = 8 , 𝜆 = −2.
system has a non-trivial solution if and only if the Now putting the value of 𝜆 in the matrix
determinant of the coefficient is zero. [
7−𝜆 3
] we obtain.
3 −1−𝜆
det(𝐴 − 𝜆𝐼) =
𝑎11 − 𝜆 𝑎12 ⋯ ⋯ ⋯ 𝑎1𝑛 7−8 3 −1 3
[ ]=[ ]
𝑎 𝑎22 − 𝜆 ⋯ ⋯ ⋯ 𝑎2𝑛 3 −1−8 3 9
| 21 |=0 (5)
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
Now we solve the equation 𝐵𝑥̅ = 0̅,
𝑎𝑛1 𝑎𝑛2 ⋯ ⋯ ⋯ 𝑎𝑛𝑛 − 𝜆
−1 3 𝑥1 0
Here (𝐴 − 𝜆𝐼) is known as characteristic matrix and [ ][ ] = [ ]
3 − 9 𝑥2 0
𝑝(𝜆) is the characteristic determinant of 𝐴. Equation
(5) is known as the characteristic equation of 𝐴, by −1 3 0 −1 3 0
[ | ]→[ | ]
developing 𝑝(𝜆) we obtain a polynomial of nth 3 −9 0 0 0 0
degree in 𝜆. this is known as the characteristic −𝑥1 + 3𝑥2 = 0 ⇒ 3𝑥2 = 𝑥1
polynomial of A. 𝑥1 = 1 , 𝑥2 = 3

Theorem. 3
Hence the eigenvector is [ ]. And eigenvalue is 8.
If 𝑤 and 𝑥 are eigenvector of a matrix A 1
corresponding to the same eigenvalue 𝜆, os are 𝑤 + 𝑥 Now we obtain the eigenvector for 𝜆 = −2. Now
(provided 𝑥 ≠ −𝑤) and 𝑘𝑥 for any 𝑘 ≠ 0. 7−𝜆 3
putting the value of 𝜆 in the matrix [ ] we
Hence the Eigen-vectors corresponding to one and the 3 −1−𝜆
obtain.
same Eigen-value 𝜆 of 𝐴, together with 0, form a 7+2 3 9 3
victor space called the eigenvalue space of 𝐴 [ ]=[ ]
3 −1+2 3 1
corresponding to that 𝜆. [7]
Now we solve the equation 𝐵𝑥̅ = 0̅,
Proof.
𝐴𝑤 = 𝜆𝑤 And 𝐴𝑥 = 𝜆𝑥 imply 9 3 𝑥1 0
[ ][ ] = [ ]
𝐴(𝑤 + 𝑥) = 𝐴𝑤 + 𝐴𝑥 = 𝜆𝑤 + 𝜆𝑥 = 𝜆(𝑤 + 𝑥) 3 1 𝑥2 0
And 9 3 0 0 0 0
𝐴(𝑘𝑤+= 𝑘((𝐴𝑤) = 𝑘(𝜆𝑤) = 𝜆(𝑘𝑤) [ | ]→[ | ]
3 1 0 3 1 0
Hence
𝐴(𝑘𝑤 + 𝑙𝑥) = 𝜆(𝑘𝑤 + 𝑙𝑥). 3𝑥1 + 𝑥2 = 0 ⇒ 𝑥2 = −3𝑥1
𝑥1 = 1 , 𝑥2 = −3
In particular, an Eigen-vector x is determined only up
to a constant factor. Hence we can normalize𝑥, that 1
Hence the eigenvector is [ ]. And eigenvalue is −2.
is, multiply it by a scalar to get a unite vector. The −3
following problems will illustrate that a 𝑛 × 𝑛 matrix
Problem: determine the eigenvalue and eigenvector
may have 𝑛 linearly independent eigenvector or it 8 −8 −2
may have fewer than 𝑛. of 𝐴 = (4 −3 − 2). [2]
3 −4 1
Problem: determine the eigenvalue and eigenvector Solution: the characteristic polynomial is
7 3
of 𝐴 = ( ). [9]
3 − 1 8−𝜆 −8 −2
det(𝐴 − 𝜆𝐼) = | 4 − 3 − 𝜆 − 2|
Solution: the characteristic polynomial is 3 −4 1−𝜆
7−𝜆 3 𝜆3 − [𝑠𝑢𝑚 𝑜𝑓 𝑑𝑖𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑛𝑡𝑠]𝜆2
det(𝐴 − 𝜆𝐼) = | |
3 −1−𝜆 + [𝑠𝑢𝑚 𝑑𝑖𝑔𝑜𝑛𝑎𝑙 𝑚𝑖𝑛𝑜𝑟𝑠] − |𝐴|
= (7 − 𝜆)(−1 − 𝜆) − (3)(3)
=0
= 𝜆2 − 6𝜆 − 16
𝜆3 − [8 − 3 + 1]𝜆2 + [−11 + 14 + 8] − 6 = 0
𝜆3 − 6𝜆2 + 11 − 6 = 0
So the characteristic equation is 𝜆2 − 2𝜆 + 2 = 0
from this equation we obtain the velue of 𝜆 as
So the characteristic equation is 𝜆2 − 2𝜆 + 2 = 0
from this equation we obtain the velue of 𝜆 as
𝜆2 − 6𝜆 − 16 = (𝜆 − 8)(𝜆 + 2) = 0
𝜆 = 1, 2, 3
𝜆 = 8 , 𝜆 = −2
121
 Anale. Seria Informatică. Vol. XIX fasc. 1 – 2021
Annals. Computer Science Series. 19th Tome 1st Fasc. – 2021

So that the required eigenvalues are 𝜆 = 1, 2,3. Now The characteristic equation is
putting the value of 𝜆 in the matrix 5−𝜆 3
| | = (5 − 𝜆)2 − 9 = 0
8−𝜆 −8 −2 3 5−𝜆
[ 4 − 3 − 𝜆 − 2 ] we obtain 𝜆 = 8, 2
3 −4 1−𝜆
For 𝜆 = 8 system (7) becomes
8−1 −8 −2 7 −8 −2 −3𝑥1 + 3𝑥2 = 0
|
[ 4 − 3 − 1 − 2 ] = [4 −4 − 2] 3𝑥1 + 3𝑥2 = 0
3 −4 1−1 3 −4 0 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥2 = 𝑥1 , 𝑥1 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦. 𝑓𝑜𝑟 𝑖𝑛𝑠𝑡𝑎𝑛𝑐𝑒, 𝑥1
= 𝑥2 = 1
Now we solve the equation 𝐵𝑥̅ = 0̅,
7 −8 − 2 𝑥1 0 For 𝜆 = 2 system (7) becomes
[4 − 4 − 2] [𝑥2 ] = [0] 3𝑥1 + 3𝑥2 = 0
|
3 −4 0 𝑥3 0 3𝑥1 − 3𝑥2 = 0
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥2
Now we solve the above system by Crammer’s rule = −𝑥1 , 𝑥1 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦. 𝑓𝑜𝑟 𝑖𝑛𝑠𝑡𝑎𝑛𝑐𝑒, 𝑥1 = 1, 𝑥2
= −1
7𝑥1 − 8𝑥2 − 2𝑥3 = 0
} We thus obtain as eigenvector of 𝐴, for instance,
4𝑥1 − 4𝑥2 − 2𝑥3 = 0
𝑥1 𝑥2 𝑥3 [1 1]𝑇 corresponding to 𝜆1 and
= =
−8 −2 7 −2 7 −8 [1 − 1]𝑇 corresponding to 𝜆2 .
| | | | | |
−4 −2 4 −2
𝑥1 𝑥2 𝑥3
4 −4 One of the vectors make 45𝑜 degree angle and the
= = other vector make 135𝑜 degree angle with the
8 6 4 positive direction of 𝑥1 . So we say that the
𝑥1 8 4
𝑥 = [𝑥2 ] = [6] = [3] eigenvalues in this problem shows that the principal
𝑥3 4 2 direction of the given membrane is expand by factors
8 and 2 respectively.
4 Now let us consider the principal direction as
Hence the eigenvector is [3]. And eigenvalue is 1. direction of a new Cartesian coordinate system, in
2 which positive 𝑢1 -semi-axis in the 1st quadrant and
Similarly we can find the eigenvectors for 𝜆 = 2,3. the positive 𝑢2 - semi-axis in the 2nd quadrant of the
old system, let suppose that 𝑢1 = 𝑟 cos 𝜃 , 𝑢2 =
3. APPLICATION OF EIGENVALUE AND 𝑟 sin 𝜑, then the boundary before the expansion of the
EIGENVECTORS circular membrane has coordinates cos 𝜑 , sin 𝜑 and
after the expansion the coordinate is
Eigenvalues problems have greatest application in
daily life. Some of the application which belonge to 𝑧1 = 8 cos 𝜑 𝑧2 = 2 sin 𝜑
engineering, pahysics and mathematice we stydied
here. Since cos2 𝜑 + sin2 𝜑 = 1, which show that the
shape of an elastic membrane is ellipse.
3.1. Expansion of Elastic Membrane
Let us consider an elastic membrane in with boundary 𝑧12 𝑧22
+ =1
circle 𝑥12 + 𝑥22 = 1 is expand from the point 𝑃(𝑥1 , 𝑥2 ) 64 4
to 𝑄(𝑦1 , 𝑦2 ) which given by
𝑦1 5 3 𝑥1
𝑦 = [𝑦 ] = 𝐴𝑥 = [ ][ ] (6)
2 3 5 𝑥2

Obtain the principal direction of the position vector 𝑥

of 𝑃 for which the f the position vector 𝑦 of 𝑄 have
the similar or exactly opposite direction, find the
shape of the circle after the instability? [6]
Solution. We are looking for a vector 𝑥 such that 𝑦 =
𝜆𝑥. Since 𝑦 = 𝐴𝑥, this gives 𝐴𝑥 = 𝜆𝑥, the equation
of an eigenvalue problem. In component, 𝐴𝑥 = 𝜆𝑥 is

(5 − 𝜆)𝑥1 + 3𝑥2 = 0 Figure 1. Undeformed and deformed membrane

} (7)
3𝑥1 + (5 − 𝜆)𝑥2 = 0 in Example 1
122
 Anale. Seria Informatică. Vol. XIX fasc. 1 – 2021
Annals. Computer Science Series. 19th Tome 1st Fasc. – 2021

3.2. Eigenvalue Problems Arising from from −1.2𝑥1 + 2.3𝑥2 + 0.4𝑥3 = 0. For the
Population Model 1200
population we multiply 𝑥 by 1+0.5+0.125 = 738.
The Leslie model describes age-specified population
Proportional growth of the numbers of females in the
growth, as follows. Let age attained by the females in
three classes will occur if the initial values are
some animal population be 9 years. Divide the
738, 369, 92 in classes 1, 2, 3 respectively. The
population into three age classes of 3 years each. Let
growth rate will be 1.2 per 3 years.
the “Leslie matrix” be
0 2.3 0.4 3.3. Vibrating System of Two Masses on Two
𝑙 = [𝑙𝑗𝑘 ] = [0.6 0 0] (8) Springs
0 0.3 0 Mas-spring system involving several masses and
spring can be treated as eigenvalue problems. For
Where 𝑙1𝑘 is the average numbers of daughter borne instance, the mechanical system in Fig. 2 is governed
to a single female during the time she is in age class 𝑘, by the system of ODEs
and 𝑙𝑗,𝑗−1 (𝑗 = 2,3) is the fraction of females in the
age class 𝑗 − 1 that will survive and pass into class 𝑗? 𝑦1′′ = −5𝑦1 + 2𝑦2
} (9)
a) Find the number of females in each class after 𝑦2′′ = 2𝑦1 − 2𝑦2
3,6,9 years if each class initially consists of 400 Where 𝑦1 and 𝑦2 are the displacements of the masses
females? from rest, as shown in the figure, and prime denote
b) In which initial distribution the number of derivatives with respect to time. In vector form, this
females in each class change by the same proportion becomes [6]
and also find the rate of change? [3]
𝑦1′′ −5 2 𝑦1
𝑦 ′′ = [ ] = 𝐴𝑦 = [ ][ ] (10)
Solution: 𝑦2′′ 2 −2 𝑦2
(a) initially 𝑥0𝑇 = [400 400 400], after three
years,
0 2.3 0.4 400 1080
𝑥(3) = 𝐿𝑥(0) = [0.6 0 0 ] [400] = [ 240 ]
0 0.3 0 400 120

On the same way the number of females after six

𝑇 𝑇
years is 𝑥(6) = (𝐿𝑥(3) ) = [600 648 72], and
after nine years this value equal to
𝑇 𝑇
𝑥(9) = (𝐿𝑥(6) ) = [1519.2 360 195.4]

(b) proportional change means that we are looking for

a distribution vector 𝑥 such that 𝐿𝑥 = 𝜆𝑥, where 𝜆 is
the rate of change Fig. 2. Masses on spring in example 4
(growth if 𝜆 > 1, decrease if 𝜆 < 1).
The characteristic equation is We try a vector solution of the form
det(𝐿 − 𝜆𝐼) = −𝜆3 = 0.6(−2.3𝜆 − 0.3 ∙ 0.4)
𝑦 = 𝑒 𝜔𝑡 (11)
= −𝜆3 + 1.38𝜆 + 0.072 = 0
A positive root is find to be (for instance, by This is suggested by a mechanical system of a single
Newton’s method) 𝜆 = 1.2. A corresponding mass on a spring. Whose motion is given by
eigenvector 𝑥 can be determined from the exponential function (and sines and cosines).
characteristic matrix. Substitution into (7) gives

−1.2 2.3 0.4 𝜔2 𝑥𝑒 𝜔𝑡 = 𝐴𝑥𝑒 𝑤𝑡

𝐴 − 1.2𝐼 = [ 0.6 −1.2 0 ], Dividing by 𝑒 𝑤𝑡 and writing 𝜔2 = 𝜆, we see that our
0 0.3 −1.2 mechanical system leads to the eigenvalue problem
1 𝐴𝑥 = 𝜆𝑥 (12)
𝑠𝑎𝑦, 𝑥 = [ 0.5 ]
2
0.125 𝑤 here 𝜆 = 𝜔
Here the eigenvalues are 𝜆 = −1, −6
Where 𝑥3 = 0.125 is selected, then we obtain 𝑥2 =
0.5 from 0.3𝑥2 − 1.2𝑥3 = 0, and 𝑥1 = 1 consequently 𝜔 = ±𝐼 𝑎𝑛𝑑 ± 𝑖√6.

123
 Anale. Seria Informatică. Vol. XIX fasc. 1 – 2021
Annals. Computer Science Series. 19th Tome 1st Fasc. – 2021

Corresponding eigenvectors are

Conflicts of interest
1 2 The author declare no conflict of interest
𝑥 = [ ] ,[ ] (13)
2 −1 regarding the publication of this paper.
From equation (12) we obtain the following REFERENCE
complex solution.
[1]. Abreu Nair, Cardoso Domingos, França
𝑥1 𝑒 ±𝑖𝑡 = 𝑥1 (𝑐𝑜𝑠 𝑡 ± 𝑠𝑖𝑛 𝑡) Francisca, Vinagre Cybele. (2016). On main
eigenvalues of certain graphs. 10.13140/
𝑥2 𝑒 ±𝑖√6𝑡 = 𝑥2 (𝑐𝑜𝑠 √6𝑡 ± 𝑠𝑖𝑛 √6𝑡) RG.2.1.2863.1926.
[2]. Himmatul Mursyidah, Subiono Eigenvalue,
By addition and subtraction we get the following four Eigenvector, eigenmode of reducible matrix and
real solutions. its application, AIP conference proceeding
1867, 020044 (2017), https://doi.org/10.1063/
𝑥1 cos 𝑡 , 𝑥1 𝑠𝑖𝑛𝑡 , 𝑥2 cos √6𝑡 , 𝑥2 sin √6𝑡
1.4994447
[3]. Brouwer A, Haemers W. (2005) Eigenvalues
A general solution is obtained by taking a linear
and perfect matchings. Lin Algebra Appl
combination of these.
395:155–162.
𝑦 = 𝑥1 (𝑎1 𝑐𝑜𝑠 𝑡 + 𝑏1 𝑠𝑖𝑛 𝑡) + 𝑥2 (𝑎2 𝑐𝑜𝑠 √6𝑡 [4]. Cioaba Sebastian. (2011). Some Applications
+ 𝑏2 𝑠𝑖𝑛 √6𝑡) of Eigenvalues of Graphs. Structural Analysis of
Complex Networks. 10.1007/978-0-8176-4789-
With arbitrary constants 𝑎1 , 𝑏1 , 𝑎2 , 𝑏2 (to which value 6_14.
can be assigned by prescribing initial displacement [5]. Fernandes Rafael, Judice Joaquim, Trevisan
and initial velocity of each of the two mass), the Vilmar. (2017). Complementary eigenvalues of
component of 𝑦 are graphs. Linear Algebra and its Applications.
𝑦1 = 𝑎1 𝑐𝑜𝑠 𝑡 + 𝑏1 𝑠𝑖𝑛 𝑡 + 2𝑎2 𝑐𝑜𝑠 √6𝑡 527. 10.1016/j.laa. 2017.03.029.
[6]. Keryszig, Erwin. 2006. Advance engineering
+ 2𝑏2 𝑠𝑖𝑛 √6𝑡 mathematics. 9th edition. Willy India.
𝑦 = 2𝑎1 cos 𝑡 + 2𝑏1 sin 𝑡 − 𝑎2 cos √6𝑡 − 𝑏2 sin √6𝑡 [7]. Li Jianxi, Chang An. (2008). Some applications
These function describe harmonic oscillation of the on the method of eigenvalue interlacing for
two masses. Physically, this hold to be expected graphs [J]. Journal of Mathematical Research &
because we have neglected damping. Exposition. 28. 10.3770/j.issn:1000-
341X.2008.02.002.
4. CONCLUSIONS [8]. Prince Tanvir, Angulo Nieves. (2014).
Application of Eigenvalues and Eigenvectors
Eigenvalue problem are of the greatest interest to the and Diagonalization to Environmental Science.
real life. Such that we have seen in that the application Applied Ecology and Environmental Sciences.
of the problem are useful in mathematics and physics 2. 106-109. 10.12691/aees-2-4-4.
for instance in analyzing problems involving [9]. Ramane Harish Chandra, Revankar Deepak,
Stretching of Elastic Membrane problems, Patil J.B. (2013). Bounds for the degree sum
Eigenvalue Problems Arising from Population Model eigenvalues and degree sum energy of a graph.
problems and Vibrating System on Two Masses of International Journal of Pure and Applied
Two Springs problems. This procedure also used for Mathematical Sciences. 6. 161-167
calculating some other physical and biological
problem like arising of Markov processes problem.
Beside the application of the eigenvalue and
eigenvector problem we find the eigenvalue and
eigenvector of the given square matrix.

124