In this presentation we had discussed how to determine eigenvalues and eigenvectors with example and MATLAB Simulink

1. SEMINAR ON EIGENVALUESSEMINAR ON EIGENVALUES
AND EIGENVECTORSAND EIGENVECTORS
By
Vinod Srivastava
M.E. Modular I & C
Roll No.151522

2. CONTENTSCONTENTS
 Introduction to Eigenvalues and Eigenvectors
 Examples
 Two-dimensional matrix
 Three-dimensional matrix
• Example using MATLAB
• References

3. INTRODUCTIONINTRODUCTION
Eigen Vector-
 In linear algebra , an eigenvector or characteristic vector of a square
matrix is a vector that does not changes its direction under the
associated linear transformation.
 In other words – If V is a vector that is not zero, than it is an
eigenvector of a square matrix A if Av is a scalar multiple of v.
This condition should be written as the equation:
AV= λv

4. Contd….Contd….
Eigen Value-
• In above equation λ is a scalar known as the eigenvalue or
characteristic value associated with eigenvector v.
• We can find the eigenvalues by determining the roots of the
characteristic equation-
0=− IA λ

5. ExamplesExamples
Two-dimensional matrix example-
Ex.1 Find the eigenvalues and eigenvectors of matrix A.
Taking the determinant to find characteristic polynomial A-
It has roots at λ = 1 and λ = 3, which are the two eigenvalues of A.






=
21
12
A
⇒=− 0IA λ 0
21
12
=
−
−
λ
λ
043 2
=+−⇒ λλ

6. Eigenvectors v of this transformation satisfy the equation,
Av= λv
Rearrange this equation to obtain-
For λ = 1, Equation becomes,
which has the solution,
( ) 0=− vIA λ
( ) 0=− vIA






=











0
0
11
11
2
1
v
v






−
=
1
1
v

7. For λ = 3, Equation becomes,
which has the solution-
Thus, the vectors vλ=1 and vλ=3 are eigenvectors of A associated with the
eigenvalues λ = 1 and λ = 3, respectively.
( ) 03 =− uIA






=











−
−
0
0
11
11
2
1
u
u






=
1
1
u

8. Three-dimensional matrix example-
Ex.2 Find the eigenvalue and eigenvector of matrix A.
the matrix has the characteristics equation-










−
−
−
=
200
130
014
A
( )( )( ) 0234
200
130
014
=+++=
+
−+
−+
=−
λλλ
λ
λ
λ
λ AI

9. therefore the eigen values of A are-
For λ = -2, Equation becomes,
which has the solution-
4,3,2 321 −=−=−= λλλ
( )










=




















−
−
=−
0
0
0
000
110
012
0
3
2
1
11
v
v
v
vAIλ










=
2
2
1
v

10. Similarly for λ = -3 and λ = -4 the corresponding eigenvectors u and x are-










=










=
0
0
2
,
0
1
1
xu

11. Example using MATLABExample using MATLAB

12. REFERENCESREFERENCES
 http://www.slideshare.net/shimireji
 Digital Control and State Variable methods by M.Gopal
 https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors

13. THANKU