Eigen value

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

2. CONTENTS
● Introduction to Eigenvalues and Eigenvectors
● Examples
➢ Two-dimensional matrix
➢ Three-dimensional matrix
•
•
Example using MATLAB
References

3. INTRODUCTION
● 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….
● 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-
A  I  0

5. Examples
● Two-dimensional matrix example-
Ex.1 Find the eigenvalues and eigenvectors of matrix A.
Taking the determinant to find characteristic polynomial A-
 
2
1
1
A
2
AI  0  0
1
2
2
1
 3 4  2
 0
It has roots at λ = 1 and λ = 3, which are the two eigenvalues of A.

6. For λ = 1, Equation becomes,
which has the solution,
Eigenvectors v of this transformation satisfy the equation,
Av= λv
Rearrange this equation to obtain-
AIv 0
A  I v  0
1
  2   
1 1v  0
1v1 

0
 
 1 
v  1

7. For λ = 3, Equation becomes,
Thus, the vectors vλ=1 and vλ=3 are eigenvectors of A associated with the
eigenvalues λ = 1 and λ = 3, respectively.
 1 0
A 3I u  0
1
  2   
1u 
1 u1 

0
 
  1 
 1 
which has the solution-
u

8. ● Three-dimensional matrix example-
Ex.2 Find the eigenvalue and eigenvector of matrix A.
the matrix has the characteristics equation-

2

 0
1 
0 
1
 3
0
  4
A   0
  4 3  2 0
0
1
  2
1
  3
0
  4
I  A  0
0

9. therefore the eigen values of A are-
1  2,2 3,3  4
   

0
0
 1
1
0
2
2 
0 v3  0
1v    0 
0 v1   0 
For λ = -2, Equation becomes,
1I  Av1  0
 
2
which has the solution-
1
v  2

10. Similarly for λ = -3 and λ = -4 the corresponding eigenvectors u and x are-
 
0
 2 
 
0
 1 
u   1  , x   0 

11. Example using MATLAB

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

13. THANKU