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SEMINAR ON EIGENVALUESSEMINAR ON EIGENVALUES
AND EIGENVECTORSAND EIGENVECTORS
By
Vinod Srivastava
M.E. Modular I & C
Roll No.151522
CONTENTSCONTENTS
 Introduction to Eigenvalues and Eigenvectors
 Examples
 Two-dimensional matrix
 Three-dimensional matrix
• Example using MATLAB
• References
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
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
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
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
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
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
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
Similarly for λ = -3 and λ = -4 the corresponding eigenvectors u and x are-










=










=
0
0
2
,
0
1
1
xu
Example using MATLABExample using MATLAB
REFERENCESREFERENCES
 http://www.slideshare.net/shimireji
 Digital Control and State Variable methods by M.Gopal
 https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors
THANKU

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Eigenvalues and Eigenvectors

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