Eigenvalue problems .ppt

Eigenvalue problems .ppt

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  Shree S’ad VidhyaMandal Institude of Technology Eigenvalue Problems (Group No.11) SUBJECT : VCLA (2110015) Department of electrical engineering second semester Regards:N.M.Patel
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  NO. NAME ENROLLMENTNO. 1 PATEL SHIVAM 150450109036 2 RAJPUT DHRMENDRA 150450109043 3 DESAI VISHAL 150450109058
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  Eigenvalue Problems A matrixeigenvalue problem considers the vector equation (1) Here A is a given square matrix, an unknown scalar, and an unknown vector is called as the eigen value or characteristic value or latent value or proper roots or root of the matrix A, and is called as eigen vector or charecteristic vector or latent vector or real vector. xxA   x  x
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  What are Eigenvalues? If A is an n x n matrix and λ is a scalar for which has a nontrivial solution , then is an eigenvalue of A and x is a corresponding eigenvector of A. is called the eigenvalue problem for A. Note that we can rewrite the equation as follows: or is the trivial solution. But our solutions must be nonzero vectors called eigenvectors that correspond to each of the distinct eigenvalues. xlA nxx   0 xn Axl xxA  xxA   0.0)(  xxAxln n Rx
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  Eigenvalue Problems How toFind Eigenvalues EXAMPLE 1 Determination of Eigenvalues We illustrate all the steps in terms of the matrix 5 2 . 2 2       A
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  EXAMPLE 1 Solution. (a) Eigenvalues.These must be determined first. Equation (1) is in components 1 1 2 2 5 2 ; 2 2 x x x x                   Ax
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  EXAMPLE 1 Solution. (continued1) (a)Eigenvalues. (continued 1) Transferring the terms on the right to the left, we get 1 2 1 1 2 2 5 2 2 2 . x x x x x x        1 2 1 2 ( 5 ) 2 0 2 ( 2 ) 0 x x x x          
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  EXAMPLE 1 This canbe written in matrix Because (1) is n.. which gives ( ) 0 A I x ,0)(  xIAIAA xxxx 
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  EXAMPLE 1 Solution. (continued2) (a) Eigenvalues. (continued 2) We see that this is a homogeneous linear system. it has a nontrivial solution (an eigenvector of A we are looking for) if and only if its coefficient determinant is zero, that is, 2 5 2 ( ) det( ) 2 2 ( 5 )( 2 ) 4 7 6 0. D                          A I
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  EXAMPLE 1 Solution. (continued3) (a) Eigenvalues. (continued 3) We call the characteristic determinant or, if expanded, the characteristic polynomial, and the characteristic equation of A. The solutions of this quadratic equation are and . These are the eigenvalues of A. Eigen vector corresponding to eigen value is )(D 0)( D 11  62  11  01  IA 
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  EXAMPLE 1   202 1024 0 0 12 24                       yx yx y x Solving these two equations we get, and There for eigenvector corresponding to eigenvalue is Eigen vector space is 0x 0y 11     0,0,1  yxE   .0,01 E
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  EXAMPLE 2 Example 2:Find the eigenvalues A. Step 1: Find the eigenvalues for A. The determinant of a triangular matrix is the product of the elements at the diagonal. Thus, the characteristic equation of A is A  3 4 0 0 3 0 0 0 1          
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  EXAMPLE 2 has algebraicmultiplicity 1 and has algebraic multiplicity 2. p()  det(I  A)  I  A  0    3 4 0 0   3 0 0 0  1  (  3)2 ( 1)1  0. 11  62 
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  EXAMPLE 2 Eigen vectorcorresponding to is Here, and There for eigen vector corrous ponding to eigen value Eigen space is 31    01  IA  02 04 , 100 000 040 3 2 2                x x Rrrx 02 x 03 x       RrrrxxxE  ,0,0,10,0,,, 3213   RrrE  /0,0,13
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  EXAMPLE 2 Now, eigenvector corresponding to Here, and .32                 200 000 040 02IA  02 04 , 3 2 1    x x Rrrx 02 x 03 x
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  EXAMPLE 2 Therefore eigenvector corresponding to eigen value Eigen space is       .,0,0,10,0,,, 3213 RrrrxxxE    RrrE  0,0,13 Now,eigen vector corresponding to 13    0 02 042 000 020 042 0 2 2 21 3                x x xI IA 
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  EXAMPLE 2 Here Therefor eigenvector of is Eigen space is 0 ; 1 3   x Rrrx 13     0,,0,, 321 rxxx    .0,1,01 RrrE 
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  Thank you