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Eigen Values Jocobi Method.ppt

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RajendraKhapre1

The document describes the Jacobi method for finding the eigenvalues and eigenvectors of a matrix. The method works by repeatedly transforming the matrix using orthogonal matrices until it is in diagonal form, with the diagonal elements being the eigenvalues. Each transformation zeros out one off-diagonal element, and the process is repeated until the matrix is fully diagonal. The eigenvectors can then be found from the resulting diagonalization matrix. An example is worked through to find the eigenvalues and eigenvectors of a 3x3 matrix using this method.

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Eigen Values by Jacobi Method by Dr. Rajendra N. Khapre Civil Engineering Department Shri Ramdeobaba College of Engineering and Management, Nagpur
In the Jacobi method, the given matrix is converted to a diagonal matrix. The diagonal elements of this diagonal matrix are nothing but the Eigen values. In order to convert the given matrix into diagonal matrix, transformation method is used. In this method, the given matrix is multiplied by orthogonal matrix in the following fashion Eigen Values by Jacobi Method ] [P ] ][ ][ [ ] [ 1 1 1 P A P A T  Here is the orthogonal matrix such that . The shape of this matrix depends on the element of matrix that has to be transformed. For instance, matrix is of size and we wish to reduce its element to zero, then the matrix will be as follows ] [ 1 P 1 1 1 ] [ ] [   P P T ] [A ] [A 3 3 12 a ] [ 1 P             1 0 0 0 cos sin 1 sin cos ] [ 1     P
Eigen Values by Jacobi Method Once matrix is obtained, it will contain elements involving terms like and . The unknown entity can be obtained by equating the term at location with zero. After substituting the value of in matrices and , we get them in their numerical form. This process transformation continues for all non-diagonal elements. Newly generated matrix is used for further calculations. In order to convert element to zero, following expression is used. Where ] [ 1 A  cos  sin  12 a  ] [ 1 A ] [ 1 P ] [ 1 A 13 a ] ][ ][ [ ] [ 2 1 2 2 P A P A T              1 0 0 cos sin 0 sin cos 1 ] [ 2     P
Eigen Values by Jacobi Method Once transformation of all non-diagonal elements is over, matrix becomes a diagonal matrix. The diagonal elements of this matrix are the Eigen values. To compute their corresponding Eigen vectors, product of matrices , , , … is obtained. For matrix of size , six transformation matrices are required in order to convert it to a diagonal element. Hence we need to find the product of The matrix will be of size . Each column of this matrix represents the Eigen vector corresponding to the Eigen value belongs to the same column in matrix . ] [A ] [ 1 P ] [ 2 P ] [ 3 P ] [A 3 3 ] ][ ][ ][ ][ ][ [ ] [ 6 5 4 3 2 1 P P P P P P P  ] [P 3 3 ] [A
Eigen Values by Jacobi Method Problem Find out the Eigen values and Eigen vectors for the following matrix using Jacobi method Solution In the Jacobi method, the given matrix must be converted into a diagonal matrix. This process is sequential and hence converting the first element of matrix to zero.                2 1 0 1 2 1 0 1 2 ] [A 12 a ] [A
Eigen Values by Jacobi Method Step 1: Converting where Above equation becomes 0 12  a ] ][ ][ [ ] [ 1 1 1 P A P A T              1 0 0 0 cos sin 0 sin cos ] [ 1     P                                      1 0 0 0 cos sin 0 sin cos 2 1 0 1 2 1 0 1 2 1 0 0 0 cos sin 0 sin cos ] [ 1         A                                1 0 0 0 cos sin 0 sin cos 2 1 0 cos cos 2 sin cos sin 2 sin sin 2 cos sin cos 2 ] [ 1               A
Eigen Values by Jacobi Method                              2 cos sin cos cos 2 cos sin cos sin sin 2 cos sin 2 sin cos cos sin 2 sin cos sin 2 cos sin cos sin 2 sin 2 cos sin cos sin cos 2 ] [ 2 2 2 2 2 2 2 2 1                             A       cos sin 2 cos sin cos sin 2 0 2 2 12       a Solving above equation for , we get Hence matrix takes following form and matrix takes following form  45   ] [ 1 A                2 7071 . 0 7071 . 0 7071 . 0 3 0 7071 . 0 0 1 ] [ 1 A ] [ 1 P             1 0 0 0 7071 . 0 7071 . 0 0 7071 . 0 7071 . 0 ] [ 1 P
Eigen Values by Jacobi Method Step 2: Converting where Above equation becomes 0 13  a ] ][ ][ [ ] [ 2 1 2 2 P A P A T                  cos 0 sin 0 1 0 sin 0 cos ] [ 2 P                                              cos 0 sin 0 1 0 sin 0 cos 2 7071 . 0 7071 . 0 7071 . 0 3 0 7071 . 0 0 1 cos 0 sin 0 1 0 sin 0 cos ] [ 2 A                                              cos 0 sin 0 1 0 sin 0 cos cos 2 sin 7071 . 0 cos 7071 . 0 cos 7071 . 0 sin 7071 . 0 3 0 sin 2 cos 7071 . 0 sin 7071 . 0 sin 7071 . 0 cos ] [ 2 A
Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 2 A ] [ 2 P                                                          2 2 2 2 2 2 2 2 2 cos 2 cos sin 7071 . 0 cos sin 7071 . 0 sin cos 7071 . 0 cos sin 2 sin 7071 . 0 cos 7071 . 0 cos sin cos 7071 . 0 3 sin 7071 . 0 cos sin 2 cos 7071 . 0 sin 7071 . 0 cos sin sin 7071 . 0 sin 2 cos sin 7071 . 0 cos sin 7071 . 0 cos ] [A       cos sin 2 cos 7071 . 0 sin 7071 . 0 cos sin 0 2 2 13       a 367 . 27                  366 . 2 628 . 0 0 628 . 0 3 325 . 0 0 325 . 0 634 . 0 ] [ 2 A             888 . 0 0 459 . 0 0 1 0 459 . 0 0 888 . 0 ] [ 2 P
Eigen Values by Jacobi Method Step 3: Converting where Above equation becomes 0 23  a ] ][ ][ [ ] [ 3 2 3 3 P A P A T                  cos sin 0 sin cos 0 0 0 1 ] [ 3 P                                              cos sin 0 sin cos 0 0 0 1 366 . 2 628 . 0 0 628 . 0 3 325 . 0 0 325 . 0 634 . 0 cos sin 0 sin cos 0 0 0 1 ] [ 3 A                                             cos sin 0 sin cos 0 0 0 1 cos 366 . 2 sin 628 . 0 cos 628 . 0 sin 3 sin 325 . 0 sin 366 . 2 cos 628 . 0 sin 628 . 0 cos 3 cos 325 . 0 0 325 . 0 634 . 0 ] [ 3 A
Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 3 A ] [ 3 P                                                        2 2 2 2 2 2 2 2 3 cos 366 . 2 cos sin 628 . 0 cos sin 628 . 0 sin 3 cos sin 366 . 2 sin 628 . 0 cos 628 . 0 cos sin 3 sin 325 . 0 cos sin 366 . 2 cos 628 . 0 sin 628 . 0 cos sin 3 sin 366 . 2 cos sin 628 . 0 cos sin 628 . 0 cos 3 cos 325 . 0 sin 325 . 0 cos 325 . 0 634 . 0 ] [A       cos sin 366 . 2 cos 628 . 0 sin 628 . 0 cos sin 3 0 2 2 23       a 392 . 58                386 . 3 0 277 . 0 0 979 . 1 170 . 0 277 . 0 170 . 0 634 . 0 ] [ 3 A             524 . 0 851 . 0 0 851 . 0 524 . 0 0 0 0 1 ] [ 3 P
Eigen Values by Jacobi Method Step 4: Converting where Above equation becomes 0 21  a ] ][ ][ [ ] [ 4 3 4 4 P A P A T              1 0 0 0 cos sin 0 sin cos ] [ 4     P                                    1 0 0 0 cos sin 0 sin cos 386 . 3 0 277 . 0 0 979 . 1 170 . 0 277 . 0 170 . 0 634 . 0 1 0 0 0 cos sin 0 sin cos ] [ 4         A                             1 0 0 0 cos sin 0 sin cos 386 . 3 0 277 . 0 sin 277 . 0 cos 979 . 1 sin 17 . 0 cos 170 . 0 sin 634 . 0 cos 277 . 0 sin 979 . 1 cos 170 . 0 sin 170 . 0 cos 634 . 0 ] [ 4               A
Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 4 A ] [ 4 P                        386 . 3 sin 277 . 0 cos 277 . 0 sin 277 . 0 cos 979 . 1 cos sin 17 . 0 cos sin 17 . 0 sin 634 . 0 cos sin 979 . 1 sin 17 . 0 cos 170 . 0 cos sin 634 . 0 cos 277 . 0 cos sin 979 . 1 cos 17 . 0 sin 17 . 0 cos sin 634 . 0 sin 979 . 1 cos sin 17 . 0 cos sin 17 . 0 cos 634 . 0 ] [ 2 2 2 2 2 2 2 2 4                             A       cos sin 979 . 1 sin 17 . 0 cos 170 . 0 cos sin 634 . 0 0 2 2 21      a 738 . 93              386 . 3 275 . 0 034 . 0 275 . 0 613 . 0 0 034 . 0 0 2 ] [ 4 A             1 0 0 0 123 . 0 992 . 0 0 992 . 0 123 . 0 ] [ 4 P
Eigen Values by Jacobi Method Step 5: Converting where Above equation becomes 0 31  a ] ][ ][ [ ] [ 5 4 5 5 P A P A T                  cos 0 sin 0 1 0 sin 0 cos ] [ 5 P                                          cos 0 sin 0 1 0 sin 0 cos 386 . 3 275 . 0 034 . 0 275 . 0 613 . 0 0 034 . 0 0 2 cos 0 sin 0 1 0 sin 0 cos ] [ 5 A                                          cos 0 sin 0 1 0 sin 0 cos cos 386 . 3 sin 034 . 0 cos 275 . 0 cos 034 . 0 sin 2 275 . 0 613 . 0 0 sin 386 . 3 cos 034 . 0 sin 275 . 0 sin 034 . 0 cos 2 ] [ 5 A
Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 5 A ] [ 5 P                                                      2 2 2 2 2 2 2 2 5 cos 386 . 3 cos sin 034 . 0 cos sin 034 . 0 sin 2 cos 275 . 0 cos sin 386 . 3 sin 034 . 0 cos 034 . 0 cos sin 2 cos 275 . 0 613 . 0 sin 275 . 0 cos sin 386 . 3 cos 034 . 0 sin 034 . 0 cos sin 2 sin 275 . 0 sin 386 . 3 cos sin 034 . 0 cos sin 034 . 0 cos 2 ] [A       cos sin 386 . 3 sin 034 . 0 cos 034 . 0 cos sin 2 0 2 2 31      a 4044 . 1              387 . 3 275 . 0 0 275 . 0 613 . 0 0 0 0 2 ] [ 5 A             999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 ] [ 5 P
Eigen Values by Jacobi Method Step 6: Converting where Above equation becomes 0 32  a ] ][ ][ [ ] [ 6 5 6 6 P A P A T                  cos sin 0 sin cos 0 0 0 1 ] [ 6 P                                          cos sin 0 sin cos 0 0 0 1 387 . 3 275 . 0 0 275 . 0 613 . 0 0 0 0 2 cos sin 0 sin cos 0 0 0 1 ] [ 6 A                                       cos sin 0 sin cos 0 0 0 1 cos 387 . 3 sin 275 . 0 cos 275 . 0 sin 613 . 0 0 sin 387 . 3 cos 275 . 0 sin 275 . 0 cos 613 . 0 0 0 0 2 ] [ 6 A
Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 6 A ] [ 6 P                                                2 2 2 2 2 2 2 2 6 cos 387 . 3 cos sin 275 . 0 cos sin 275 . 0 sin 613 . 0 cos sin 387 . 3 sin 275 . 0 cos 275 . 0 cos sin 613 . 0 0 cos sin 387 . 3 cos 275 . 0 sin 275 . 0 cos sin 613 . 0 sin 387 . 3 cos sin 275 . 0 cos sin 275 . 0 cos 613 . 0 0 0 0 2 ] [A       cos sin 387 . 3 sin 275 . 0 cos 275 . 0 cos sin 613 . 0 0 2 2 32      a 4044 . 1              414 . 3 0 0 0 586 . 0 0 0 0 2 ] [ 6 A             995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 ] [ 6 P
Now, the given matrix has been converted to the diagonal matrix. The diagonal elements of this matrix are nothing but the Eigen values. To find out their respective Eigen vectors, we need to compute Eigen Values by Jacobi Method ] [A ] ][ ][ ][ ][ ][ [ ] [ 6 5 4 3 2 1 P P P P P P P                                                                     995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 1 0 0 0 123 . 0 992 . 0 0 992 . 0 123 . 0 524 . 0 851 . 0 0 851 . 0 524 . 0 0 0 0 1 888 . 0 0 459 . 0 0 1 0 459 . 0 0 888 . 0 1 0 0 0 7071 . 0 7071 . 0 0 7071 . 0 7071 . 0 ] [P                                                           995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 1 0 0 0 123 . 0 992 . 0 0 992 . 0 123 . 0 524 . 0 851 . 0 0 851 . 0 524 . 0 0 0 0 1 8880 . 0 0 4590 . 0 3246 . 0 7071 . 0 6279 . 0 3246 . 0 7071 . 0 6279 . 0 ] [P                                               995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 1 0 0 0 123 . 0 992 . 0 0 992 . 0 123 . 0 4653 . 0 7557 . 0 4590 . 0 7718 . 0 0943 . 0 6279 . 0 4317 . 0 6467 . 0 6279 . 0 ] [P
The Eigen vectors for the Eigen values are the elements in each columns of matrix . Hence for the given matrix Eigen values are found out to be 2, 0.586, 3.414. Eigen Values by Jacobi Method                                     995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 4653 . 0 5483 . 0 6932 . 0 7718 . 0 6345 . 0 0163 . 0 4317 . 0 5433 . 0 7188 . 0 ] [P                         995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 4475 . 0 5483 . 0 7041 . 0 7714 . 0 6345 . 0 0030 . 0 4492 . 0 5433 . 0 7073 . 0 ] [P              4990 . 0 5017 . 0 7041 . 0 7054 . 0 7069 . 0 0030 . 0 5002 . 0 4966 . 0 7073 . 0 ] [P ] [P
Eigen Values by Jacobi Method Problem Find out the Eigen values and Eigen vectors for the following matrices using Jacobi method 1) 2)        8 6 6 3 ] [A        6 2 2 3 ] [A

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Eigen Values Jocobi Method.ppt

  • 1. Eigen Values by Jacobi Method by Dr. Rajendra N. Khapre Civil Engineering Department Shri Ramdeobaba College of Engineering and Management, Nagpur
  • 2. In the Jacobi method, the given matrix is converted to a diagonal matrix. The diagonal elements of this diagonal matrix are nothing but the Eigen values. In order to convert the given matrix into diagonal matrix, transformation method is used. In this method, the given matrix is multiplied by orthogonal matrix in the following fashion Eigen Values by Jacobi Method ] [P ] ][ ][ [ ] [ 1 1 1 P A P A T  Here is the orthogonal matrix such that . The shape of this matrix depends on the element of matrix that has to be transformed. For instance, matrix is of size and we wish to reduce its element to zero, then the matrix will be as follows ] [ 1 P 1 1 1 ] [ ] [   P P T ] [A ] [A 3 3 12 a ] [ 1 P             1 0 0 0 cos sin 1 sin cos ] [ 1     P
  • 3. Eigen Values by Jacobi Method Once matrix is obtained, it will contain elements involving terms like and . The unknown entity can be obtained by equating the term at location with zero. After substituting the value of in matrices and , we get them in their numerical form. This process transformation continues for all non-diagonal elements. Newly generated matrix is used for further calculations. In order to convert element to zero, following expression is used. Where ] [ 1 A  cos  sin  12 a  ] [ 1 A ] [ 1 P ] [ 1 A 13 a ] ][ ][ [ ] [ 2 1 2 2 P A P A T              1 0 0 cos sin 0 sin cos 1 ] [ 2     P
  • 4. Eigen Values by Jacobi Method Once transformation of all non-diagonal elements is over, matrix becomes a diagonal matrix. The diagonal elements of this matrix are the Eigen values. To compute their corresponding Eigen vectors, product of matrices , , , … is obtained. For matrix of size , six transformation matrices are required in order to convert it to a diagonal element. Hence we need to find the product of The matrix will be of size . Each column of this matrix represents the Eigen vector corresponding to the Eigen value belongs to the same column in matrix . ] [A ] [ 1 P ] [ 2 P ] [ 3 P ] [A 3 3 ] ][ ][ ][ ][ ][ [ ] [ 6 5 4 3 2 1 P P P P P P P  ] [P 3 3 ] [A
  • 5. Eigen Values by Jacobi Method Problem Find out the Eigen values and Eigen vectors for the following matrix using Jacobi method Solution In the Jacobi method, the given matrix must be converted into a diagonal matrix. This process is sequential and hence converting the first element of matrix to zero.                2 1 0 1 2 1 0 1 2 ] [A 12 a ] [A
  • 6. Eigen Values by Jacobi Method Step 1: Converting where Above equation becomes 0 12  a ] ][ ][ [ ] [ 1 1 1 P A P A T              1 0 0 0 cos sin 0 sin cos ] [ 1     P                                      1 0 0 0 cos sin 0 sin cos 2 1 0 1 2 1 0 1 2 1 0 0 0 cos sin 0 sin cos ] [ 1         A                                1 0 0 0 cos sin 0 sin cos 2 1 0 cos cos 2 sin cos sin 2 sin sin 2 cos sin cos 2 ] [ 1               A
  • 7. Eigen Values by Jacobi Method                              2 cos sin cos cos 2 cos sin cos sin sin 2 cos sin 2 sin cos cos sin 2 sin cos sin 2 cos sin cos sin 2 sin 2 cos sin cos sin cos 2 ] [ 2 2 2 2 2 2 2 2 1                             A       cos sin 2 cos sin cos sin 2 0 2 2 12       a Solving above equation for , we get Hence matrix takes following form and matrix takes following form  45   ] [ 1 A                2 7071 . 0 7071 . 0 7071 . 0 3 0 7071 . 0 0 1 ] [ 1 A ] [ 1 P             1 0 0 0 7071 . 0 7071 . 0 0 7071 . 0 7071 . 0 ] [ 1 P
  • 8. Eigen Values by Jacobi Method Step 2: Converting where Above equation becomes 0 13  a ] ][ ][ [ ] [ 2 1 2 2 P A P A T                  cos 0 sin 0 1 0 sin 0 cos ] [ 2 P                                              cos 0 sin 0 1 0 sin 0 cos 2 7071 . 0 7071 . 0 7071 . 0 3 0 7071 . 0 0 1 cos 0 sin 0 1 0 sin 0 cos ] [ 2 A                                              cos 0 sin 0 1 0 sin 0 cos cos 2 sin 7071 . 0 cos 7071 . 0 cos 7071 . 0 sin 7071 . 0 3 0 sin 2 cos 7071 . 0 sin 7071 . 0 sin 7071 . 0 cos ] [ 2 A
  • 9. Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 2 A ] [ 2 P                                                          2 2 2 2 2 2 2 2 2 cos 2 cos sin 7071 . 0 cos sin 7071 . 0 sin cos 7071 . 0 cos sin 2 sin 7071 . 0 cos 7071 . 0 cos sin cos 7071 . 0 3 sin 7071 . 0 cos sin 2 cos 7071 . 0 sin 7071 . 0 cos sin sin 7071 . 0 sin 2 cos sin 7071 . 0 cos sin 7071 . 0 cos ] [A       cos sin 2 cos 7071 . 0 sin 7071 . 0 cos sin 0 2 2 13       a 367 . 27                  366 . 2 628 . 0 0 628 . 0 3 325 . 0 0 325 . 0 634 . 0 ] [ 2 A             888 . 0 0 459 . 0 0 1 0 459 . 0 0 888 . 0 ] [ 2 P
  • 10. Eigen Values by Jacobi Method Step 3: Converting where Above equation becomes 0 23  a ] ][ ][ [ ] [ 3 2 3 3 P A P A T                  cos sin 0 sin cos 0 0 0 1 ] [ 3 P                                              cos sin 0 sin cos 0 0 0 1 366 . 2 628 . 0 0 628 . 0 3 325 . 0 0 325 . 0 634 . 0 cos sin 0 sin cos 0 0 0 1 ] [ 3 A                                             cos sin 0 sin cos 0 0 0 1 cos 366 . 2 sin 628 . 0 cos 628 . 0 sin 3 sin 325 . 0 sin 366 . 2 cos 628 . 0 sin 628 . 0 cos 3 cos 325 . 0 0 325 . 0 634 . 0 ] [ 3 A
  • 11. Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 3 A ] [ 3 P                                                        2 2 2 2 2 2 2 2 3 cos 366 . 2 cos sin 628 . 0 cos sin 628 . 0 sin 3 cos sin 366 . 2 sin 628 . 0 cos 628 . 0 cos sin 3 sin 325 . 0 cos sin 366 . 2 cos 628 . 0 sin 628 . 0 cos sin 3 sin 366 . 2 cos sin 628 . 0 cos sin 628 . 0 cos 3 cos 325 . 0 sin 325 . 0 cos 325 . 0 634 . 0 ] [A       cos sin 366 . 2 cos 628 . 0 sin 628 . 0 cos sin 3 0 2 2 23       a 392 . 58                386 . 3 0 277 . 0 0 979 . 1 170 . 0 277 . 0 170 . 0 634 . 0 ] [ 3 A             524 . 0 851 . 0 0 851 . 0 524 . 0 0 0 0 1 ] [ 3 P
  • 12. Eigen Values by Jacobi Method Step 4: Converting where Above equation becomes 0 21  a ] ][ ][ [ ] [ 4 3 4 4 P A P A T              1 0 0 0 cos sin 0 sin cos ] [ 4     P                                    1 0 0 0 cos sin 0 sin cos 386 . 3 0 277 . 0 0 979 . 1 170 . 0 277 . 0 170 . 0 634 . 0 1 0 0 0 cos sin 0 sin cos ] [ 4         A                             1 0 0 0 cos sin 0 sin cos 386 . 3 0 277 . 0 sin 277 . 0 cos 979 . 1 sin 17 . 0 cos 170 . 0 sin 634 . 0 cos 277 . 0 sin 979 . 1 cos 170 . 0 sin 170 . 0 cos 634 . 0 ] [ 4               A
  • 13. Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 4 A ] [ 4 P                        386 . 3 sin 277 . 0 cos 277 . 0 sin 277 . 0 cos 979 . 1 cos sin 17 . 0 cos sin 17 . 0 sin 634 . 0 cos sin 979 . 1 sin 17 . 0 cos 170 . 0 cos sin 634 . 0 cos 277 . 0 cos sin 979 . 1 cos 17 . 0 sin 17 . 0 cos sin 634 . 0 sin 979 . 1 cos sin 17 . 0 cos sin 17 . 0 cos 634 . 0 ] [ 2 2 2 2 2 2 2 2 4                             A       cos sin 979 . 1 sin 17 . 0 cos 170 . 0 cos sin 634 . 0 0 2 2 21      a 738 . 93              386 . 3 275 . 0 034 . 0 275 . 0 613 . 0 0 034 . 0 0 2 ] [ 4 A             1 0 0 0 123 . 0 992 . 0 0 992 . 0 123 . 0 ] [ 4 P
  • 14. Eigen Values by Jacobi Method Step 5: Converting where Above equation becomes 0 31  a ] ][ ][ [ ] [ 5 4 5 5 P A P A T                  cos 0 sin 0 1 0 sin 0 cos ] [ 5 P                                          cos 0 sin 0 1 0 sin 0 cos 386 . 3 275 . 0 034 . 0 275 . 0 613 . 0 0 034 . 0 0 2 cos 0 sin 0 1 0 sin 0 cos ] [ 5 A                                          cos 0 sin 0 1 0 sin 0 cos cos 386 . 3 sin 034 . 0 cos 275 . 0 cos 034 . 0 sin 2 275 . 0 613 . 0 0 sin 386 . 3 cos 034 . 0 sin 275 . 0 sin 034 . 0 cos 2 ] [ 5 A
  • 15. Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 5 A ] [ 5 P                                                      2 2 2 2 2 2 2 2 5 cos 386 . 3 cos sin 034 . 0 cos sin 034 . 0 sin 2 cos 275 . 0 cos sin 386 . 3 sin 034 . 0 cos 034 . 0 cos sin 2 cos 275 . 0 613 . 0 sin 275 . 0 cos sin 386 . 3 cos 034 . 0 sin 034 . 0 cos sin 2 sin 275 . 0 sin 386 . 3 cos sin 034 . 0 cos sin 034 . 0 cos 2 ] [A       cos sin 386 . 3 sin 034 . 0 cos 034 . 0 cos sin 2 0 2 2 31      a 4044 . 1              387 . 3 275 . 0 0 275 . 0 613 . 0 0 0 0 2 ] [ 5 A             999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 ] [ 5 P
  • 16. Eigen Values by Jacobi Method Step 6: Converting where Above equation becomes 0 32  a ] ][ ][ [ ] [ 6 5 6 6 P A P A T                  cos sin 0 sin cos 0 0 0 1 ] [ 6 P                                          cos sin 0 sin cos 0 0 0 1 387 . 3 275 . 0 0 275 . 0 613 . 0 0 0 0 2 cos sin 0 sin cos 0 0 0 1 ] [ 6 A                                       cos sin 0 sin cos 0 0 0 1 cos 387 . 3 sin 275 . 0 cos 275 . 0 sin 613 . 0 0 sin 387 . 3 cos 275 . 0 sin 275 . 0 cos 613 . 0 0 0 0 2 ] [ 6 A
  • 17. Eigen Values by Jacobi Method Solving above equation for , we get Hence matrix takes following form and matrix takes following form  ] [ 6 A ] [ 6 P                                                2 2 2 2 2 2 2 2 6 cos 387 . 3 cos sin 275 . 0 cos sin 275 . 0 sin 613 . 0 cos sin 387 . 3 sin 275 . 0 cos 275 . 0 cos sin 613 . 0 0 cos sin 387 . 3 cos 275 . 0 sin 275 . 0 cos sin 613 . 0 sin 387 . 3 cos sin 275 . 0 cos sin 275 . 0 cos 613 . 0 0 0 0 2 ] [A       cos sin 387 . 3 sin 275 . 0 cos 275 . 0 cos sin 613 . 0 0 2 2 32      a 4044 . 1              414 . 3 0 0 0 586 . 0 0 0 0 2 ] [ 6 A             995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 ] [ 6 P
  • 18. Now, the given matrix has been converted to the diagonal matrix. The diagonal elements of this matrix are nothing but the Eigen values. To find out their respective Eigen vectors, we need to compute Eigen Values by Jacobi Method ] [A ] ][ ][ ][ ][ ][ [ ] [ 6 5 4 3 2 1 P P P P P P P                                                                     995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 1 0 0 0 123 . 0 992 . 0 0 992 . 0 123 . 0 524 . 0 851 . 0 0 851 . 0 524 . 0 0 0 0 1 888 . 0 0 459 . 0 0 1 0 459 . 0 0 888 . 0 1 0 0 0 7071 . 0 7071 . 0 0 7071 . 0 7071 . 0 ] [P                                                           995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 1 0 0 0 123 . 0 992 . 0 0 992 . 0 123 . 0 524 . 0 851 . 0 0 851 . 0 524 . 0 0 0 0 1 8880 . 0 0 4590 . 0 3246 . 0 7071 . 0 6279 . 0 3246 . 0 7071 . 0 6279 . 0 ] [P                                               995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 1 0 0 0 123 . 0 992 . 0 0 992 . 0 123 . 0 4653 . 0 7557 . 0 4590 . 0 7718 . 0 0943 . 0 6279 . 0 4317 . 0 6467 . 0 6279 . 0 ] [P
  • 19. The Eigen vectors for the Eigen values are the elements in each columns of matrix . Hence for the given matrix Eigen values are found out to be 2, 0.586, 3.414. Eigen Values by Jacobi Method                                     995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 999 . 0 0 025 . 0 0 1 0 025 . 0 0 999 . 0 4653 . 0 5483 . 0 6932 . 0 7718 . 0 6345 . 0 0163 . 0 4317 . 0 5433 . 0 7188 . 0 ] [P                         995 . 0 098 . 0 0 098 . 0 995 . 0 0 0 0 1 4475 . 0 5483 . 0 7041 . 0 7714 . 0 6345 . 0 0030 . 0 4492 . 0 5433 . 0 7073 . 0 ] [P              4990 . 0 5017 . 0 7041 . 0 7054 . 0 7069 . 0 0030 . 0 5002 . 0 4966 . 0 7073 . 0 ] [P ] [P
  • 20. Eigen Values by Jacobi Method Problem Find out the Eigen values and Eigen vectors for the following matrices using Jacobi method 1) 2)        8 6 6 3 ] [A        6 2 2 3 ] [A