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.

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