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Home > Matrix & Vector calculators > Power Method for dominant eigenvalue example
19. Power Method for finding dominant eigenvalue example ( Enter your problem )
( Enter your problem )
  1. Example `[[2,3],[4,10]]`
  2. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
  3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
  4. Example `[[2,3],[4,10]]`
 		Other related methods
  1. Transforming matrix to Row Echelon Form
  2. Transforming matrix to Reduced Row Echelon Form
  3. Rank of matrix
  4. Characteristic polynomial of matrix
  5. Eigenvalues
  6. Eigenvectors (Eigenspace)
  7. Triangular Matrix
  8. LU decomposition using Gauss Elimination method of matrix
  9. LU decomposition using Doolittle's method of matrix
  10. LU decomposition using Crout's method of matrix
  11. Diagonal Matrix
  12. Cholesky Decomposition
  13. QR Decomposition (Gram Schmidt Method)
  14. QR Decomposition (Householder Method)
  15. LQ Decomposition
  16. Pivots
  17. Singular Value Decomposition (SVD)
  18. Moore-Penrose Pseudoinverse
  19. Power Method for dominant eigenvalue
  20. determinants using Sarrus Rule
  21. determinants using properties of determinants
  22. Row Space
  23. Column Space
  24. Null Space
3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
(Previous example)		20. determinants using Sarrus Rule
(Next method)

4. Example `[[2,3],[4,10]]`


Find Power Method for finding dominant eigenvalue ...
`[[2,3],[4,10]]`

Solution:
 `A=` 
23
410


 `x_0=` 
1
1


`1^(st)` Iteration

 `A x_0 =` 
23
410
 
1
1
 `=` 
5
14


and by scaling we obtain the approximation
`x_1=``1/14`
5
14
`=`
0.3571
1


`2^(nd)` Iteration

 `A x_1 =` 
23
410
 
0.3571
1
 `=` 
3.7143
11.4286


and by scaling we obtain the approximation
`x_2=``1/11.4286`
3.7143
11.4286
`=`
0.325
1


`3^(rd)` Iteration

 `A x_2 =` 
23
410
 
0.325
1
 `=` 
3.65
11.3


and by scaling we obtain the approximation
`x_3=``1/11.3`
3.65
11.3
`=`
0.323
1


`4^(th)` Iteration

 `A x_3 =` 
23
410
 
0.323
1
 `=` 
3.646
11.292


and by scaling we obtain the approximation
`x_4=``1/11.292`
3.646
11.292
`=`
0.3229
1


`5^(th)` Iteration

 `A x_4 =` 
23
410
 
0.3229
1
 `=` 
3.6458
11.2915


and by scaling we obtain the approximation
`x_5=``1/11.2915`
3.6458
11.2915
`=`
0.3229
1


`:.` The dominant eigenvalue `lamda=11.2915~=11.29`

and the dominant eigenvector is :
`=`
0.3229
1
`~=`
0.32
1


This material is intended as a summary. Use your textbook for detail explanation.
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3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
(Previous example)		20. determinants using Sarrus Rule
(Next method)


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