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Home > Matrix & Vector calculators > Inverse Power Method for dominant eigenvalue example
20. Inverse Power Method for finding dominant eigenvalue example ( Enter your problem )
( Enter your problem )
  1. Example `[[2,3],[5,4]]`
  2. Example `[[1,6,1],[1,2,0],[0,0,3]]`
  3. Example `[[1,2,0],[-2,1,2],[1,3,1]]`
  4. Example `[[3,2],[1,4]]`
 		Other related methods
  1. Transforming matrix to Row Echelon Form (ref)
  2. Transforming matrix to Reduced Row Echelon Form (rref)
  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. Inverse Power Method for dominant eigenvalue
  21. Determinant by gaussian elimination
  22. Expanding determinant along row / column
  23. Determinants using montante (bareiss algorithm)
  24. Leibniz formula for determinant
  25. determinants using Sarrus Rule
  26. determinants using properties of determinants
  27. Row Space
  28. Column Space
  29. Null Space
1. Example `[[2,3],[5,4]]`
(Previous example)		3. Example `[[1,2,0],[-2,1,2],[1,3,1]]`
(Next example)

2. Example `[[1,6,1],[1,2,0],[0,0,3]]`


Find Inverse Power Method for finding dominant eigenvalue ...
`[[1,6,1],[1,2,0],[0,0,3]]`
`x_0` =

Solution:
`A` = 
`1``6``1`
`1``2``0`
`0``0``3`


1. Find `A^-1`


`|A|` = 
 `1`  `6`  `1` 
 `1`  `2`  `0` 
 `0`  `0`  `3` 


 =
 `1` `xx` 
 `2`  `0` 
 `0`  `3` 
 `-6` `xx` 
 `1`  `0` 
 `0`  `3` 
 `+1` `xx` 
 `1`  `2` 
 `0`  `0` 


`=1 xx (2 xx 3 - 0 xx 0) -6 xx (1 xx 3 - 0 xx 0) +1 xx (1 xx 0 - 2 xx 0)`

`=1 xx (6 +0) -6 xx (3 +0) +1 xx (0 +0)`

`=1 xx (6) -6 xx (3) +1 xx (0)`

`= 6 -18 +0`

`=-12`


`Adj(A)` = 
Adj
`1``6``1`
`1``2``0`
`0``0``3`


 = 
 + 
 `2`  `0` 
 `0`  `3` 
 - 
 `1`  `0` 
 `0`  `3` 
 + 
 `1`  `2` 
 `0`  `0` 
 - 
 `6`  `1` 
 `0`  `3` 
 + 
 `1`  `1` 
 `0`  `3` 
 - 
 `1`  `6` 
 `0`  `0` 
 + 
 `6`  `1` 
 `2`  `0` 
 - 
 `1`  `1` 
 `1`  `0` 
 + 
 `1`  `6` 
 `1`  `2` 
T


 = 
`+(2 xx 3 - 0 xx 0)``-(1 xx 3 - 0 xx 0)``+(1 xx 0 - 2 xx 0)`
`-(6 xx 3 - 1 xx 0)``+(1 xx 3 - 1 xx 0)``-(1 xx 0 - 6 xx 0)`
`+(6 xx 0 - 1 xx 2)``-(1 xx 0 - 1 xx 1)``+(1 xx 2 - 6 xx 1)`
T


 = 
`+(6 +0)``-(3 +0)``+(0 +0)`
`-(18 +0)``+(3 +0)``-(0 +0)`
`+(0 -2)``-(0 -1)``+(2 -6)`
T


 = 
`6``-3``0`
`-18``3``0`
`-2``1``-4`
T


 = 
`6``-18``-2`
`-3``3``1`
`0``0``-4`


`"Now, "A^(-1)=1/|A| xx Adj(A)`

 = `1/(-12) xx`
`6``-18``-2`
`-3``3``1`
`0``0``-4`


 = 
`-0.5``1.5``0.1667`
`0.25``-0.25``-0.0833`
`0``0``0.3333`


 `A^-1=` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333


 `x_0=` 
1
1
1


`1^(st)` iteration :

Multiply the matrix by the vector
 `A^-1 x_0 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
1
1
1
 `=` 
1.1667
-0.0833
0.3333


Normalize the resulting vector
To normalize, divide each element of vector by its largest absolute value, which is `1.1667`

`x_1=``1/1.1667`
1.1667
-0.0833
0.3333
`=`
1
-0.0714
0.2857


`2^(nd)` iteration :

Repeat the multiplication
 `A^-1 x_1 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
1
-0.0714
0.2857
 `=` 
-0.5595
0.244
0.0952


Normalize again
The largest absolute value is `0.5595`

`x_2=``1/0.5595`
-0.5595
0.244
0.0952
`=`
-1
0.4362
0.1702


`3^(rd)` iteration :

Repeat the multiplication
 `A^-1 x_2 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
-1
0.4362
0.1702
 `=` 
1.1826
-0.3732
0.0567


Normalize again
The largest absolute value is `1.1826`

`x_3=``1/1.1826`
1.1826
-0.3732
0.0567
`=`
1
-0.3156
0.048


`4^(th)` iteration :

Repeat the multiplication
 `A^-1 x_3 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
1
-0.3156
0.048
 `=` 
-0.9654
0.3249
0.016


Normalize again
The largest absolute value is `0.9654`

`x_4=``1/0.9654`
-0.9654
0.3249
0.016
`=`
-1
0.3365
0.0166


`5^(th)` iteration :

Repeat the multiplication
 `A^-1 x_4 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
-1
0.3365
0.0166
 `=` 
1.0076
-0.3355
0.0055


Normalize again
The largest absolute value is `1.0076`

`x_5=``1/1.0076`
1.0076
-0.3355
0.0055
`=`
1
-0.333
0.0055


`6^(th)` iteration :

Repeat the multiplication
 `A^-1 x_5 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
1
-0.333
0.0055
 `=` 
-0.9986
0.3328
0.0018


Normalize again
The largest absolute value is `0.9986`

`x_6=``1/0.9986`
-0.9986
0.3328
0.0018
`=`
-1
0.3333
0.0018


`7^(th)` iteration :

Repeat the multiplication
 `A^-1 x_6 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
-1
0.3333
0.0018
 `=` 
1.0002
-0.3335
0.0006


Normalize again
The largest absolute value is `1.0002`

`x_7=``1/1.0002`
1.0002
-0.3335
0.0006
`=`
1
-0.3334
0.0006


`8^(th)` iteration :

Repeat the multiplication
 `A^-1 x_7 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
1
-0.3334
0.0006
 `=` 
-1
0.3333
0.0002


Normalize again
The largest absolute value is `1`

`x_8=``1/1`
-1
0.3333
0.0002
`=`
-1
0.3333
0.0002


`9^(th)` iteration :

Repeat the multiplication
 `A^-1 x_8 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
-1
0.3333
0.0002
 `=` 
1
-0.3333
0.0001


Normalize again
The largest absolute value is `1`

`x_9=``1/1`
1
-0.3333
0.0001
`=`
1
-0.3333
0.0001


`10^(th)` iteration :

Repeat the multiplication
 `A^-1 x_9 =` 
-0.51.50.1667
0.25-0.25-0.0833
000.3333
 
1
-0.3333
0.0001
 `=` 
-1
0.3333
0


Normalize again
The largest absolute value is `1`

`x_10=``1/1`
-1
0.3333
0
`=`
-1
0.3333
0


`:.` The dominant eigenvalue `lamda=1`

and the dominant eigenvector is :
`=`
-1
0.3333
0
`~=`
-1
0.33
0



This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here


1. Example `[[2,3],[5,4]]`
(Previous example)		3. Example `[[1,2,0],[-2,1,2],[1,3,1]]`
(Next example)


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