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Home > Matrix & Vector calculators > Inverse Power Method for dominant eigenvalue example
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20. Inverse Power Method for finding dominant eigenvalue example
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- Example `[[2,3],[5,4]]`
- Example `[[1,6,1],[1,2,0],[0,0,3]]`
- Example `[[1,2,0],[-2,1,2],[1,3,1]]`
- Example `[[3,2],[1,4]]`
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Other related methods
- Transforming matrix to Row Echelon Form (ref)
- Transforming matrix to Reduced Row Echelon Form (rref)
- Rank of matrix
- Characteristic polynomial of matrix
- Eigenvalues
- Eigenvectors (Eigenspace)
- Triangular Matrix
- LU decomposition using Gauss Elimination method of matrix
- LU decomposition using Doolittle's method of matrix
- LU decomposition using Crout's method of matrix
- Diagonal Matrix
- Cholesky Decomposition
- QR Decomposition (Gram Schmidt Method)
- QR Decomposition (Householder Method)
- LQ Decomposition
- Pivots
- Singular Value Decomposition (SVD)
- Moore-Penrose Pseudoinverse
- Power Method for dominant eigenvalue
- Inverse Power Method for dominant eigenvalue
- Determinant by gaussian elimination
- Expanding determinant along row / column
- Determinants using montante (bareiss algorithm)
- Leibniz formula for determinant
- determinants using Sarrus Rule
- determinants using properties of determinants
- Row Space
- Column Space
- Null Space
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2. Example `[[1,6,1],[1,2,0],[0,0,3]]`
(Previous example) | 4. Example `[[3,2],[1,4]]`
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3. Example `[[1,2,0],[-2,1,2],[1,3,1]]`
Find Inverse Power Method for finding dominant eigenvalue ...
`[[1,2,0],[-2,1,2],[1,3,1]]`
`x_0` = Solution:| `A` | = | | `1` | `2` | `0` | | | `-2` | `1` | `2` | | | `1` | `3` | `1` | |
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1. Find `A^-1`| `|A|` | = | | `1` | `2` | `0` | | | `-2` | `1` | `2` | | | `1` | `3` | `1` | |
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`=1 xx (1 xx 1 - 2 xx 3) -2 xx (-2 xx 1 - 2 xx 1) +0 xx (-2 xx 3 - 1 xx 1)` `=1 xx (1 -6) -2 xx (-2 -2) +0 xx (-6 -1)` `=1 xx (-5) -2 xx (-4) +0 xx (-7)` `= -5 +8 +0` `=3` | `Adj(A)` | = | | Adj | | `1` | `2` | `0` | | | `-2` | `1` | `2` | | | `1` | `3` | `1` | |
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| = | | `+(1 xx 1 - 2 xx 3)` | `-(-2 xx 1 - 2 xx 1)` | `+(-2 xx 3 - 1 xx 1)` | | | `-(2 xx 1 - 0 xx 3)` | `+(1 xx 1 - 0 xx 1)` | `-(1 xx 3 - 2 xx 1)` | | | `+(2 xx 2 - 0 xx 1)` | `-(1 xx 2 - 0 xx (-2))` | `+(1 xx 1 - 2 xx (-2))` | |
| T |
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| = | | `+(1 -6)` | `-(-2 -2)` | `+(-6 -1)` | | | `-(2 +0)` | `+(1 +0)` | `-(3 -2)` | | | `+(4 +0)` | `-(2 +0)` | `+(1 +4)` | |
| T |
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| = | | `-5` | `4` | `-7` | | | `-2` | `1` | `-1` | | | `4` | `-2` | `5` | |
| T |
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| = | | `-5` | `-2` | `4` | | | `4` | `1` | `-2` | | | `-7` | `-1` | `5` | |
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`"Now, "A^(-1)=1/|A| xx Adj(A)` | = | `1/(3) xx` | | `-5` | `-2` | `4` | | | `4` | `1` | `-2` | | | `-7` | `-1` | `5` | |
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| = | | `-1.6667` | `-0.6667` | `1.3333` | | | `1.3333` | `0.3333` | `-0.6667` | | | `-2.3333` | `-0.3333` | `1.6667` | |
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| `A^-1=` | | -1.6667 | -0.6667 | 1.3333 | | | 1.3333 | 0.3333 | -0.6667 | | | -2.3333 | -0.3333 | 1.6667 | |
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`1^(st)` iteration :Multiply the matrix by the vector| `A^-1 x_0 =` | | -1.6667 | -0.6667 | 1.3333 | | | 1.3333 | 0.3333 | -0.6667 | | | -2.3333 | -0.3333 | 1.6667 | |
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Normalize the resulting vectorTo normalize, divide each element of vector by its largest absolute value, which is `1` `:.` The dominant eigenvalue `lamda=1` and the dominant eigenvector is :
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
2. Example `[[1,6,1],[1,2,0],[0,0,3]]`
(Previous example) | 4. Example `[[3,2],[1,4]]`
(Next example) |
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