6. Eigenvectors (Eigenspace) example
( Enter your problem )
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- Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
- Example `[[3,2,4],[2,0,2],[4,2,3]]`
- Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
- Example `[[2,3],[4,10]]`
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Other related methods
- Transforming matrix to Row Echelon Form
- Transforming matrix to Reduced Row Echelon Form
- 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
- 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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1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
Find Matrix Eigenvectors (Eigenspace) ... `[[8,-6,2],[-6,7,-4],[2,-4,3]]`Solution:`|A-lamdaI|=0` | `(8-lamda)` | `-6` | `2` | | | `-6` | `(7-lamda)` | `-4` | | | `2` | `-4` | `(3-lamda)` | |
| = 0 |
`:.(8-lamda)((7-lamda) × (3-lamda) - (-4) × (-4))-(-6)((-6) × (3-lamda) - (-4) × 2)+2((-6) × (-4) - (7-lamda) × 2)=0` `:.(8-lamda)((21-10lamda+lamda^2)-16)+6((-18+6lamda)-(-8))+2(24-(14-2lamda))=0` `:.(8-lamda)(5-10lamda+lamda^2)+6(-10+6lamda)+2(10+2lamda)=0` `:. (40-85lamda+18lamda^2-lamda^3)+(-60+36lamda)+(20+4lamda)=0` `:.(-lamda^3+18lamda^2-45lamda)=0` `:.-lamda(lamda-3)(lamda-15)=0` `:.lamda=0 or (lamda-3)=0 or (lamda-15)=0` `:.lamda=0 or lamda=3 or lamda=15` `:.` The eigenvalues of the matrix `A` are given by `lamda=0,3,15` 1. Eigenvectors for `lamda=0`
1. Eigenvectors for `lamda=0` = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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Now, reduce this matrix `R_1 larr R_1-:8` = | | `1` | `-0.75` | `0.25` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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`R_2 larr R_2+6xx R_1` = | | `1` | `-0.75` | `0.25` | | | `0` | `2.5` | `-2.5` | | | `2` | `-4` | `3` | |
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`R_3 larr R_3-2xx R_1` = | | `1` | `-0.75` | `0.25` | | | `0` | `2.5` | `-2.5` | | | `0` | `-2.5` | `2.5` | |
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`R_2 larr R_2-:2.5` = | | `1` | `-0.75` | `0.25` | | | `0` | `1` | `-1` | | | `0` | `-2.5` | `2.5` | |
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`R_1 larr R_1+0.75xx R_2` = | | `1` | `0` | `-0.5` | | | `0` | `1` | `-1` | | | `0` | `-2.5` | `2.5` | |
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`R_3 larr R_3+2.5xx R_2` = | | `1` | `0` | `-0.5` | | | `0` | `1` | `-1` | | | `0` | `0` | `0` | |
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The system associated with the eigenvalue `lamda=0` `(A-0I)` | | = | | `1` | `0` | `-0.5` | | | `0` | `1` | `-1` | | | `0` | `0` | `0` | |
| | | = | |
`=>x_1-0.5x_3=0,x_2-x_3=0` `=>x_1=0.5x_3,x_2=x_3` `:.` eigenvectors corresponding to the eigenvalue `lamda=0` is Let `x_3=1` 2. Eigenvectors for `lamda=3`
2. Eigenvectors for `lamda=3` = | | `5` | `-6` | `2` | | | `-6` | `4` | `-4` | | | `2` | `-4` | `0` | |
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Now, reduce this matrix `R_1 larr R_1-:5` = | | `1` | `-1.2` | `0.4` | | | `-6` | `4` | `-4` | | | `2` | `-4` | `0` | |
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`R_2 larr R_2+6xx R_1` = | | `1` | `-1.2` | `0.4` | | | `0` | `-3.2` | `-1.6` | | | `2` | `-4` | `0` | |
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`R_3 larr R_3-2xx R_1` = | | `1` | `-1.2` | `0.4` | | | `0` | `-3.2` | `-1.6` | | | `0` | `-1.6` | `-0.8` | |
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`R_2 larr R_2-:(-3.2)` = | | `1` | `-1.2` | `0.4` | | | `0` | `1` | `0.5` | | | `0` | `-1.6` | `-0.8` | |
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`R_1 larr R_1+1.2xx R_2` = | | `1` | `0` | `1` | | | `0` | `1` | `0.5` | | | `0` | `-1.6` | `-0.8` | |
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`R_3 larr R_3+1.6xx R_2` = | | `1` | `0` | `1` | | | `0` | `1` | `0.5` | | | `0` | `0` | `0` | |
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The system associated with the eigenvalue `lamda=3` `(A-3I)` | | = | | `1` | `0` | `1` | | | `0` | `1` | `0.5` | | | `0` | `0` | `0` | |
| | | = | |
`=>x_1+x_3=0,x_2+0.5x_3=0` `=>x_1=-x_3,x_2=-0.5x_3` `:.` eigenvectors corresponding to the eigenvalue `lamda=3` is Let `x_3=1` 3. Eigenvectors for `lamda=15`
3. Eigenvectors for `lamda=15` = | | `-7` | `-6` | `2` | | | `-6` | `-8` | `-4` | | | `2` | `-4` | `-12` | |
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Now, reduce this matrix `R_1 larr R_1-:(-7)` = | | `1` | `0.8571` | `-0.2857` | | | `-6` | `-8` | `-4` | | | `2` | `-4` | `-12` | |
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`R_2 larr R_2+6xx R_1` = | | `1` | `0.8571` | `-0.2857` | | | `0` | `-2.8571` | `-5.7143` | | | `2` | `-4` | `-12` | |
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`R_3 larr R_3-2xx R_1` = | | `1` | `0.8571` | `-0.2857` | | | `0` | `-2.8571` | `-5.7143` | | | `0` | `-5.7143` | `-11.4286` | |
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`R_2 larr R_2-:(-2.8571)` = | | `1` | `0.8571` | `-0.2857` | | | `0` | `1` | `2` | | | `0` | `-5.7143` | `-11.4286` | |
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`R_1 larr R_1-0.8571xx R_2` = | | `1` | `0` | `-2` | | | `0` | `1` | `2` | | | `0` | `-5.7143` | `-11.4286` | |
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`R_3 larr R_3+5.7143xx R_2` = | | `1` | `0` | `-2` | | | `0` | `1` | `2` | | | `0` | `0` | `0` | |
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The system associated with the eigenvalue `lamda=15` `(A-15I)` | | = | | `1` | `0` | `-2` | | | `0` | `1` | `2` | | | `0` | `0` | `0` | |
| | | = | |
`=>x_1-2x_3=0,x_2+2x_3=0` `=>x_1=2x_3,x_2=-2x_3` `:.` eigenvectors corresponding to the eigenvalue `lamda=15` is Let `x_3=1`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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