Find Matrix Diagonalization ...
`[[2,3],[4,10]]`Solution:A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such that `A=PDP^-1`
Find eigenvalues of the matrix `A`
`|A-lamdaI|=0`
| `(2-lamda)` | `3` | | | `4` | `(10-lamda)` | |
| = 0 |
`:.(2-lamda) × (10-lamda) - 3 × 4=0`
`:.(20-12lamda+lamda^2)-12=0`
`:.(lamda^2-12lamda+8)=0`
`:.(lamda-0.7084973779)(lamda-11.2915026221)=0`
`:.(lamda-0.7084973779)=0 or (lamda-11.2915026221)=0`
`:.lamda=0.7084973779 or lamda=11.2915026221`
`:.` The eigenvalues of the matrix `A` are given by `lamda=0.7084973779,11.2915026221`
1. Eigenvectors for `lamda=0.7084973779`
1. Eigenvectors for `lamda=0.7084973779`
`A-lamdaI = ` | | - `0.7084973779` | |
= | | - | | 0.7084973779 | 0 | | | 0 | 0.7084973779 | |
|
= | | `1.2915026221` | `3` | | | `4` | `9.2915026221` | |
|
Now, reduce this matrix
`R_1 larr R_1-:1.2915026221`
= | | `1` | `2.3228756555` | | | `4` | `9.2915026221` | |
|
`R_2 larr R_2-4xx R_1`
The system associated with the eigenvalue `lamda=0.7084973779`
`=>x_1+2.3228756555x_2=0`
`=>x_1=-2.3228756555x_2`
`:.` eigenvectors corresponding to the eigenvalue `lamda=0.7084973779` is
Let `x_2=1`
2. Eigenvectors for `lamda=11.2915026221`
2. Eigenvectors for `lamda=11.2915026221`
`A-lamdaI = ` | | - `11.2915026221` | |
= | | - | | 11.2915026221 | 0 | | | 0 | 11.2915026221 | |
|
= | | `-9.2915026221` | `3` | | | `4` | `-1.2915026221` | |
|
Now, reduce this matrix
`R_1 larr R_1-:(-9.2915026221)`
= | | `1` | `-0.3228756555` | | | `4` | `-1.2915026221` | |
|
`R_2 larr R_2-4xx R_1`
The system associated with the eigenvalue `lamda=11.2915026221`
`=>x_1-0.3228756555x_2=0`
`=>x_1=0.3228756555x_2`
`:.` eigenvectors corresponding to the eigenvalue `lamda=11.2915026221` is
Let `x_2=1`
2. The eigenvectors compose the columns of matrix P
`:.P` | = | | `-2.3228756555` | `0.3228756555` | | | `1` | `1` | |
|
1. The diagonal matrix D is composed of the eigenvalues
`:.D` | = | | `0.7084973779` | `0` | | | `0` | `11.2915026221` | |
|
3. Now find `P^-1`
`|P|` | = | | `-2.3228756555` | `0.3228756555` | | | `1` | `1` | |
|
`=(-2.3228756555) × 1 - 0.3228756555 × 1`
`=(-2.3228756555) - 0.3228756555`
`=-2.6457513111`
`Adj(P)` | = | Adj | | `-2.3228756555` | `0.3228756555` | | | `1` | `1` | |
|
|
= | | `+(1)` | `-(1)` | | | `-(0.3228756555)` | `+(-2.3228756555)` | |
| T |
|
= | | `1` | `-1` | | | `-0.3228756555` | `-2.3228756555` | |
| T |
|
= | | `1` | `-0.3228756555` | | | `-1` | `-2.3228756555` | |
|
`"Now, "P^(-1)=1/|P| × Adj(P)`
= | `1/(-2.6457513111)` × | | `1` | `-0.3228756555` | | | `-1` | `-2.3228756555` | |
|
= | | `-0.377964473` | `0.122035527` | | | `0.377964473` | `0.877964473` | |
|
`:.P^-1` | = | | `-0.377964473` | `0.122035527` | | | `0.377964473` | `0.877964473` | |
|
4. Now checking `A=PDP^(-1)` ?
`P×D` | = | | `-2.3228756555` | `0.3228756555` | | | `1` | `1` | |
| × | | `0.7084973779` | `0` | | | `0` | `11.2915026221` | |
|
= | | `-2.3228756555×0.7084973779+0.3228756555×0` | `-2.3228756555×0+0.3228756555×11.2915026221` | | | `1×0.7084973779+1×0` | `1×0+1×11.2915026221` | |
|
= | | `-1.6457513111+0` | `0+3.6457513111` | | | `0.7084973779+0` | `0+11.2915026221` | |
|
= | | `-1.6457513111` | `3.6457513111` | | | `0.7084973779` | `11.2915026221` | |
|
`(P × D)×(P^-1)` | = | | `-1.6457513111` | `3.6457513111` | | | `0.7084973779` | `11.2915026221` | |
| × | | `-0.377964473` | `0.122035527` | | | `0.377964473` | `0.877964473` | |
|
= | | `-1.6457513111×(-0.377964473)+3.6457513111×0.377964473` | `-1.6457513111×0.122035527+3.6457513111×0.877964473` | | | `0.7084973779×(-0.377964473)+11.2915026221×0.377964473` | `0.7084973779×0.122035527+11.2915026221×0.877964473` | |
|
= | | `0.622035527+1.377964473` | `-0.2008401285+3.2008401285` | | | `-0.2677868381+4.2677868381` | `0.0864618509+9.9135381491` | |
|
Solution is possible.
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then