Diagonal Matrix Example [[2,3],[4,10]]

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`

Here `A`

	`2`	`3`
	`4`	`10`

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`

`A-lamdaI = `

	2	3
	4	10

- `0.7084973779`

	1	0
	0	1

	2	3
	4	10

	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`

	`1`	`2.3228756555`
	`0`	`0`

The system associated with the eigenvalue `lamda=0.7084973779`

`(A-0.7084973779I)`

	`x_1`
	`x_2`

	`1`	`2.3228756555`
	`0`	`0`

	`x_1`
	`x_2`

	`0`
	`0`

`=>x_1+2.3228756555x_2=0`

`=>x_1=-2.3228756555x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.7084973779` is

`v=`

	`-2.3228756555x_2`
	`x_2`

Let `x_2=1`

`v_1=`

	`-2.3228756555`
	`1`

`v_1=`

	`-2.3228756555`
	`1`

2. Eigenvectors for `lamda=11.2915026221`

`A-lamdaI = `

	2	3
	4	10

- `11.2915026221`

	1	0
	0	1

	2	3
	4	10

	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`

	`1`	`-0.3228756555`
	`0`	`0`

The system associated with the eigenvalue `lamda=11.2915026221`

`(A-11.2915026221I)`

	`x_1`
	`x_2`

	`1`	`-0.3228756555`
	`0`	`0`

	`x_1`
	`x_2`

	`0`
	`0`

`=>x_1-0.3228756555x_2=0`

`=>x_1=0.3228756555x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=11.2915026221` is

`v=`

	`0.3228756555x_2`
	`x_2`

Let `x_2=1`

`v_2=`

	`0.3228756555`
	`1`

`v_2=`

	`0.3228756555`
	`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)`

	`1`	`-1`
	`-0.3228756555`	`-2.3228756555`

	`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`

	`2`	`3`
	`4`	`10`

`:.P*D*P^-1`

	`2`	`3`
	`4`	`10`

And `A`

	`2`	`3`
	`4`	`10`

Solution is possible.

This material is intended as a summary. Use your textbook for detail explanation.
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4. Example `[[2,3],[4,10]]`