Diagonal Matrix Example [[3,2,4],[2,0,2],[4,2,3]]

Find Matrix Diagonalization ...
`[[3,2,4],[2,0,2],[4,2,3]]`

Solution:

A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such that `A=PDP^-1`

Here `A`

`3`	`2`	`4`
`2`	`0`	`2`
`4`	`2`	`3`

Find eigenvalues of the matrix `A`

`|A-lamdaI|=0`

`(3-lamda)`	`2`	`4`
`2`	`(-lamda)`	`2`
`4`	`2`	`(3-lamda)`

= 0

`:.(3-lamda)((-lamda) × (3-lamda) - 2 × 2)-2(2 × (3-lamda) - 2 × 4)+4(2 × 2 - (-lamda) × 4)=0`

`:.(3-lamda)((-3lamda+lamda^2)-4)-2((6-2lamda)-8)+4(4-(-4lamda))=0`

`:.(3-lamda)(-4-3lamda+lamda^2)-2(-2-2lamda)+4(4+4lamda)=0`

`:. (-12-5lamda+6lamda^2-lamda^3)-(-4-4lamda)+(16+16lamda)=0`

`:.(-lamda^3+6lamda^2+15lamda+8)=0`

`:.-(lamda+1)(lamda+1)(lamda-8)=0`

`:.(lamda+1)=0 or (lamda+1)=0 or (lamda-8)=0`

`:.lamda=-1 or lamda=-1 or lamda=8`

`:.` The eigenvalues of the matrix `A` are given by `lamda=-1,8`

1. Eigenvectors for `lamda=-1`

`A-lamdaI = `

3	2	4
2	0	2
4	2	3

- `(-1)`

1	0	0
0	1	0
0	0	1

3	2	4
2	0	2
4	2	3

-1	0	0
0	-1	0
0	0	-1

`4`	`2`	`4`
`2`	`1`	`2`
`4`	`2`	`4`

Now, reduce this matrix
`R_1 larr R_1-:4`

`1`	`0.5`	`1`
`2`	`1`	`2`
`4`	`2`	`4`

`R_2 larr R_2-2xx R_1`

`1`	`0.5`	`1`
`0`	`0`	`0`
`4`	`2`	`4`

`R_3 larr R_3-4xx R_1`

`1`	`0.5`	`1`
`0`	`0`	`0`
`0`	`0`	`0`

The system associated with the eigenvalue `lamda=-1`

`(A-(-1)I)`

	`x_1`
	`x_2`
	`x_3`

`1`	`0.5`	`1`
`0`	`0`	`0`
`0`	`0`	`0`

	`x_1`
	`x_2`
	`x_3`

	`0`
	`0`
	`0`

`=>x_1+0.5x_2+x_3=0`

`=>x_1=-0.5x_2-x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=-1` is

`v=`

	`-0.5x_2-x_3`
	`x_2`
	`x_3`

Let `x_2=1,x_3=0`

`v_1=`

	`-0.5`
	`1`
	`0`

Let `x_2=0,x_3=1`

`v_2=`

	`-1`
	`0`
	`1`

`v_1=`

	`-0.5`
	`1`
	`0`

`v_2=`

	`-1`
	`0`
	`1`

1. Orthogonal Eigenvectors for `lamda=-1`

`v_1=`

	`-0.5`
	`1`
	`0`

`v_2=`

	`-0.8`
	`-0.4`
	`1`

3. Eigenvectors for `lamda=8`

`A-lamdaI = `

3	2	4
2	0	2
4	2	3

- `8`

1	0	0
0	1	0
0	0	1

3	2	4
2	0	2
4	2	3

8	0	0
0	8	0
0	0	8

`-5`	`2`	`4`
`2`	`-8`	`2`
`4`	`2`	`-5`

Now, reduce this matrix
`R_1 larr R_1-:(-5)`

`1`	`-0.4`	`-0.8`
`2`	`-8`	`2`
`4`	`2`	`-5`

`R_2 larr R_2-2xx R_1`

`1`	`-0.4`	`-0.8`
`0`	`-7.2`	`3.6`
`4`	`2`	`-5`

`R_3 larr R_3-4xx R_1`

`1`	`-0.4`	`-0.8`
`0`	`-7.2`	`3.6`
`0`	`3.6`	`-1.8`

`R_2 larr R_2-:(-7.2)`

`1`	`-0.4`	`-0.8`
`0`	`1`	`-0.5`
`0`	`3.6`	`-1.8`

`R_1 larr R_1+0.4xx R_2`

`1`	`0`	`-1`
`0`	`1`	`-0.5`
`0`	`3.6`	`-1.8`

`R_3 larr R_3-3.6xx R_2`

`1`	`0`	`-1`
`0`	`1`	`-0.5`
`0`	`0`	`0`

The system associated with the eigenvalue `lamda=8`

`(A-8I)`

	`x_1`
	`x_2`
	`x_3`

`1`	`0`	`-1`
`0`	`1`	`-0.5`
`0`	`0`	`0`

	`x_1`
	`x_2`
	`x_3`

	`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=8` is

`v=`

	`x_3`
	`0.5x_3`
	`x_3`

Let `x_3=1`

`v_3=`

	`1`
	`0.5`
	`1`

`v_3=`

	`1`
	`0.5`
	`1`

2. The eigenvectors compose the columns of matrix P

`:.P`

`-0.5`	`-0.8`	`1`
`1`	`-0.4`	`0.5`
`0`	`1`	`1`

1. The diagonal matrix D is composed of the eigenvalues

`:.D`

`-1`	`0`	`0`
`0`	`-1`	`0`
`0`	`0`	`8`

3. Now find `P^-1`

`|P|`

`-0.5`	`-0.8`	`1`
`1`	`-0.4`	`0.5`
`0`	`1`	`1`

`(-0.5)`

	`-0.4`	`0.5`
	`1`	`1`

`-(-0.8)`

	`1`	`0.5`
	`0`	`1`

`+1`

	`1`	`-0.4`
	`0`	`1`

`=(-0.5) xx ((-0.4) × 1 - 0.5 × 1)-(-0.8) xx (1 × 1 - 0.5 × 0)+1 xx (1 × 1 - (-0.4) × 0)`

`=(-0.5) xx (-0.4 +(-0.5))-(-0.8) xx (1 +0)+1 xx (1 +0)`

`=(-0.5) xx (-0.9)-(-0.8) xx (1)+1 xx (1)`

`= 0.45 +0.8 +1`

`=2.25`

`Adj(P)`

Adj

`-0.5`	`-0.8`	`1`
`1`	`-0.4`	`0.5`
`0`	`1`	`1`

	`-0.4`	`0.5`
	`1`	`1`

	`1`	`0.5`
	`0`	`1`

	`1`	`-0.4`
	`0`	`1`

	`-0.8`	`1`
	`1`	`1`

	`-0.5`	`1`
	`0`	`1`

	`-0.5`	`-0.8`
	`0`	`1`

	`-0.8`	`1`
	`-0.4`	`0.5`

	`-0.5`	`1`
	`1`	`0.5`

	`-0.5`	`-0.8`
	`1`	`-0.4`

`+((-0.4) × 1 - 0.5 × 1)`	`-(1 × 1 - 0.5 × 0)`	`+(1 × 1 - (-0.4) × 0)`
`-((-0.8) × 1 - 1 × 1)`	`+((-0.5) × 1 - 1 × 0)`	`-((-0.5) × 1 - (-0.8) × 0)`
`+((-0.8) × 0.5 - 1 × (-0.4))`	`-((-0.5) × 0.5 - 1 × 1)`	`+((-0.5) × (-0.4) - (-0.8) × 1)`

`+((-0.4) - 0.5)`	`-(1 - 0)`	`+(1 - 0)`
`-((-0.8) - 1)`	`+((-0.5) - 0)`	`-((-0.5) - 0)`
`+((-0.4) - -0.4)`	`-((-0.25) - 1)`	`+(0.2 - -0.8)`

`-0.9`	`-1`	`1`
`1.8`	`-0.5`	`0.5`
`0`	`1.25`	`1`

`-0.9`	`1.8`	`0`
`-1`	`-0.5`	`1.25`
`1`	`0.5`	`1`

`"Now, "P^(-1)=1/|P| × Adj(P)`

`1/(2.25)` ×

`-0.9`	`1.8`	`0`
`-1`	`-0.5`	`1.25`
`1`	`0.5`	`1`

`-0.4`	`0.8`	`0`
`-0.4444444444`	`-0.2222222222`	`0.5555555556`
`0.4444444444`	`0.2222222222`	`0.4444444444`

`:.P^-1`

`-0.4`	`0.8`	`0`
`-0.4444444444`	`-0.2222222222`	`0.5555555556`
`0.4444444444`	`0.2222222222`	`0.4444444444`

4. Now checking `A=PDP^(-1)` ?

`P×D`

`-0.5`	`-0.8`	`1`
`1`	`-0.4`	`0.5`
`0`	`1`	`1`

`-1`	`0`	`0`
`0`	`-1`	`0`
`0`	`0`	`8`

`-0.5×(-1)+(-0.8)×0+1×0`	`-0.5×0+(-0.8)×(-1)+1×0`	`-0.5×0+(-0.8)×0+1×8`
`1×(-1)+(-0.4)×0+0.5×0`	`1×0+(-0.4)×(-1)+0.5×0`	`1×0+(-0.4)×0+0.5×8`
`0×(-1)+1×0+1×0`	`0×0+1×(-1)+1×0`	`0×0+1×0+1×8`

`0.5+0+0`	`0+0.8+0`	`0+0+8`
`-1+0+0`	`0+0.4+0`	`0+0+4`
`0+0+0`	`0+(-1)+0`	`0+0+8`

`0.5`	`0.8`	`8`
`-1`	`0.4`	`4`
`0`	`-1`	`8`

`(P × D)×(P^-1)`

`0.5`	`0.8`	`8`
`-1`	`0.4`	`4`
`0`	`-1`	`8`

`-0.4`	`0.8`	`0`
`-0.4444444444`	`-0.2222222222`	`0.5555555556`
`0.4444444444`	`0.2222222222`	`0.4444444444`

`0.5×(-0.4)+0.8×(-0.4444444444)+8×0.4444444444`	`0.5×0.8+0.8×(-0.2222222222)+8×0.2222222222`	`0.5×0+0.8×0.5555555556+8×0.4444444444`
`-1×(-0.4)+0.4×(-0.4444444444)+4×0.4444444444`	`-1×0.8+0.4×(-0.2222222222)+4×0.2222222222`	`-1×0+0.4×0.5555555556+4×0.4444444444`
`0×(-0.4)+(-1)×(-0.4444444444)+8×0.4444444444`	`0×0.8+(-1)×(-0.2222222222)+8×0.2222222222`	`0×0+(-1)×0.5555555556+8×0.4444444444`

`-0.2+(-0.3555555556)+3.5555555556`	`0.4+(-0.1777777778)+1.7777777778`	`0+0.4444444444+3.5555555556`
`0.4+(-0.1777777778)+1.7777777778`	`-0.8+(-0.0888888889)+0.8888888889`	`0+0.2222222222+1.7777777778`
`0+0.4444444444+3.5555555556`	`0+0.2222222222+1.7777777778`	`0+(-0.5555555556)+3.5555555556`

`3`	`2`	`4`
`2`	`0`	`2`
`4`	`2`	`3`

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

`3`	`2`	`4`
`2`	`0`	`2`
`4`	`2`	`3`

And `A`

`3`	`2`	`4`
`2`	`0`	`2`
`4`	`2`	`3`

Solution is possible.

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

2. Example `[[3,2,4],[2,0,2],[4,2,3]]`