Diagonal Matrix Example [[1,1,1],[-1,-3,-3],[2,4,4]]

Find Matrix Diagonalization ...
`[[1,1,1],[-1,-3,-3],[2,4,4]]`

Solution:

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

Here `A`

`1`	`1`	`1`
`-1`	`-3`	`-3`
`2`	`4`	`4`

Find eigenvalues of the matrix `A`

`|A-lamdaI|=0`

`(1-lamda)`	`1`	`1`
`-1`	`(-3-lamda)`	`-3`
`2`	`4`	`(4-lamda)`

= 0

`:.(1-lamda)((-3-lamda) × (4-lamda) - (-3) × 4)-1((-1) × (4-lamda) - (-3) × 2)+1((-1) × 4 - (-3-lamda) × 2)=0`

`:.(1-lamda)((-12-lamda+lamda^2)-(-12))-1((-4+lamda)-(-6))+1((-4)-(-6-2lamda))=0`

`:.(1-lamda)(-lamda+lamda^2)-1(2+lamda)+1(2+2lamda)=0`

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

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

`:.-lamda^2(lamda-2)=0`

`:.lamda^2=0 or (lamda-2)=0`

`:.lamda=0 or lamda=2`

`:.` The eigenvalues of the matrix `A` are given by `lamda=0,2`

1. Eigenvectors for `lamda=0`

`A-lamdaI = `

1	1	1
-1	-3	-3
2	4	4

- `0`

1	0	0
0	1	0
0	0	1

`1`	`1`	`1`
`-1`	`-3`	`-3`
`2`	`4`	`4`

Now, reduce this matrix
`R_2 larr R_2+ R_1`

`1`	`1`	`1`
`0`	`-2`	`-2`
`2`	`4`	`4`

`R_3 larr R_3-2xx R_1`

`1`	`1`	`1`
`0`	`-2`	`-2`
`0`	`2`	`2`

`R_2 larr R_2-:(-2)`

`1`	`1`	`1`
`0`	`1`	`1`
`0`	`2`	`2`

`R_1 larr R_1- R_2`

`1`	`0`	`0`
`0`	`1`	`1`
`0`	`2`	`2`

`R_3 larr R_3-2xx R_2`

`1`	`0`	`0`
`0`	`1`	`1`
`0`	`0`	`0`

The system associated with the eigenvalue `lamda=0`

`(A-0I)`

	`x_1`
	`x_2`
	`x_3`

`1`	`0`	`0`
`0`	`1`	`1`
`0`	`0`	`0`

	`x_1`
	`x_2`
	`x_3`

	`0`
	`0`
	`0`

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

`=>x_1=0,x_2=-x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0` is

`v=`

	`0`
	`-x_3`
	`x_3`

Let `x_3=1`

`v_1=`

	`0`
	`-1`
	`1`

`v_1=`

	`0`
	`-1`
	`1`

3. Eigenvectors for `lamda=2`

`A-lamdaI = `

1	1	1
-1	-3	-3
2	4	4

- `2`

1	0	0
0	1	0
0	0	1

1	1	1
-1	-3	-3
2	4	4

2	0	0
0	2	0
0	0	2

`-1`	`1`	`1`
`-1`	`-5`	`-3`
`2`	`4`	`2`

Now, reduce this matrix
`R_1 larr R_1-:(-1)`

`1`	`-1`	`-1`
`-1`	`-5`	`-3`
`2`	`4`	`2`

`R_2 larr R_2+ R_1`

`1`	`-1`	`-1`
`0`	`-6`	`-4`
`2`	`4`	`2`

`R_3 larr R_3-2xx R_1`

`1`	`-1`	`-1`
`0`	`-6`	`-4`
`0`	`6`	`4`

`R_2 larr R_2-:(-6)`

`1`	`-1`	`-1`
`0`	`1`	`0.6666666667`
`0`	`6`	`4`

`R_1 larr R_1+ R_2`

`1`	`0`	`-0.3333333333`
`0`	`1`	`0.6666666667`
`0`	`6`	`4`

`R_3 larr R_3-6xx R_2`

`1`	`0`	`-0.3333333333`
`0`	`1`	`0.6666666667`
`0`	`0`	`0`

The system associated with the eigenvalue `lamda=2`

`(A-2I)`

	`x_1`
	`x_2`
	`x_3`

`1`	`0`	`-0.3333333333`
`0`	`1`	`0.6666666667`
`0`	`0`	`0`

	`x_1`
	`x_2`
	`x_3`

	`0`
	`0`
	`0`

`=>x_1-0.3333333333x_3=0,x_2+0.6666666667x_3=0`

`=>x_1=0.3333333333x_3,x_2=-0.6666666667x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=2` is

`v=`

	`0.3333333333x_3`
	`-0.6666666667x_3`
	`x_3`

Let `x_3=1`

`v_2=`

	`0.3333333333`
	`-0.6666666667`
	`1`

`v_2=`

	`0.3333333333`
	`-0.6666666667`
	`1`

To allow diagonalization, the number of eigenvectors must equal the matrix dimentions.
There are 2 eigenvectors which is less then 3 and therefore the matix cannot be diagonalized.
= not diagonalizable

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

3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`