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Home > Matrix & Vector > Matrix operation > Matrix Diagonalization example
11. Diagonal Matrix example ( Enter your problem )
( Enter your problem )
  1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
  2. Example `[[3,2,4],[2,0,2],[4,2,3]]`
  3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
  4. Example `[[2,3],[4,10]]`

1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`


1. Find Matrix Diagonalization ...
`[[8,-6,2],[-6,7,-4],[2,-4,3]]`

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


Here `A` = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`



Find eigenvalues of the matrix `A`

`|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`

`A-lamdaI = `
8-62
-67-4
2-43
 - `0` 
100
010
001


 = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


Now, reduce this matrix
`R_1 larr R_1-:8`

 = 
`1``-0.75``0.25`
`-6``7``-4`
`2``-4``3`


`R_2 larr R_2+6xx R_1`

 = 
`1``-0.75``0.25`
`0``2.5``-2.5`
`2``-4``3`


`R_3 larr R_3-2xx R_1`

 = 
`1``-0.75``0.25`
`0``2.5``-2.5`
`0``-2.5``2.5`


`R_2 larr R_2-:2.5`

 = 
`1``-0.75``0.25`
`0``1``-1`
`0``-2.5``2.5`


`R_1 larr R_1+0.75xx R_2`

 = 
`1``0``-0.5`
`0``1``-1`
`0``-2.5``2.5`


`R_3 larr R_3+2.5xx R_2`

 = 
`1``0``-0.5`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=0`

`(A-0I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-0.5`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`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

`v=`
`0.5x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_1=`
`0.5`
`1`
`1`
`v_1=`
`0.5`
`1`
`1`


2. Eigenvectors for `lamda=3`



2. Eigenvectors for `lamda=3`

`A-lamdaI = `
8-62
-67-4
2-43
 - `3` 
100
010
001


 = 
8-62
-67-4
2-43
 - 
300
030
003

 = 
`5``-6``2`
`-6``4``-4`
`2``-4``0`


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

 = 
`1``-1.2``0.4`
`-6``4``-4`
`2``-4``0`


`R_2 larr R_2+6xx R_1`

 = 
`1``-1.2``0.4`
`0``-3.2``-1.6`
`2``-4``0`


`R_3 larr R_3-2xx R_1`

 = 
`1``-1.2``0.4`
`0``-3.2``-1.6`
`0``-1.6``-0.8`


`R_2 larr R_2-:(-3.2)`

 = 
`1``-1.2``0.4`
`0``1``0.5`
`0``-1.6``-0.8`


`R_1 larr R_1+1.2xx R_2`

 = 
`1``0``1`
`0``1``0.5`
`0``-1.6``-0.8`


`R_3 larr R_3+1.6xx R_2`

 = 
`1``0``1`
`0``1``0.5`
`0``0``0`


The system associated with the eigenvalue `lamda=3`

`(A-3I)`
`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=3` is

`v=`
`-x_3`
`-0.5x_3`
`x_3`


Let `x_3=1`

`v_2=`
`-1`
`-0.5`
`1`
`v_2=`
`-1`
`-0.5`
`1`


3. Eigenvectors for `lamda=15`



3. Eigenvectors for `lamda=15`

`A-lamdaI = `
8-62
-67-4
2-43
 - `15` 
100
010
001


 = 
8-62
-67-4
2-43
 - 
1500
0150
0015

 = 
`-7``-6``2`
`-6``-8``-4`
`2``-4``-12`


Now, reduce this matrix
`R_1 larr R_1-:(-7)`

 = 
`1``0.8571428571``-0.2857142857`
`-6``-8``-4`
`2``-4``-12`


`R_2 larr R_2+6xx R_1`

 = 
`1``0.8571428571``-0.2857142857`
`0``-2.8571428571``-5.7142857143`
`2``-4``-12`


`R_3 larr R_3-2xx R_1`

 = 
`1``0.8571428571``-0.2857142857`
`0``-2.8571428571``-5.7142857143`
`0``-5.7142857143``-11.4285714286`


`R_2 larr R_2-:(-2.8571428571)`

 = 
`1``0.8571428571``-0.2857142857`
`0``1``2`
`0``-5.7142857143``-11.4285714286`


`R_1 larr R_1-0.8571428571xx R_2`

 = 
`1``0``-2`
`0``1``2`
`0``-5.7142857143``-11.4285714286`


`R_3 larr R_3+5.7142857143xx R_2`

 = 
`1``0``-2`
`0``1``2`
`0``0``0`


The system associated with the eigenvalue `lamda=15`

`(A-15I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-2`
`0``1``2`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`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

`v=`
`2x_3`
`-2x_3`
`x_3`


Let `x_3=1`

`v_3=`
`2`
`-2`
`1`
`v_3=`
`2`
`-2`
`1`



2. The eigenvectors compose the columns of matrix P
`:.P` = 
`0.5``-1``2`
`1``-0.5``-2`
`1``1``1`



1. The diagonal matrix D is composed of the eigenvalues
`:.D` = 
`0``0``0`
`0``3``0`
`0``0``15`


3. Now find `P^-1`

`|P|` = 
 `0.5`  `-1`  `2` 
 `1`  `-0.5`  `-2` 
 `1`  `1`  `1` 


 =
 `0.5` × 
 `-0.5`  `-2` 
 `1`  `1` 
 `-(-1)` × 
 `1`  `-2` 
 `1`  `1` 
 `+2` × 
 `1`  `-0.5` 
 `1`  `1` 


`=0.5 xx ((-0.5) × 1 - (-2) × 1)-(-1) xx (1 × 1 - (-2) × 1)+2 xx (1 × 1 - (-0.5) × 1)`

`=0.5 xx (-0.5 +2)-(-1) xx (1 +2)+2 xx (1 +0.5)`

`=0.5 xx (1.5)-(-1) xx (3)+2 xx (1.5)`

`= 0.75 +3 +3`

`=6.75`


`Adj(P)` = 
Adj
`0.5``-1``2`
`1``-0.5``-2`
`1``1``1`


 = 
 + 
 `-0.5`  `-2` 
 `1`  `1` 
 - 
 `1`  `-2` 
 `1`  `1` 
 + 
 `1`  `-0.5` 
 `1`  `1` 
 - 
 `-1`  `2` 
 `1`  `1` 
 + 
 `0.5`  `2` 
 `1`  `1` 
 - 
 `0.5`  `-1` 
 `1`  `1` 
 + 
 `-1`  `2` 
 `-0.5`  `-2` 
 - 
 `0.5`  `2` 
 `1`  `-2` 
 + 
 `0.5`  `-1` 
 `1`  `-0.5` 
T


 = 
`+((-0.5) × 1 - (-2) × 1)``-(1 × 1 - (-2) × 1)``+(1 × 1 - (-0.5) × 1)`
`-((-1) × 1 - 2 × 1)``+(0.5 × 1 - 2 × 1)``-(0.5 × 1 - (-1) × 1)`
`+((-1) × (-2) - 2 × (-0.5))``-(0.5 × (-2) - 2 × 1)``+(0.5 × (-0.5) - (-1) × 1)`
T


 = 
`+((-0.5) - -2)``-(1 - -2)``+(1 - -0.5)`
`-((-1) - 2)``+(0.5 - 2)``-(0.5 - -1)`
`+(2 - -1)``-((-1) - 2)``+((-0.25) - -1)`
T


 = 
`1.5``-3``1.5`
`3``-1.5``-1.5`
`3``3``0.75`
T


 = 
`1.5``3``3`
`-3``-1.5``3`
`1.5``-1.5``0.75`


`"Now, "P^(-1)=1/|P| × Adj(P)`

 = `1/(6.75)` ×
`1.5``3``3`
`-3``-1.5``3`
`1.5``-1.5``0.75`


 = 
`0.2222222222``0.4444444444``0.4444444444`
`-0.4444444444``-0.2222222222``0.4444444444`
`0.2222222222``-0.2222222222``0.1111111111`


`:.P^-1` = 
`0.2222222222``0.4444444444``0.4444444444`
`-0.4444444444``-0.2222222222``0.4444444444`
`0.2222222222``-0.2222222222``0.1111111111`



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

`P×D`=
`0.5``-1``2`
`1``-0.5``-2`
`1``1``1`
×
`0``0``0`
`0``3``0`
`0``0``15`


=
`0.5×0+(-1)×0+2×0``0.5×0+(-1)×3+2×0``0.5×0+(-1)×0+2×15`
`1×0+(-0.5)×0+(-2)×0``1×0+(-0.5)×3+(-2)×0``1×0+(-0.5)×0+(-2)×15`
`1×0+1×0+1×0``1×0+1×3+1×0``1×0+1×0+1×15`


=
`0+0+0``0+(-3)+0``0+0+30`
`0+0+0``0+(-1.5)+0``0+0+(-30)`
`0+0+0``0+3+0``0+0+15`


=
`0``-3``30`
`0``-1.5``-30`
`0``3``15`


`(P × D)×(P^-1)`=
`0``-3``30`
`0``-1.5``-30`
`0``3``15`
×
`0.2222222222``0.4444444444``0.4444444444`
`-0.4444444444``-0.2222222222``0.4444444444`
`0.2222222222``-0.2222222222``0.1111111111`


=
`0×0.2222222222+(-3)×(-0.4444444444)+30×0.2222222222``0×0.4444444444+(-3)×(-0.2222222222)+30×(-0.2222222222)``0×0.4444444444+(-3)×0.4444444444+30×0.1111111111`
`0×0.2222222222+(-1.5)×(-0.4444444444)+(-30)×0.2222222222``0×0.4444444444+(-1.5)×(-0.2222222222)+(-30)×(-0.2222222222)``0×0.4444444444+(-1.5)×0.4444444444+(-30)×0.1111111111`
`0×0.2222222222+3×(-0.4444444444)+15×0.2222222222``0×0.4444444444+3×(-0.2222222222)+15×(-0.2222222222)``0×0.4444444444+3×0.4444444444+15×0.1111111111`


=
`0+1.3333333333+6.6666666667``0+0.6666666667+(-6.6666666667)``0+(-1.3333333333)+3.3333333333`
`0+0.6666666667+(-6.6666666667)``0+0.3333333333+6.6666666667``0+(-0.6666666667)+(-3.3333333333)`
`0+(-1.3333333333)+3.3333333333``0+(-0.6666666667)+(-3.3333333333)``0+1.3333333333+1.6666666667`


=
`8``-6``2`
`-6``7``-4`
`2``-4``3`


`:.P*D*P^-1` = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


And `A` = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


Solution is possible.


This material is intended as a summary. Use your textbook for detail explanation.
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