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Home > Matrix & Vector calculators > Diagonal Matrix 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]]`
 		Other related methods
  1. Transforming matrix to Row Echelon Form
  2. Transforming matrix to Reduced Row Echelon Form
  3. Rank of matrix
  4. Characteristic polynomial of matrix
  5. Eigenvalues
  6. Eigenvectors (Eigenspace)
  7. Triangular Matrix
  8. LU decomposition using Gauss Elimination method of matrix
  9. LU decomposition using Doolittle's method of matrix
  10. LU decomposition using Crout's method of matrix
  11. Diagonal Matrix
  12. Cholesky Decomposition
  13. QR Decomposition (Gram Schmidt Method)
  14. QR Decomposition (Householder Method)
  15. LQ Decomposition
  16. Pivots
  17. Singular Value Decomposition (SVD)
  18. Moore-Penrose Pseudoinverse
  19. Power Method for dominant eigenvalue
  20. determinants using Sarrus Rule
  21. determinants using properties of determinants
  22. Row Space
  23. Column Space
  24. Null Space
10. LU decomposition using Crout's method of matrix
(Previous method)		2. Example `[[3,2,4],[2,0,2],[4,2,3]]`
(Next example)

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


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

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


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`

 `(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`

`:.` The eigenvalues of the matrix A are given by `lamda=0,3,15`


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



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` `1=8-:8`
`R_1 larr R_1-:8`		 `-3/4` `-3/4=-6-:8`
`R_1 larr R_1-:8`		 `1/4` `1/4=2-:8`
`R_1 larr R_1-:8`
`-6``7``-4`
`2``-4``3`


`R_2 larr R_2+6xx R_1`

 = 
`1``-3/4``1/4`
 `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1`		 `5/2` `5/2=7+6xx-3/4`
`R_2 larr R_2+6xx R_1`		 `-5/2` `-5/2=-4+6xx1/4`
`R_2 larr R_2+6xx R_1`
`2``-4``3`


`R_3 larr R_3-2xx R_1`

 = 
`1``-3/4``1/4`
`0``5/2``-5/2`
 `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1`		 `-5/2` `-5/2=-4-2xx-3/4`
`R_3 larr R_3-2xx R_1`		 `5/2` `5/2=3-2xx1/4`
`R_3 larr R_3-2xx R_1`


`R_2 larr R_2xx2/5`

 = 
`1``-3/4``1/4`
 `0` `0=0xx2/5`
`R_2 larr R_2xx2/5`		 `1` `1=5/2xx2/5`
`R_2 larr R_2xx2/5`		 `-1` `-1=-5/2xx2/5`
`R_2 larr R_2xx2/5`
`0``-5/2``5/2`


`R_1 larr R_1+3/4xx R_2`

 = 
 `1` `1=1+3/4xx0`
`R_1 larr R_1+3/4xx R_2`		 `0` `0=-3/4+3/4xx1`
`R_1 larr R_1+3/4xx R_2`		 `-1/2` `-1/2=1/4+3/4xx-1`
`R_1 larr R_1+3/4xx R_2`
`0``1``-1`
`0``-5/2``5/2`


`R_3 larr R_3+5/2xx R_2`

 = 
`1``0``-1/2`
`0``1``-1`
 `0` `0=0+5/2xx0`
`R_3 larr R_3+5/2xx R_2`		 `0` `0=-5/2+5/2xx1`
`R_3 larr R_3+5/2xx R_2`		 `0` `0=5/2+5/2xx-1`
`R_3 larr R_3+5/2xx R_2`


The system associated with the eigenvalue `lamda=0`

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


`=>x_1-1/2x_3=0,x_2-x_3=0`

`=>x_1=1/2x_3,x_2=x_3`

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

`v=`
`1/2x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_1=`
`1/2`
`1`
`1`



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=5-:5`
`R_1 larr R_1-:5`		 `-6/5` `-6/5=-6-:5`
`R_1 larr R_1-:5`		 `2/5` `2/5=2-:5`
`R_1 larr R_1-:5`
`-6``4``-4`
`2``-4``0`


`R_2 larr R_2+6xx R_1`

 = 
`1``-6/5``2/5`
 `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1`		 `-16/5` `-16/5=4+6xx-6/5`
`R_2 larr R_2+6xx R_1`		 `-8/5` `-8/5=-4+6xx2/5`
`R_2 larr R_2+6xx R_1`
`2``-4``0`


`R_3 larr R_3-2xx R_1`

 = 
`1``-6/5``2/5`
`0``-16/5``-8/5`
 `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1`		 `-8/5` `-8/5=-4-2xx-6/5`
`R_3 larr R_3-2xx R_1`		 `-4/5` `-4/5=0-2xx2/5`
`R_3 larr R_3-2xx R_1`


`R_2 larr R_2xx-5/16`

 = 
`1``-6/5``2/5`
 `0` `0=0xx-5/16`
`R_2 larr R_2xx-5/16`		 `1` `1=-16/5xx-5/16`
`R_2 larr R_2xx-5/16`		 `1/2` `1/2=-8/5xx-5/16`
`R_2 larr R_2xx-5/16`
`0``-8/5``-4/5`


`R_1 larr R_1+6/5xx R_2`

 = 
 `1` `1=1+6/5xx0`
`R_1 larr R_1+6/5xx R_2`		 `0` `0=-6/5+6/5xx1`
`R_1 larr R_1+6/5xx R_2`		 `1` `1=2/5+6/5xx1/2`
`R_1 larr R_1+6/5xx R_2`
`0``1``1/2`
`0``-8/5``-4/5`


`R_3 larr R_3+8/5xx R_2`

 = 
`1``0``1`
`0``1``1/2`
 `0` `0=0+8/5xx0`
`R_3 larr R_3+8/5xx R_2`		 `0` `0=-8/5+8/5xx1`
`R_3 larr R_3+8/5xx R_2`		 `0` `0=-4/5+8/5xx1/2`
`R_3 larr R_3+8/5xx R_2`


The system associated with the eigenvalue `lamda=3`

`(A-3I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``1`
`0``1``1/2`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1+x_3=0,x_2+1/2x_3=0`

`=>x_1=-x_3,x_2=-1/2x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=3` is

`v=`
`-x_3`
`-1/2x_3`
`x_3`


Let `x_3=1`

`v_2=`
`-1`
`-1/2`
`1`



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` `1=-7-:-7`
`R_1 larr R_1-:-7`		 `6/7` `6/7=-6-:-7`
`R_1 larr R_1-:-7`		 `-2/7` `-2/7=2-:-7`
`R_1 larr R_1-:-7`
`-6``-8``-4`
`2``-4``-12`


`R_2 larr R_2+6xx R_1`

 = 
`1``6/7``-2/7`
 `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1`		 `-20/7` `-20/7=-8+6xx6/7`
`R_2 larr R_2+6xx R_1`		 `-40/7` `-40/7=-4+6xx-2/7`
`R_2 larr R_2+6xx R_1`
`2``-4``-12`


`R_3 larr R_3-2xx R_1`

 = 
`1``6/7``-2/7`
`0``-20/7``-40/7`
 `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1`		 `-40/7` `-40/7=-4-2xx6/7`
`R_3 larr R_3-2xx R_1`		 `-80/7` `-80/7=-12-2xx-2/7`
`R_3 larr R_3-2xx R_1`


`R_2 larr R_2xx-7/20`

 = 
`1``6/7``-2/7`
 `0` `0=0xx-7/20`
`R_2 larr R_2xx-7/20`		 `1` `1=-20/7xx-7/20`
`R_2 larr R_2xx-7/20`		 `2` `2=-40/7xx-7/20`
`R_2 larr R_2xx-7/20`
`0``-40/7``-80/7`


`R_1 larr R_1-6/7xx R_2`

 = 
 `1` `1=1-6/7xx0`
`R_1 larr R_1-6/7xx R_2`		 `0` `0=6/7-6/7xx1`
`R_1 larr R_1-6/7xx R_2`		 `-2` `-2=-2/7-6/7xx2`
`R_1 larr R_1-6/7xx R_2`
`0``1``2`
`0``-40/7``-80/7`


`R_3 larr R_3+40/7xx R_2`

 = 
`1``0``-2`
`0``1``2`
 `0` `0=0+40/7xx0`
`R_3 larr R_3+40/7xx R_2`		 `0` `0=-40/7+40/7xx1`
`R_3 larr R_3+40/7xx R_2`		 `0` `0=-80/7+40/7xx2`
`R_3 larr R_3+40/7xx R_2`


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`



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



3. Now find `P^-1`

`|P|` = 
 `1/2`  `-1`  `2` 
 `1`  `-1/2`  `-2` 
 `1`  `1`  `1` 


 =
 `1/2` × 
 `-1/2`  `-2` 
 `1`  `1` 
 `+1` × 
 `1`  `-2` 
 `1`  `1` 
 `+2` × 
 `1`  `-1/2` 
 `1`  `1` 


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

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

`=1/2 xx (3/2) - +1 xx (3) +2 xx (3/2)`

`= 3/4 +3 +3`

`=27/4`


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


 = 
 + 
 `-1/2`  `-2` 
 `1`  `1` 
 - 
 `1`  `-2` 
 `1`  `1` 
 + 
 `1`  `-1/2` 
 `1`  `1` 
 - 
 `-1`  `2` 
 `1`  `1` 
 + 
 `1/2`  `2` 
 `1`  `1` 
 - 
 `1/2`  `-1` 
 `1`  `1` 
 + 
 `-1`  `2` 
 `-1/2`  `-2` 
 - 
 `1/2`  `2` 
 `1`  `-2` 
 + 
 `1/2`  `-1` 
 `1`  `-1/2` 
T


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


 = 
`+(-1/2 +2)``-(1 +2)``+(1 +1/2)`
`-(-1 -2)``+(1/2 -2)``-(1/2 +1)`
`+(2 +1)``-(-1 -2)``+(-1/4 +1)`
T


 = 
`3/2``-3``3/2`
`3``-3/2``-3/2`
`3``3``3/4`
T


 = 
`3/2``3``3`
`-3``-3/2``3`
`3/2``-3/2``3/4`


`"Now, "P^(-1)=1/|P| × Adj(P)`

 = `1/27/4` ×
`3/2``3``3`
`-3``-3/2``3`
`3/2``-3/2``3/4`


 = 
`2/9``4/9``4/9`
`-4/9``-2/9``4/9`
`2/9``-2/9``1/9`


`:.P^-1` = 
`2/9``4/9``4/9`
`-4/9``-2/9``4/9`
`2/9``-2/9``1/9`



4. Now verify that `A=PDP^(-1)`

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


=
`1/2×0-1×0+2×0``1/2×0-1×3+2×0``1/2×0-1×0+2×15`
`1×0-1/2×0-2×0``1×0-1/2×3-2×0``1×0-1/2×0-2×15`
`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``0-3/2+0``0+0-30`
`0+0+0``0+3+0``0+0+15`


=
`0``-3``30`
`0``-3/2``-30`
`0``3``15`


`(P × D)×(P^-1)`=
`0``-3``30`
`0``-3/2``-30`
`0``3``15`
×
`2/9``4/9``4/9`
`-4/9``-2/9``4/9`
`2/9``-2/9``1/9`


=
`0×2/9-3×-4/9+30×2/9``0×4/9-3×-2/9+30×-2/9``0×4/9-3×4/9+30×1/9`
`0×2/9-3/2×-4/9-30×2/9``0×4/9-3/2×-2/9-30×-2/9``0×4/9-3/2×4/9-30×1/9`
`0×2/9+3×-4/9+15×2/9``0×4/9+3×-2/9+15×-2/9``0×4/9+3×4/9+15×1/9`


=
`0+4/3+20/3``0+2/3-20/3``0-4/3+10/3`
`0+2/3-20/3``0+1/3+20/3``0-2/3-10/3`
`0-4/3+10/3``0-2/3-10/3``0+4/3+5/3`


=
`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`


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


10. LU decomposition using Crout's method of matrix
(Previous method)		2. Example `[[3,2,4],[2,0,2],[4,2,3]]`
(Next example)


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