We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more   

AtoZmath.com
Support us
(New) All problem can be solved using search box
I want to sell my website www.AtoZmath.com with complete code
Home What's new College Algebra Games Feedback About us
Algebra Matrix & Vector Numerical Methods Statistical Methods Operation Research Word Problems Calculus Geometry Pre-Algebra


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
1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
(Previous example)		3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
(Next example)

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


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

Solution:
Here `A` = 
`3``2``4`
`2``0``2`
`4``2``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`

 `(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`

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


1. The diagonal matrix D is composed of the eigenvalues
`:.D` = 
`-1``0``0`
`0``-1``0`
`0``0``8`



1. Eigenvectors for `lamda=-1`

`A-lamdaI = `
324
202
423
 + `1` 
100
010
001


 = 
324
202
423
 + 
100
010
001

 = 
`4``2``4`
`2``1``2`
`4``2``4`


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

 = 
 `1` `1=4-:4`
`R_1 larr R_1-:4`		 `1/2` `1/2=2-:4`
`R_1 larr R_1-:4`		 `1` `1=4-:4`
`R_1 larr R_1-:4`
`2``1``2`
`4``2``4`


`R_2 larr R_2-2xx R_1`

 = 
`1``1/2``1`
 `0` `0=2-2xx1`
`R_2 larr R_2-2xx R_1`		 `0` `0=1-2xx1/2`
`R_2 larr R_2-2xx R_1`		 `0` `0=2-2xx1`
`R_2 larr R_2-2xx R_1`
`4``2``4`


`R_3 larr R_3-4xx R_1`

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


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

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


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

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

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

`v=`
`-1/2x_2-x_3`
`x_2`
`x_3`


Let `x_2=1,x_3=0`

`v_1=`
`-1/2`
`1`
`0`


Let `x_2=0,x_3=1`

`v_2=`
`-1`
`0`
`1`



3. Eigenvectors for `lamda=8`

`A-lamdaI = `
324
202
423
 - `8` 
100
010
001


 = 
324
202
423
 - 
800
080
008

 = 
`-5``2``4`
`2``-8``2`
`4``2``-5`


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

 = 
 `1` `1=-5-:-5`
`R_1 larr R_1-:-5`		 `-2/5` `-2/5=2-:-5`
`R_1 larr R_1-:-5`		 `-4/5` `-4/5=4-:-5`
`R_1 larr R_1-:-5`
`2``-8``2`
`4``2``-5`


`R_2 larr R_2-2xx R_1`

 = 
`1``-2/5``-4/5`
 `0` `0=2-2xx1`
`R_2 larr R_2-2xx R_1`		 `-36/5` `-36/5=-8-2xx-2/5`
`R_2 larr R_2-2xx R_1`		 `18/5` `18/5=2-2xx-4/5`
`R_2 larr R_2-2xx R_1`
`4``2``-5`


`R_3 larr R_3-4xx R_1`

 = 
`1``-2/5``-4/5`
`0``-36/5``18/5`
 `0` `0=4-4xx1`
`R_3 larr R_3-4xx R_1`		 `18/5` `18/5=2-4xx-2/5`
`R_3 larr R_3-4xx R_1`		 `-9/5` `-9/5=-5-4xx-4/5`
`R_3 larr R_3-4xx R_1`


`R_2 larr R_2xx-5/36`

 = 
`1``-2/5``-4/5`
 `0` `0=0xx-5/36`
`R_2 larr R_2xx-5/36`		 `1` `1=-36/5xx-5/36`
`R_2 larr R_2xx-5/36`		 `-1/2` `-1/2=18/5xx-5/36`
`R_2 larr R_2xx-5/36`
`0``18/5``-9/5`


`R_1 larr R_1+2/5xx R_2`

 = 
 `1` `1=1+2/5xx0`
`R_1 larr R_1+2/5xx R_2`		 `0` `0=-2/5+2/5xx1`
`R_1 larr R_1+2/5xx R_2`		 `-1` `-1=-4/5+2/5xx-1/2`
`R_1 larr R_1+2/5xx R_2`
`0``1``-1/2`
`0``18/5``-9/5`


`R_3 larr R_3-18/5xx R_2`

 = 
`1``0``-1`
`0``1``-1/2`
 `0` `0=0-18/5xx0`
`R_3 larr R_3-18/5xx R_2`		 `0` `0=18/5-18/5xx1`
`R_3 larr R_3-18/5xx R_2`		 `0` `0=-9/5-18/5xx-1/2`
`R_3 larr R_3-18/5xx R_2`


The system associated with the eigenvalue `lamda=8`

`(A-8I)`
`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=8` is

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


Let `x_3=1`

`v_3=`
`1`
`1/2`
`1`



2. The eigenvectors compose the columns of matrix P
`:.P` = 
`-1/2``-1``1`
`1``0``1/2`
`0``1``1`



3. Now find `P^-1`

`|P|` = 
 `-1/2`  `-1`  `1` 
 `1`  `0`  `1/2` 
 `0`  `1`  `1` 


 =
 `-1/2` × 
 `0`  `1/2` 
 `1`  `1` 
 `+1` × 
 `1`  `1/2` 
 `0`  `1` 
 `+1` × 
 `1`  `0` 
 `0`  `1` 


`=-1/2 xx (0 × 1 - 1/2 × 1) +1 xx (1 × 1 - 1/2 × 0) +1 xx (1 × 1 - 0 × 0)`

`=-1/2 xx (0 -1/2) +1 xx (1 +0) +1 xx (1 +0)`

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

`= 1/4 +1 +1`

`=9/4`


`Adj(P)` = 
Adj
`-1/2``-1``1`
`1``0``1/2`
`0``1``1`


 = 
 + 
 `0`  `1/2` 
 `1`  `1` 
 - 
 `1`  `1/2` 
 `0`  `1` 
 + 
 `1`  `0` 
 `0`  `1` 
 - 
 `-1`  `1` 
 `1`  `1` 
 + 
 `-1/2`  `1` 
 `0`  `1` 
 - 
 `-1/2`  `-1` 
 `0`  `1` 
 + 
 `-1`  `1` 
 `0`  `1/2` 
 - 
 `-1/2`  `1` 
 `1`  `1/2` 
 + 
 `-1/2`  `-1` 
 `1`  `0` 
T


 = 
`+(0 × 1 - 1/2 × 1)``-(1 × 1 - 1/2 × 0)``+(1 × 1 - 0 × 0)`
`-(-1 × 1 - 1 × 1)``+(-1/2 × 1 - 1 × 0)``-(-1/2 × 1 - (-1) × 0)`
`+(-1 × 1/2 - 1 × 0)``-(-1/2 × 1/2 - 1 × 1)``+(-1/2 × 0 - (-1) × 1)`
T


 = 
`+(0 -1/2)``-(1 +0)``+(1 +0)`
`-(-1 -1)``+(-1/2 +0)``-(-1/2 +0)`
`+(-1/2 +0)``-(-1/4 -1)``+(0 +1)`
T


 = 
`-1/2``-1``1`
`2``-1/2``1/2`
`-1/2``5/4``1`
T


 = 
`-1/2``2``-1/2`
`-1``-1/2``5/4`
`1``1/2``1`


`"Now, "P^(-1)=1/|P| × Adj(P)`

 = `1/(9/4)` ×
`-1/2``2``-1/2`
`-1``-1/2``5/4`
`1``1/2``1`


 = 
`-2/9``8/9``-2/9`
`-4/9``-2/9``5/9`
`4/9``2/9``4/9`


`:.P^-1` = 
`-2/9``8/9``-2/9`
`-4/9``-2/9``5/9`
`4/9``2/9``4/9`



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

`P×D`=
`-1/2``-1``1`
`1``0``1/2`
`0``1``1`
×
`-1``0``0`
`0``-1``0`
`0``0``8`


=
`-1/2×-1-1×0+1×0``-1/2×0-1×-1+1×0``-1/2×0-1×0+1×8`
`1×-1+0×0+1/2×0``1×0+0×-1+1/2×0``1×0+0×0+1/2×8`
`0×-1+1×0+1×0``0×0+1×-1+1×0``0×0+1×0+1×8`


=
`1/2+0+0``0+1+0``0+0+8`
`-1+0+0``0+0+0``0+0+4`
`0+0+0``0-1+0``0+0+8`


=
`1/2``1``8`
`-1``0``4`
`0``-1``8`


`(P × D)×(P^-1)`=
`1/2``1``8`
`-1``0``4`
`0``-1``8`
×
`-2/9``8/9``-2/9`
`-4/9``-2/9``5/9`
`4/9``2/9``4/9`


=
`1/2×-2/9+1×-4/9+8×4/9``1/2×8/9+1×-2/9+8×2/9``1/2×-2/9+1×5/9+8×4/9`
`-1×-2/9+0×-4/9+4×4/9``-1×8/9+0×-2/9+4×2/9``-1×-2/9+0×5/9+4×4/9`
`0×-2/9-1×-4/9+8×4/9``0×8/9-1×-2/9+8×2/9``0×-2/9-1×5/9+8×4/9`


=
`-1/9-4/9+32/9``4/9-2/9+16/9``-1/9+5/9+32/9`
`2/9+0+16/9``-8/9+0+8/9``2/9+0+16/9`
`0+4/9+32/9``0+2/9+16/9``0-5/9+32/9`


=
`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`


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


1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
(Previous example)		3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
(Next example)


Share this solution or page with your friends.


Home What's new College Algebra Games Feedback About us
 
Copyright © 2024. All rights reserved. Terms, Privacy
 
 

.