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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. Determinant by gaussian elimination
  21. Expanding determinant along row / column
  22. Determinants using montante (bareiss algorithm)
  23. Leibniz formula for determinant
  24. determinants using Sarrus Rule
  25. determinants using properties of determinants
  26. Row Space
  27. Column Space
  28. 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]]`


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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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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