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Home > Matrix & Vector calculators > SVD - Singular Value Decomposition example
17. SVD - Singular Value Decomposition example ( Enter your problem )
( Enter your problem )
  1. Example `[[4,0],[3,-5]]`
  2. Example `[[1,0,1,0],[0,1,0,1]]`
  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 `[[4,0],[3,-5]]`
(Previous example)		3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
(Next example)

2. Example `[[1,0,1,0],[0,1,0,1]]`


Find Singular Value Decomposition (SVD) of a Matrix ...
`[[1,0,1,0],[0,1,0,1]]`

Solution:
`A = `
`1``0``1``0`
`0``1``0``1`


`A * A'`
`A^T` = 
`1``0``1``0`
`0``1``0``1`
T
 = 
`1``0`
`0``1`
`1``0`
`0``1`


`A×(A^T)`=
`1``0``1``0`
`0``1``0``1`
×
`1``0`
`0``1`
`1``0`
`0``1`


=
`1×1+0×0+1×1+0×0``1×0+0×1+1×0+0×1`
`0×1+1×0+0×1+1×0``0×0+1×1+0×0+1×1`


=
`1+0+1+0``0+0+0+0`
`0+0+0+0``0+1+0+1`


=
`2``0`
`0``2`




`:. A * A' = `
`2``0`
`0``2`


Find Eigen vector for `A * A'`

`|A * A'-lamdaI|=0`

 `(2-lamda)`  `0` 
 `0`  `(2-lamda)` 
 = 0


`:.(2-lamda) × (2-lamda) - 0 × 0=0`

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

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

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

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

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

1. Eigenvectors for `lamda=2`



1. Eigenvectors for `lamda=2`

`A * A'-lamdaI = `
20
02
 - `2` 
10
01


 = 
20
02
 - 
20
02

 = 
`0``0`
`0``0`


Now, reduce this matrix
The system associated with the eigenvalue `lamda=2`

`(A * A'-2I)`
`x_1`
`x_2`
 = 
`0``0`
`0``0`
 
`x_1`
`x_2`
 = 
`0`
`0`


`=>`

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

`v=`
`x_1`
`x_2`


Let `x_1=1,x_2=0`

`v_1=`
`1`
`0`


Let `x_1=0,x_2=1`

`v_2=`
`0`
`1`



For Eigenvector-1 `(1,0)`, Length L = `sqrt(1^2+0^2)=1`

So, normalizing gives `u_1=(1/1,0/1)=(1,0)`

For Eigenvector-2 `(0,1)`, Length L = `sqrt(0^2+1^2)=1`

So, normalizing gives `u_2=(0/1,1/1)=(0,1)`


`A' * A`
`A^T` = 
`1``0``1``0`
`0``1``0``1`
T
 = 
`1``0`
`0``1`
`1``0`
`0``1`


`(A^T)×A`=
`1``0`
`0``1`
`1``0`
`0``1`
×
`1``0``1``0`
`0``1``0``1`


=
`1×1+0×0``1×0+0×1``1×1+0×0``1×0+0×1`
`0×1+1×0``0×0+1×1``0×1+1×0``0×0+1×1`
`1×1+0×0``1×0+0×1``1×1+0×0``1×0+0×1`
`0×1+1×0``0×0+1×1``0×1+1×0``0×0+1×1`


=
`1+0``0+0``1+0``0+0`
`0+0``0+1``0+0``0+1`
`1+0``0+0``1+0``0+0`
`0+0``0+1``0+0``0+1`


=
`1``0``1``0`
`0``1``0``1`
`1``0``1``0`
`0``1``0``1`




`:. A' * A = `
`1``0``1``0`
`0``1``0``1`
`1``0``1``0`
`0``1``0``1`


Find Eigen vector for `A' * A`

`|A' * A-lamdaI|=0`

 `(1-lamda)`  `0`  `1`  `0` 
 `0`  `(1-lamda)`  `0`  `1` 
 `1`  `0`  `(1-lamda)`  `0` 
 `0`  `1`  `0`  `(1-lamda)` 
 = 0


`:.(1-lamda) × |[(1-lamda),0,1],[0,(1-lamda),0],[1,0,(1-lamda)]|+0 × |[0,0,1],[1,(1-lamda),0],[0,0,(1-lamda)]|+1 × |[0,(1-lamda),1],[1,0,0],[0,1,(1-lamda)]|+0 × |[0,(1-lamda),0],[1,0,(1-lamda)],[0,1,0]|=0`

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

`:.(-2lamda+5lamda^2-4lamda^3+lamda^4)+0+(2lamda-lamda^2)+0=0`

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

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

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

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

1. Eigenvectors for `lamda=2`



1. Eigenvectors for `lamda=2`

`A' * A-lamdaI = `
1010
0101
1010
0101
 - `2` 
1000
0100
0010
0001


 = 
1010
0101
1010
0101
 - 
2000
0200
0020
0002

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


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

 = 
 `1` `1=-1-:-1`
`R_1 larr R_1-:-1`		 `0` `0=0-:-1`
`R_1 larr R_1-:-1`		 `-1` `-1=1-:-1`
`R_1 larr R_1-:-1`		 `0` `0=0-:-1`
`R_1 larr R_1-:-1`
`0``-1``0``1`
`1``0``-1``0`
`0``1``0``-1`


`R_3 larr R_3- R_1`

 = 
`1``0``-1``0`
`0``-1``0``1`
 `0` `0=1-1`
`R_3 larr R_3- R_1`		 `0` `0=0-0`
`R_3 larr R_3- R_1`		 `0` `0=-1--1`
`R_3 larr R_3- R_1`		 `0` `0=0-0`
`R_3 larr R_3- R_1`
`0``1``0``-1`


`R_2 larr R_2-:-1`

 = 
`1``0``-1``0`
 `0` `0=0-:-1`
`R_2 larr R_2-:-1`		 `1` `1=-1-:-1`
`R_2 larr R_2-:-1`		 `0` `0=0-:-1`
`R_2 larr R_2-:-1`		 `-1` `-1=1-:-1`
`R_2 larr R_2-:-1`
`0``0``0``0`
`0``1``0``-1`


`R_4 larr R_4- R_2`

 = 
`1``0``-1``0`
`0``1``0``-1`
`0``0``0``0`
 `0` `0=0-0`
`R_4 larr R_4- R_2`		 `0` `0=1-1`
`R_4 larr R_4- R_2`		 `0` `0=0-0`
`R_4 larr R_4- R_2`		 `0` `0=-1--1`
`R_4 larr R_4- R_2`


The system associated with the eigenvalue `lamda=2`

`(A' * A-2I)`
`x_1`
`x_2`
`x_3`
`x_4`
 = 
`1``0``-1``0`
`0``1``0``-1`
`0``0``0``0`
`0``0``0``0`
 
`x_1`
`x_2`
`x_3`
`x_4`
 = 
`0`
`0`
`0`
`0`


`=>x_1-x_3=0,x_2-x_4=0`

`=>x_1=x_3,x_2=x_4`

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

`v=`
`x_3`
`x_4`
`x_3`
`x_4`


Let `x_3=1,x_4=0`

`v_1=`
`1`
`0`
`1`
`0`


Let `x_3=0,x_4=1`

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



3. Eigenvectors for `lamda=0`



3. Eigenvectors for `lamda=0`

`A' * A-lamdaI = `
1010
0101
1010
0101
 - `0` 
1000
0100
0010
0001


 = 
`1``0``1``0`
`0``1``0``1`
`1``0``1``0`
`0``1``0``1`


Now, reduce this matrix
`R_3 larr R_3- R_1`

 = 
`1``0``1``0`
`0``1``0``1`
 `0` `0=1-1`
`R_3 larr R_3- R_1`		 `0` `0=0-0`
`R_3 larr R_3- R_1`		 `0` `0=1-1`
`R_3 larr R_3- R_1`		 `0` `0=0-0`
`R_3 larr R_3- R_1`
`0``1``0``1`


`R_4 larr R_4- R_2`

 = 
`1``0``1``0`
`0``1``0``1`
`0``0``0``0`
 `0` `0=0-0`
`R_4 larr R_4- R_2`		 `0` `0=1-1`
`R_4 larr R_4- R_2`		 `0` `0=0-0`
`R_4 larr R_4- R_2`		 `0` `0=1-1`
`R_4 larr R_4- R_2`


The system associated with the eigenvalue `lamda=0`

`(A' * A-0I)`
`x_1`
`x_2`
`x_3`
`x_4`
 = 
`1``0``1``0`
`0``1``0``1`
`0``0``0``0`
`0``0``0``0`
 
`x_1`
`x_2`
`x_3`
`x_4`
 = 
`0`
`0`
`0`
`0`


`=>x_1+x_3=0,x_2+x_4=0`

`=>x_1=-x_3,x_2=-x_4`

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

`v=`
`-x_3`
`-x_4`
`x_3`
`x_4`


Let `x_3=1,x_4=0`

`v_3=`
`-1`
`0`
`1`
`0`


Let `x_3=0,x_4=1`

`v_4=`
`0`
`-1`
`0`
`1`



For Eigenvector-1 `(1,0,1,0)`, Length L = `sqrt(1^2+0^2+1^2+0^2)=1.41421`

So, normalizing gives `v_1=(1/1.41421,0/1.41421,1/1.41421,0/1.41421)=(0.7071,0,0.7071,0)`

For Eigenvector-2 `(0,1,0,1)`, Length L = `sqrt(0^2+1^2+0^2+1^2)=1.41421`

So, normalizing gives `v_2=(0/1.41421,1/1.41421,0/1.41421,1/1.41421)=(0,0.7071,0,0.7071)`

For Eigenvector-3 `(-1,0,1,0)`, Length L = `sqrt((-1)^2+0^2+1^2+0^2)=1.41421`

So, normalizing gives `v_3=((-1)/1.41421,0/1.41421,1/1.41421,0/1.41421)=(-0.7071,0,0.7071,0)`

For Eigenvector-4 `(0,-1,0,1)`, Length L = `sqrt(0^2+(-1)^2+0^2+1^2)=1.41421`

So, normalizing gives `v_4=(0/1.41421,(-1)/1.41421,0/1.41421,1/1.41421)=(0,-0.7071,0,0.7071)`

`1^"st"` Solution

`:. Sigma = `
`sqrt(2)``0``0``0`
`0``sqrt(2)``0``0`
`=`
`1.41421``0``0``0`
`0``1.41421``0``0`


`:. U = ``[u_1,u_2]``=`
`1``0`
`0``1`


`V` is found using formula `v_i=1/sigma_i A^T*u_i`

`:. V = `
`0.70711``0``-0.70711``0`
`0``0.70711``0``-0.70711`
`0.70711``0``0.70711``0`
`0``0.70711``0``0.70711`


Or
`2^"nd"` Solution

`:. Sigma = `
`sqrt(2)``0``0``0`
`0``sqrt(2)``0``0`
`=`
`1.41421``0``0``0`
`0``1.41421``0``0`


`:. V = ``[v_1,v_2,v_3,v_4]``=`
`0.70711``0``-0.70711``0`
`0``0.70711``0``-0.70711`
`0.70711``0``0.70711``0`
`0``0.70711``0``0.70711`


`U` is found using formula `u_i=1/sigma_i A*v_i`

`:. U = `
`1``0`
`0``1`


Verify `1^"st"` Solution `A = U Sigma V^T`


`U×Sigma`=
`1``0`
`0``1`
×
`1.4142``0``0``0`
`0``1.4142``0``0`


=
`1×1.4142+0×0``1×0+0×1.4142``1×0+0×0``1×0+0×0`
`0×1.4142+1×0``0×0+1×1.4142``0×0+1×0``0×0+1×0`


=
`1.4142+0``0+0``0+0``0+0`
`0+0``0+1.4142``0+0``0+0`


=
`1.4142``0``0``0`
`0``1.4142``0``0`


`(U × Sigma)×(V^T)`=
`1.4142``0``0``0`
`0``1.4142``0``0`
×
`0.7071``0``0.7071``0`
`0``0.7071``0``0.7071`
`-0.7071``0``0.7071``0`
`0``-0.7071``0``0.7071`


=
`1.4142×0.7071+0×0+0×-0.7071+0×0``1.4142×0+0×0.7071+0×0+0×-0.7071``1.4142×0.7071+0×0+0×0.7071+0×0``1.4142×0+0×0.7071+0×0+0×0.7071`
`0×0.7071+1.4142×0+0×-0.7071+0×0``0×0+1.4142×0.7071+0×0+0×-0.7071``0×0.7071+1.4142×0+0×0.7071+0×0``0×0+1.4142×0.7071+0×0+0×0.7071`


=
`1+0+0+0``0+0+0+0``1+0+0+0``0+0+0+0`
`0+0+0+0``0+1+0+0``0+0+0+0``0+1+0+0`


=
`1``0``1``0`
`0``1``0``1`



Verify `2^"nd"` Solution `A = U Sigma V^T`


`U×Sigma`=
`1``0`
`0``1`
×
`1.4142``0``0``0`
`0``1.4142``0``0`


=
`1×1.4142+0×0``1×0+0×1.4142``1×0+0×0``1×0+0×0`
`0×1.4142+1×0``0×0+1×1.4142``0×0+1×0``0×0+1×0`


=
`1.4142+0``0+0``0+0``0+0`
`0+0``0+1.4142``0+0``0+0`


=
`1.4142``0``0``0`
`0``1.4142``0``0`


`(U × Sigma)×(V^T)`=
`1.4142``0``0``0`
`0``1.4142``0``0`
×
`0.7071``0``0.7071``0`
`0``0.7071``0``0.7071`
`-0.7071``0``0.7071``0`
`0``-0.7071``0``0.7071`


=
`1.4142×0.7071+0×0+0×-0.7071+0×0``1.4142×0+0×0.7071+0×0+0×-0.7071``1.4142×0.7071+0×0+0×0.7071+0×0``1.4142×0+0×0.7071+0×0+0×0.7071`
`0×0.7071+1.4142×0+0×-0.7071+0×0``0×0+1.4142×0.7071+0×0+0×-0.7071``0×0.7071+1.4142×0+0×0.7071+0×0``0×0+1.4142×0.7071+0×0+0×0.7071`


=
`1+0+0+0``0+0+0+0``1+0+0+0``0+0+0+0`
`0+0+0+0``0+1+0+0``0+0+0+0``0+1+0+0`


=
`1``0``1``0`
`0``1``0``1`




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


1. Example `[[4,0],[3,-5]]`
(Previous example)		3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
(Next example)


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