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Home > Matrix & Vector > Matrix operation > 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]]`

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


Find SVD - Singular Value Decomposition ...
`[[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^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`

`:.lamda=0 or lamda=2 or lamda=2`

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

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


`R_2 larr R_2-:(-1)`

 = 
`1``0``-1``0`
`0``1``0``-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`


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`
`v_1=`
`1`
`0`
`1`
`0`
,`v_2=`
`0`
`1`
`0`
`1`


1. Orthogonal Eigenvectors for `lamda=2`

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


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`
`v_3=`
`-1`
`0`
`1`
`0`
,`v_4=`
`0`
`-1`
`0`
`1`


3. Orthogonal Eigenvectors for `lamda=0`

`v_3=`
`-1`
`0`
`1`
`0`
,`v_4=`
`0`
`-1`
`0`
`1`


For Eigenvector-1 `(1,0,1,0)`, Length L = `sqrt(|1|^2+|0|^2+|1|^2+|0|^2)=1.4142`

So, normalizing gives `v_1=((1)/(1.4142),(0)/(1.4142),(1)/(1.4142),(0)/(1.4142))=(0.7071,0,0.7071,0)`

For Eigenvector-2 `(0,1,0,1)`, Length L = `sqrt(|0|^2+|1|^2+|0|^2+|1|^2)=1.4142`

So, normalizing gives `v_2=((0)/(1.4142),(1)/(1.4142),(0)/(1.4142),(1)/(1.4142))=(0,0.7071,0,0.7071)`

For Eigenvector-3 `(-1,0,1,0)`, Length L = `sqrt(|-1|^2+|0|^2+|1|^2+|0|^2)=1.4142`

So, normalizing gives `v_3=((-1)/(1.4142),(0)/(1.4142),(1)/(1.4142),(0)/(1.4142))=(-0.7071,0,0.7071,0)`

For Eigenvector-4 `(0,-1,0,1)`, Length L = `sqrt(|0|^2+|-1|^2+|0|^2+|1|^2)=1.4142`

So, normalizing gives `v_4=((0)/(1.4142),(-1)/(1.4142),(0)/(1.4142),(1)/(1.4142))=(0,-0.7071,0,0.7071)`

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

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


`:. Sigma = `
`sqrt(2)``0``0``0`
`0``sqrt(2)``0``0`
`=`
`1.4142``0``0``0`
`0``1.4142``0``0`


`:. V = ``[v_1,v_2,v_3,v_4]``=`
`0.7071``0``-0.7071``0`
`0``0.7071``0``-0.7071`
`0.7071``0``0.7071``0`
`0``0.7071``0``0.7071`


Verify Solution `A = U Sigma V^T`


`U×Sigma`=
`0.99999``0`
`0``0.99999`
×
`1.41421``0``0``0`
`0``1.41421``0``0`


=
`0.99999×1.41421+0×0``0.99999×0+0×1.41421``0.99999×0+0×0``0.99999×0+0×0`
`0×1.41421+0.99999×0``0×0+0.99999×1.41421``0×0+0.99999×0``0×0+0.99999×0`


=
`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`
×
`0.70711``0``0.70711``0`
`0``0.70711``0``0.70711`
`-0.70711``0``0.70711``0`
`0``-0.70711``0``0.70711`


=
`1.4142×0.70711+0×0+0×(-0.70711)+0×0``1.4142×0+0×0.70711+0×0+0×(-0.70711)``1.4142×0.70711+0×0+0×0.70711+0×0``1.4142×0+0×0.70711+0×0+0×0.70711`
`0×0.70711+1.4142×0+0×(-0.70711)+0×0``0×0+1.4142×0.70711+0×0+0×(-0.70711)``0×0.70711+1.4142×0+0×0.70711+0×0``0×0+1.4142×0.70711+0×0+0×0.70711`


=
`0.99999+0+0+0``0+0+0+0``0.99999+0+0+0``0+0+0+0`
`0+0+0+0``0+0.99999+0+0``0+0+0+0``0+0.99999+0+0`


=
`0.99999``0``0.99999``0`
`0``0.99999``0``0.99999`


Solution is possible.
Solution is possible.



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