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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 (ref)
  2. Transforming matrix to Reduced Row Echelon Form (rref)
  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. Inverse Power Method for dominant eigenvalue
  21. Determinant by gaussian elimination
  22. Expanding determinant along row / column
  23. Determinants using montante (bareiss algorithm)
  24. Leibniz formula for determinant
  25. determinants using Sarrus Rule
  26. determinants using properties of determinants
  27. Row Space
  28. Column Space
  29. Null Space
3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
(Previous example)		18. Moore-Penrose Pseudoinverse
(Next method)

4. Example `[[2,3],[4,10]]`


Find SVD - Singular Value Decomposition ...
`[[2,3],[4,10]]`

Solution:
`A = `
`2``3`
`4``10`



`A' * A`
`A^T` = 
`2``3`
`4``10`
T
 = 
`2``4`
`3``10`


`(A^T)×A`=
`2``4`
`3``10`
×
`2``3`
`4``10`


=
`2×2+4×4``2×3+4×10`
`3×2+10×4``3×3+10×10`


=
`4+16``6+40`
`6+40``9+100`


=
`20``46`
`46``109`
`A' * A = `
`20``46`
`46``109`


Find Eigen vector for `A' * A`

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

 `(20-lamda)`  `46` 
 `46`  `(109-lamda)` 
 = 0


`:.(20-lamda) × (109-lamda) - 46 × 46=0`

`:.(2180-129lamda+lamda^2)-2116=0`

`:.(lamda^2-129lamda+64)=0`

`:.(lamda-0.498)(lamda-128.502)=0`

`:.(lamda-0.498)=0 or (lamda-128.502)=0`

`:.lamda=0.498 or lamda=128.502`

`:.` The eigenvalues of the matrix `A' * A` are given by `lamda=0.498,128.502`

1. Eigenvectors for `lamda=128.502`



1. Eigenvectors for `lamda=128.502`

`A' * A-lamdaI = `
2046
46109
 - `128.502` 
10
01


 = 
2046
46109
 - 
128.5020
0128.502

 = 
`-108.502``46`
`46``-19.502`


Now, reduce this matrix
`R_1 larr R_1-:(-108.502)`

 = 
`1``-0.424`
`46``-19.502`


`R_2 larr R_2-46xx R_1`

 = 
`1``-0.424`
`0``0`


The system associated with the eigenvalue `lamda=128.502`

`(A' * A-128.502I)`
`x_1`
`x_2`
 = 
`1``-0.424`
`0``0`
 
`x_1`
`x_2`
 = 
`0`
`0`


`=>x_1-0.424x_2=0`

`=>x_1=0.424x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=128.502` is

`v=`
`0.424x_2`
`x_2`


Let `x_2=1`

`v_1=`
`0.424`
`1`
`v_1=`
`0.424`
`1`


2. Eigenvectors for `lamda=0.498`



2. Eigenvectors for `lamda=0.498`

`A' * A-lamdaI = `
2046
46109
 - `0.498` 
10
01


 = 
2046
46109
 - 
0.4980
00.498

 = 
`19.502``46`
`46``108.502`


Now, reduce this matrix
`R_1 larr R_1-:19.502`

 = 
`1``2.3587`
`46``108.502`


`R_2 larr R_2-46xx R_1`

 = 
`1``2.3587`
`0``0`


The system associated with the eigenvalue `lamda=0.498`

`(A' * A-0.498I)`
`x_1`
`x_2`
 = 
`1``2.3587`
`0``0`
 
`x_1`
`x_2`
 = 
`0`
`0`


`=>x_1+2.3587x_2=0`

`=>x_1=-2.3587x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.498` is

`v=`
`-2.3587x_2`
`x_2`


Let `x_2=1`

`v_2=`
`-2.3587`
`1`
`v_2=`
`-2.3587`
`1`


For Eigenvector-1 `(0.424,1)`, Length L = `sqrt(|0.42396|^2+|1|^2)=1.0862`

So, normalizing gives `v_1=((0.424)/(1.0862),(1)/(1.0862))=(0.3903,0.9207)`

For Eigenvector-2 `(-2.3587,1)`, Length L = `sqrt(|-2.35874|^2+|1|^2)=2.562`

So, normalizing gives `v_2=((-2.3587)/(2.562),(1)/(2.562))=(-0.9207,0.3903)`

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

`:. U = `
`0.3125``-0.9501`
`0.9499``0.312`


`:. Sigma = `
`sqrt(128.502)``0`
`0``sqrt(0.498)`
`=`
`11.3359``0`
`0``0.7057`


`:. V = ``[v_1,v_2]``=`
`0.3903``-0.9207`
`0.9207``0.3903`


Verify Solution `A = U Sigma V^T`


`U×Sigma`=
`0.31252``-0.95009`
`0.94992``0.31202`
×
`11.33587``0`
`0``0.70572`


=
`0.31252×11.33587+(-0.95009)×0``0.31252×0+(-0.95009)×0.70572`
`0.94992×11.33587+0.31202×0``0.94992×0+0.31202×0.70572`


=
`3.54269+0``0+(-0.6705)`
`10.76817+0``0+0.2202`


=
`3.54269``-0.6705`
`10.76817``0.2202`


`(U × Sigma)×(V^T)`=
`3.54269``-0.6705`
`10.76817``0.2202`
×
`0.39033``0.92068`
`-0.92068``0.39033`


=
`3.54269×0.39033+(-0.6705)×(-0.92068)``3.54269×0.92068+(-0.6705)×0.39033`
`10.76817×0.39033+0.2202×(-0.92068)``10.76817×0.92068+0.2202×0.39033`


=
`1.38282+0.61731``3.26168+(-0.26172)`
`4.20314+(-0.20273)``9.91404+0.08595`


=
`2.00013``2.99996`
`4.00041``9.99999`


Solution is possible.
Solution is possible.



This material is intended as a summary. Use your textbook for detail explanation.
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3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
(Previous example)		18. Moore-Penrose Pseudoinverse
(Next method)


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