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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
2. Example `[[1,0,1,0],[0,1,0,1]]`
(Previous example)		4. Example `[[2,3],[4,10]]`
(Next example)

3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`


Find SVD - Singular Value Decomposition ...
`[[1,1,1],[-1,-3,-3],[2,4,4]]`

Solution:
`A = `
`1``1``1`
`-1``-3``-3`
`2``4``4`


Here we are trying to find out two solutions using `A*A'` and `A'*A`

`1^"st"` Solution using `A*A'` for normalized vectors `u_i`

`A * A'`
`A^T` = 
`1``1``1`
`-1``-3``-3`
`2``4``4`
T
 = 
`1``-1``2`
`1``-3``4`
`1``-3``4`


`A×(A^T)`=
`1``1``1`
`-1``-3``-3`
`2``4``4`
×
`1``-1``2`
`1``-3``4`
`1``-3``4`


=
`1×1+1×1+1×1``1×(-1)+1×(-3)+1×(-3)``1×2+1×4+1×4`
`-1×1+(-3)×1+(-3)×1``-1×(-1)+(-3)×(-3)+(-3)×(-3)``-1×2+(-3)×4+(-3)×4`
`2×1+4×1+4×1``2×(-1)+4×(-3)+4×(-3)``2×2+4×4+4×4`


=
`1+1+1``-1+(-3)+(-3)``2+4+4`
`-1+(-3)+(-3)``1+9+9``-2+(-12)+(-12)`
`2+4+4``-2+(-12)+(-12)``4+16+16`


=
`3``-7``10`
`-7``19``-26`
`10``-26``36`
`A * A' = `
`3``-7``10`
`-7``19``-26`
`10``-26``36`


Find Eigen vector for `A * A'`

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

 `(3-lamda)`  `-7`  `10` 
 `-7`  `(19-lamda)`  `-26` 
 `10`  `-26`  `(36-lamda)` 
 = 0


`:.(3-lamda)((19-lamda) × (36-lamda) - (-26) × (-26))-(-7)((-7) × (36-lamda) - (-26) × 10)+10((-7) × (-26) - (19-lamda) × 10)=0`

`:.(3-lamda)((684-55lamda+lamda^2)-676)+7((-252+7lamda)-(-260))+10(182-(190-10lamda))=0`

`:.(3-lamda)(8-55lamda+lamda^2)+7(8+7lamda)+10(-8+10lamda)=0`

`:. (24-173lamda+58lamda^2-lamda^3)+(56+49lamda)+(-80+100lamda)=0`

`:.(-lamda^3+58lamda^2-24lamda)=0`

`:.-lamda(lamda-0.4168)(lamda-57.5832)=0`

`:.lamda=0 or (lamda-0.4168)=0 or (lamda-57.5832)=0`

`:.lamda=0 or lamda=0.4168 or lamda=57.5832`

`:.` The eigenvalues of the matrix `A * A'` are given by `lamda=0,0.4168,57.5832`

1. Eigenvectors for `lamda=57.5832`



1. Eigenvectors for `lamda=57.5832`

`A * A'-lamdaI = `
3-710
-719-26
10-2636
 - `57.5832` 
100
010
001


 = 
3-710
-719-26
10-2636
 - 
57.583200
057.58320
0057.5832

 = 
`-54.5832``-7``10`
`-7``-38.5832``-26`
`10``-26``-21.5832`


Now, reduce this matrix
`R_1 larr R_1-:(-54.5832)`

 = 
`1``0.1282``-0.1832`
`-7``-38.5832``-26`
`10``-26``-21.5832`


`R_2 larr R_2+7xx R_1`

 = 
`1``0.1282``-0.1832`
`0``-37.6855``-27.2824`
`10``-26``-21.5832`


`R_3 larr R_3-10xx R_1`

 = 
`1``0.1282``-0.1832`
`0``-37.6855``-27.2824`
`0``-27.2824``-19.7511`


`R_2 larr R_2-:(-37.6855)`

 = 
`1``0.1282``-0.1832`
`0``1``0.724`
`0``-27.2824``-19.7511`


`R_1 larr R_1-0.1282xx R_2`

 = 
`1``0``-0.276`
`0``1``0.724`
`0``-27.2824``-19.7511`


`R_3 larr R_3+27.2824xx R_2`

 = 
`1``0``-0.276`
`0``1``0.724`
`0``0``0`


The system associated with the eigenvalue `lamda=57.5832`

`(A * A'-57.5832I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-0.276`
`0``1``0.724`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-0.276x_3=0,x_2+0.724x_3=0`

`=>x_1=0.276x_3,x_2=-0.724x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=57.5832` is

`v=`
`0.276x_3`
`-0.724x_3`
`x_3`


Let `x_3=1`

`v_1=`
`0.276`
`-0.724`
`1`
`v_1=`
`0.276`
`-0.724`
`1`


2. Eigenvectors for `lamda=0.4168`



2. Eigenvectors for `lamda=0.4168`

`A * A'-lamdaI = `
3-710
-719-26
10-2636
 - `0.4168` 
100
010
001


 = 
3-710
-719-26
10-2636
 - 
0.416800
00.41680
000.4168

 = 
`2.5832``-7``10`
`-7``18.5832``-26`
`10``-26``35.5832`


Now, reduce this matrix
`R_1 larr R_1-:2.5832`

 = 
`1``-2.7098``3.8711`
`-7``18.5832``-26`
`10``-26``35.5832`


`R_2 larr R_2+7xx R_1`

 = 
`1``-2.7098``3.8711`
`0``-0.3854``1.098`
`10``-26``35.5832`


`R_3 larr R_3-10xx R_1`

 = 
`1``-2.7098``3.8711`
`0``-0.3854``1.098`
`0``1.098``-3.1283`


`R_2 larr R_2-:(-0.3854)`

 = 
`1``-2.7098``3.8711`
`0``1``-2.849`
`0``1.098``-3.1283`


`R_1 larr R_1+2.7098xx R_2`

 = 
`1``0``-3.849`
`0``1``-2.849`
`0``1.098``-3.1283`


`R_3 larr R_3-1.098xx R_2`

 = 
`1``0``-3.849`
`0``1``-2.849`
`0``0``0`


The system associated with the eigenvalue `lamda=0.4168`

`(A * A'-0.4168I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-3.849`
`0``1``-2.849`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-3.849x_3=0,x_2-2.849x_3=0`

`=>x_1=3.849x_3,x_2=2.849x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.4168` is

`v=`
`3.849x_3`
`2.849x_3`
`x_3`


Let `x_3=1`

`v_2=`
`3.849`
`2.849`
`1`
`v_2=`
`3.849`
`2.849`
`1`


3. Eigenvectors for `lamda=0`



3. Eigenvectors for `lamda=0`

`A * A'-lamdaI = `
3-710
-719-26
10-2636
 - `0` 
100
010
001


 = 
`3``-7``10`
`-7``19``-26`
`10``-26``36`


Now, reduce this matrix
`R_1 larr R_1-:3`

 = 
`1``-2.3333``3.3333`
`-7``19``-26`
`10``-26``36`


`R_2 larr R_2+7xx R_1`

 = 
`1``-2.3333``3.3333`
`0``2.6667``-2.6667`
`10``-26``36`


`R_3 larr R_3-10xx R_1`

 = 
`1``-2.3333``3.3333`
`0``2.6667``-2.6667`
`0``-2.6667``2.6667`


`R_2 larr R_2-:2.6667`

 = 
`1``-2.3333``3.3333`
`0``1``-1`
`0``-2.6667``2.6667`


`R_1 larr R_1+2.3333xx R_2`

 = 
`1``0``1`
`0``1``-1`
`0``-2.6667``2.6667`


`R_3 larr R_3+2.6667xx R_2`

 = 
`1``0``1`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=0`

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


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

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

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

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


Let `x_3=1`

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


For Eigenvector-1 `(0.276,-0.724,1)`, Length L = `sqrt(|0.27605|^2+|-0.72395|^2+|1|^2)=1.265`

So, normalizing gives `u_1=((0.276)/(1.265),(-0.724)/(1.265),(1)/(1.265))=(0.2182,-0.5723,0.7905)`

For Eigenvector-2 `(3.849,2.849,1)`, Length L = `sqrt(|3.84895|^2+|2.84895|^2+|1|^2)=4.8919`

So, normalizing gives `u_2=((3.849)/(4.8919),(2.849)/(4.8919),(1)/(4.8919))=(0.7868,0.5824,0.2044)`

For Eigenvector-3 `(-1,1,1)`, Length L = `sqrt(|-1|^2+|1|^2+|1|^2)=1.7321`

So, normalizing gives `u_3=((-1)/(1.7321),(1)/(1.7321),(1)/(1.7321))=(-0.5774,0.5774,0.5774)`

`2^"nd"` Solution using `A'*A` for normalized vectors `v_i`

`A' * A`
`A^T` = 
`1``1``1`
`-1``-3``-3`
`2``4``4`
T
 = 
`1``-1``2`
`1``-3``4`
`1``-3``4`


`(A^T)×A`=
`1``-1``2`
`1``-3``4`
`1``-3``4`
×
`1``1``1`
`-1``-3``-3`
`2``4``4`


=
`1×1+(-1)×(-1)+2×2``1×1+(-1)×(-3)+2×4``1×1+(-1)×(-3)+2×4`
`1×1+(-3)×(-1)+4×2``1×1+(-3)×(-3)+4×4``1×1+(-3)×(-3)+4×4`
`1×1+(-3)×(-1)+4×2``1×1+(-3)×(-3)+4×4``1×1+(-3)×(-3)+4×4`


=
`1+1+4``1+3+8``1+3+8`
`1+3+8``1+9+16``1+9+16`
`1+3+8``1+9+16``1+9+16`


=
`6``12``12`
`12``26``26`
`12``26``26`
`A' * A = `
`6``12``12`
`12``26``26`
`12``26``26`


Find Eigen vector for `A' * A`

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

 `(6-lamda)`  `12`  `12` 
 `12`  `(26-lamda)`  `26` 
 `12`  `26`  `(26-lamda)` 
 = 0


`:.(6-lamda)((26-lamda) × (26-lamda) - 26 × 26)-12(12 × (26-lamda) - 26 × 12)+12(12 × 26 - (26-lamda) × 12)=0`

`:.(6-lamda)((676-52lamda+lamda^2)-676)-12((312-12lamda)-312)+12(312-(312-12lamda))=0`

`:.(6-lamda)(-52lamda+lamda^2)-12(-12lamda)+12(12lamda)=0`

`:. (-312lamda+58lamda^2-lamda^3)-(-144lamda)+(144lamda)=0`

`:.(-lamda^3+58lamda^2-24lamda)=0`

`:.-lamda(lamda-0.4168)(lamda-57.5832)=0`

`:.lamda=0 or (lamda-0.4168)=0 or (lamda-57.5832)=0`

`:.lamda=0 or lamda=0.4168 or lamda=57.5832`

`:.` The eigenvalues of the matrix `A' * A` are given by `lamda=0,0.4168,57.5832`

1. Eigenvectors for `lamda=57.5832`



1. Eigenvectors for `lamda=57.5832`

`A' * A-lamdaI = `
61212
122626
122626
 - `57.5832` 
100
010
001


 = 
61212
122626
122626
 - 
57.583200
057.58320
0057.5832

 = 
`-51.5832``12``12`
`12``-31.5832``26`
`12``26``-31.5832`


Now, reduce this matrix
`R_1 larr R_1-:(-51.5832)`

 = 
`1``-0.2326``-0.2326`
`12``-31.5832``26`
`12``26``-31.5832`


`R_2 larr R_2-12xx R_1`

 = 
`1``-0.2326``-0.2326`
`0``-28.7916``28.7916`
`12``26``-31.5832`


`R_3 larr R_3-12xx R_1`

 = 
`1``-0.2326``-0.2326`
`0``-28.7916``28.7916`
`0``28.7916``-28.7916`


`R_2 larr R_2-:(-28.7916)`

 = 
`1``-0.2326``-0.2326`
`0``1``-1`
`0``28.7916``-28.7916`


`R_1 larr R_1+0.2326xx R_2`

 = 
`1``0``-0.4653`
`0``1``-1`
`0``28.7916``-28.7916`


`R_3 larr R_3-28.7916xx R_2`

 = 
`1``0``-0.4653`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=57.5832`

`(A' * A-57.5832I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-0.4653`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-0.4653x_3=0,x_2-x_3=0`

`=>x_1=0.4653x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=57.5832` is

`v=`
`0.4653x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_1=`
`0.4653`
`1`
`1`
`v_1=`
`0.4653`
`1`
`1`


2. Eigenvectors for `lamda=0.4168`



2. Eigenvectors for `lamda=0.4168`

`A' * A-lamdaI = `
61212
122626
122626
 - `0.4168` 
100
010
001


 = 
61212
122626
122626
 - 
0.416800
00.41680
000.4168

 = 
`5.5832``12``12`
`12``25.5832``26`
`12``26``25.5832`


Now, reduce this matrix
`R_1 larr R_1-:5.5832`

 = 
`1``2.1493``2.1493`
`12``25.5832``26`
`12``26``25.5832`


`R_2 larr R_2-12xx R_1`

 = 
`1``2.1493``2.1493`
`0``-0.2084``0.2084`
`12``26``25.5832`


`R_3 larr R_3-12xx R_1`

 = 
`1``2.1493``2.1493`
`0``-0.2084``0.2084`
`0``0.2084``-0.2084`


`R_2 larr R_2-:(-0.2084)`

 = 
`1``2.1493``2.1493`
`0``1``-1`
`0``0.2084``-0.2084`


`R_1 larr R_1-2.1493xx R_2`

 = 
`1``0``4.2986`
`0``1``-1`
`0``0.2084``-0.2084`


`R_3 larr R_3-0.2084xx R_2`

 = 
`1``0``4.2986`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=0.4168`

`(A' * A-0.4168I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``4.2986`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1+4.2986x_3=0,x_2-x_3=0`

`=>x_1=-4.2986x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.4168` is

`v=`
`-4.2986x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_2=`
`-4.2986`
`1`
`1`
`v_2=`
`-4.2986`
`1`
`1`


3. Eigenvectors for `lamda=0`



3. Eigenvectors for `lamda=0`

`A' * A-lamdaI = `
61212
122626
122626
 - `0` 
100
010
001


 = 
`6``12``12`
`12``26``26`
`12``26``26`


Now, reduce this matrix
`R_1 larr R_1-:6`

 = 
`1``2``2`
`12``26``26`
`12``26``26`


`R_2 larr R_2-12xx R_1`

 = 
`1``2``2`
`0``2``2`
`12``26``26`


`R_3 larr R_3-12xx R_1`

 = 
`1``2``2`
`0``2``2`
`0``2``2`


`R_2 larr R_2-:2`

 = 
`1``2``2`
`0``1``1`
`0``2``2`


`R_1 larr R_1-2xx R_2`

 = 
`1``0``0`
`0``1``1`
`0``2``2`


`R_3 larr R_3-2xx R_2`

 = 
`1``0``0`
`0``1``1`
`0``0``0`


The system associated with the eigenvalue `lamda=0`

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


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

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

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

`v=`
`0`
`-x_3`
`x_3`


Let `x_3=1`

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


For Eigenvector-1 `(0.4653,1,1)`, Length L = `sqrt(|0.46527|^2+|1|^2+|1|^2)=1.4888`

So, normalizing gives `v_1=((0.4653)/(1.4888),(1)/(1.4888),(1)/(1.4888))=(0.3125,0.6717,0.6717)`

For Eigenvector-2 `(-4.2986,1,1)`, Length L = `sqrt(|-4.2986|^2+|1|^2+|1|^2)=4.5253`

So, normalizing gives `v_2=((-4.2986)/(4.5253),(1)/(4.5253),(1)/(4.5253))=(-0.9499,0.221,0.221)`

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

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

`1^"st"` SVD Solution using `A*A'`

`:. U = ``[u_1,u_2,u_3]``=`
`0.2182``0.7868``-0.5774`
`-0.5723``0.5824``0.5774`
`0.7905``0.2044``0.5774`


`:. Sigma = `
`sqrt(57.5832)``0``0`
`0``sqrt(0.4168)``0`
`0``0``sqrt(0)`
`=`
`7.5884``0``0`
`0``0.6456``0`
`0``0``0`


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

`:. V = `
`0.3125``0.9498``0`
`0.6717``-0.2212``-0.7071`
`0.6717``-0.2212``0.7071`



`2^"nd"` SVD Solution using `A'*A`

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

`:. U = `
`0.2182``-0.7867``-0.5774`
`-0.5723``-0.5826``0.5774`
`0.7905``-0.2042``0.5774`


`:. Sigma = `
`sqrt(57.5832)``0``0`
`0``sqrt(0.4168)``0`
`0``0``sqrt(0)`
`=`
`7.5884``0``0`
`0``0.6456``0`
`0``0``0`


`:. V = ``[v_1,v_2,v_3]``=`
`0.3125``-0.9499``0`
`0.6717``0.221``-0.7071`
`0.6717``0.221``0.7071`



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


`U×Sigma`=
`0.21822``0.7868``-0.57735`
`-0.57228``0.58238``0.57735`
`0.79049``0.20442``0.57735`
×
`7.58836``0``0`
`0``0.64559``0`
`0``0``0`


=
`0.21822×7.58836+0.7868×0+(-0.57735)×0``0.21822×0+0.7868×0.64559+(-0.57735)×0``0.21822×0+0.7868×0+(-0.57735)×0`
`-0.57228×7.58836+0.58238×0+0.57735×0``-0.57228×0+0.58238×0.64559+0.57735×0``-0.57228×0+0.58238×0+0.57735×0`
`0.79049×7.58836+0.20442×0+0.57735×0``0.79049×0+0.20442×0.64559+0.57735×0``0.79049×0+0.20442×0+0.57735×0`


=
`1.65593+0+0``0+0.50795+0``0+0+0`
`-4.34267+0+0``0+0.37598+0``0+0+0`
`5.99852+0+0``0+0.13197+0``0+0+0`


=
`1.65593``0.50795``0`
`-4.34267``0.37598``0`
`5.99852``0.13197``0`


`(U × Sigma)×(V^T)`=
`1.65593``0.50795``0`
`-4.34267``0.37598``0`
`5.99852``0.13197``0`
×
`0.31252``0.6717``0.6717`
`0.94983``-0.22119``-0.22119`
`0``-0.70711``0.70711`


=
`1.65593×0.31252+0.50795×0.94983+0×0``1.65593×0.6717+0.50795×(-0.22119)+0×(-0.70711)``1.65593×0.6717+0.50795×(-0.22119)+0×0.70711`
`-4.34267×0.31252+0.37598×0.94983+0×0``-4.34267×0.6717+0.37598×(-0.22119)+0×(-0.70711)``-4.34267×0.6717+0.37598×(-0.22119)+0×0.70711`
`5.99852×0.31252+0.13197×0.94983+0×0``5.99852×0.6717+0.13197×(-0.22119)+0×(-0.70711)``5.99852×0.6717+0.13197×(-0.22119)+0×0.70711`


=
`0.51751+0.48247+0``1.11229+(-0.11235)+0``1.11229+(-0.11235)+0`
`-1.35717+0.35712+0``-2.91697+(-0.08316)+0``-2.91697+(-0.08316)+0`
`1.87466+0.12535+0``4.02921+(-0.02919)+0``4.02921+(-0.02919)+0`


=
`0.99998``0.99994``0.99994`
`-1.00005``-3.00013``-3.00013`
`2.00001``4.00002``4.00002`


`1^"st"` Solution is possible.

`1^"st"` Solution is possible.


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


`U×Sigma`=
`0.21822``-0.78672``-0.57735`
`-0.57228``-0.58257``0.57735`
`0.7905``-0.20415``0.57735`
×
`7.58836``0``0`
`0``0.64559``0`
`0``0``0`


=
`0.21822×7.58836+(-0.78672)×0+(-0.57735)×0``0.21822×0+(-0.78672)×0.64559+(-0.57735)×0``0.21822×0+(-0.78672)×0+(-0.57735)×0`
`-0.57228×7.58836+(-0.58257)×0+0.57735×0``-0.57228×0+(-0.58257)×0.64559+0.57735×0``-0.57228×0+(-0.58257)×0+0.57735×0`
`0.7905×7.58836+(-0.20415)×0+0.57735×0``0.7905×0+(-0.20415)×0.64559+0.57735×0``0.7905×0+(-0.20415)×0+0.57735×0`


=
`1.65593+0+0``0+(-0.5079)+0``0+0+0`
`-4.34267+0+0``0+(-0.3761)+0``0+0+0`
`5.9986+0+0``0+(-0.1318)+0``0+0+0`


=
`1.65593``-0.5079``0`
`-4.34267``-0.3761``0`
`5.9986``-0.1318``0`


`(U × Sigma)×(V^T)`=
`1.65593``-0.5079``0`
`-4.34267``-0.3761``0`
`5.9986``-0.1318``0`
×
`0.31252``0.67169``0.67169`
`-0.94991``0.22098``0.22098`
`0``-0.70711``0.70711`


=
`1.65593×0.31252+(-0.5079)×(-0.94991)+0×0``1.65593×0.67169+(-0.5079)×0.22098+0×(-0.70711)``1.65593×0.67169+(-0.5079)×0.22098+0×0.70711`
`-4.34267×0.31252+(-0.3761)×(-0.94991)+0×0``-4.34267×0.67169+(-0.3761)×0.22098+0×(-0.70711)``-4.34267×0.67169+(-0.3761)×0.22098+0×0.70711`
`5.9986×0.31252+(-0.1318)×(-0.94991)+0×0``5.9986×0.67169+(-0.1318)×0.22098+0×(-0.70711)``5.9986×0.67169+(-0.1318)×0.22098+0×0.70711`


=
`0.51751+0.48246+0``1.11227+(-0.11224)+0``1.11227+(-0.11224)+0`
`-1.35717+0.35726+0``-2.91693+(-0.08311)+0``-2.91693+(-0.08311)+0`
`1.87468+0.1252+0``4.0292+(-0.02912)+0``4.0292+(-0.02912)+0`


=
`0.99997``1.00004``1.00004`
`-0.99991``-3.00004``-3.00004`
`1.99988``4.00007``4.00007`


`2^"nd"` Solution is possible.

`2^"nd"` Solution is possible.



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