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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
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-0.41678814)(lamda-57.58321186)=0`

`:.lamda=0 or(lamda-0.41678814)=0 or(lamda-57.58321186)=0 `

`:.` The eigenvalues of the matrix `A * A'` are given by `lamda=0,0.41678814,57.58321186`,

1. Eigenvectors for `lamda=57.58321186`



1. Eigenvectors for `lamda=57.58321186`

`A * A'-lamdaI = `
3-710
-719-26
10-2636
 - `57.58321186` 
100
010
001


 = 
3-710
-719-26
10-2636
 - 
57.5832118600
057.583211860
0057.58321186

 = 
`-54.58321186``-7``10`
`-7``-38.58321186``-26`
`10``-26``-21.58321186`


Now, reduce this matrix
`R_1 larr R_1-:-54.58321186`

 = 
`1``0.12824456``-0.18320651`
`-7``-38.58321186``-26`
`10``-26``-21.58321186`


`R_2 larr R_2+7xx R_1`

 = 
`1``0.12824456``-0.18320651`
`0``-37.68549993``-27.2824456`
`10``-26``-21.58321186`


`R_3 larr R_3-10xx R_1`

 = 
`1``0.12824456``-0.18320651`
`0``-37.68549993``-27.2824456`
`0``-27.2824456``-19.75114671`


`R_2 larr R_2-:-37.68549993`

 = 
`1``0.12824456``-0.18320651`
`0``1``0.72395074`
`0``-27.2824456``-19.75114671`


`R_1 larr R_1-0.12824456xx R_2`

 = 
`1``0``-0.27604926`
`0``1``0.72395074`
`0``-27.2824456``-19.75114671`


`R_3 larr R_3+27.2824456xx R_2`

 = 
`1``0``-0.27604926`
`0``1``0.72395074`
`0``0``0`


The system associated with the eigenvalue `lamda=57.58321186`

`(A * A'-57.58321186I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-0.27604926`
`0``1``0.72395074`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-0.27604926x_3=0,x_2+0.72395074x_3=0`

`=>x_1=0.27604926x_3,x_2=-0.72395074x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=57.58321186` is

`v=`
`0.27604926x_3`
`-0.72395074x_3`
`x_3`


Let `x_3=1`

`v_1=`
`0.27604926`
`-0.72395074`
`1`
`v_1=`
`0.27604926`
`-0.72395074`
`1`


2. Eigenvectors for `lamda=0.41678814`



2. Eigenvectors for `lamda=0.41678814`

`A * A'-lamdaI = `
3-710
-719-26
10-2636
 - `0.41678814` 
100
010
001


 = 
3-710
-719-26
10-2636
 - 
0.4167881400
00.416788140
000.41678814

 = 
`2.58321186``-7``10`
`-7``18.58321186``-26`
`10``-26``35.58321186`


Now, reduce this matrix
interchanging rows `R_1 harr R_3`

 = 
`10``-26``35.58321186`
`-7``18.58321186``-26`
`2.58321186``-7``10`


`R_1 larr R_1-:10`

 = 
`1``-2.6``3.55832119`
`-7``18.58321186``-26`
`2.58321186``-7``10`


`R_2 larr R_2+7xx R_1`

 = 
`1``-2.6``3.55832119`
`0``0.38321186``-1.0917517`
`2.58321186``-7``10`


`R_3 larr R_3-2.58321186xx R_1`

 = 
`1``-2.6``3.55832119`
`0``0.38321186``-1.0917517`
`0``-0.28364917``0.80810253`


`R_2 larr R_2-:0.38321186`

 = 
`1``-2.6``3.55832119`
`0``1``-2.84895074`
`0``-0.28364917``0.80810253`


`R_1 larr R_1+2.6xx R_2`

 = 
`1``0``-3.84895074`
`0``1``-2.84895074`
`0``-0.28364917``0.80810253`


`R_3 larr R_3+0.28364917xx R_2`

 = 
`1``0``-3.84895074`
`0``1``-2.84895074`
`0``0``0`


The system associated with the eigenvalue `lamda=0.41678814`

`(A * A'-0.41678814I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-3.84895074`
`0``1``-2.84895074`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-3.84895074x_3=0,x_2-2.84895074x_3=0`

`=>x_1=3.84895074x_3,x_2=2.84895074x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.41678814` is

`v=`
`3.84895074x_3`
`2.84895074x_3`
`x_3`


Let `x_3=1`

`v_2=`
`3.84895074`
`2.84895074`
`1`
`v_2=`
`3.84895074`
`2.84895074`
`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
interchanging rows `R_1 harr R_3`

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


`R_1 larr R_1-:10`

 = 
`1``-2.6``3.6`
`-7``19``-26`
`3``-7``10`


`R_2 larr R_2+7xx R_1`

 = 
`1``-2.6``3.6`
`0``0.8``-0.8`
`3``-7``10`


`R_3 larr R_3-3xx R_1`

 = 
`1``-2.6``3.6`
`0``0.8``-0.8`
`0``0.8``-0.8`


`R_2 larr R_2xx1.25`

 = 
`1``-2.6``3.6`
`0``1``-1`
`0``0.8``-0.8`


`R_1 larr R_1+2.6xx R_2`

 = 
`1``0``1`
`0``1``-1`
`0``0.8``-0.8`


`R_3 larr R_3-0.8xx 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 `(0.27604926,-0.72395074,1)`, Length L = `sqrt(0.27604926^2+(-0.72395074)^2+1^2)=1.26503275`

So, normalizing gives `u_1=(0.27604926/1.26503275,(-0.72395074)/1.26503275,1/1.26503275)=(0.21821511,-0.57227826,0.79049337)`

For Eigenvector-2 `(3.84895074,2.84895074,1)`, Length L = `sqrt(3.84895074^2+2.84895074^2+1^2)=4.89192622`

So, normalizing gives `u_2=(3.84895074/4.89192622,2.84895074/4.89192622,1/4.89192622)=(0.78679656,0.58237811,0.20441846)`

For Eigenvector-3 `(-1,1,1)`, Length L = `sqrt((-1)^2+1^2+1^2)=1.73205081`

So, normalizing gives `u_3=((-1)/1.73205081,1/1.73205081,1/1.73205081)=(-0.57735027,0.57735027,0.57735027)`

`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-0.41678814)(lamda-57.58321186)=0`

`:.lamda=0 or(lamda-0.41678814)=0 or(lamda-57.58321186)=0 `

`:.` The eigenvalues of the matrix `A' * A` are given by `lamda=0,0.41678814,57.58321186`,

1. Eigenvectors for `lamda=57.58321186`



1. Eigenvectors for `lamda=57.58321186`

`A' * A-lamdaI = `
61212
122626
122626
 - `57.58321186` 
100
010
001


 = 
61212
122626
122626
 - 
57.5832118600
057.583211860
0057.58321186

 = 
`-51.58321186``12``12`
`12``-31.58321186``26`
`12``26``-31.58321186`


Now, reduce this matrix
`R_1 larr R_1-:-51.58321186`

 = 
`1``-0.23263383``-0.23263383`
`12``-31.58321186``26`
`12``26``-31.58321186`


`R_2 larr R_2-12xx R_1`

 = 
`1``-0.23263383``-0.23263383`
`0``-28.79160593``28.79160593`
`12``26``-31.58321186`


`R_3 larr R_3-12xx R_1`

 = 
`1``-0.23263383``-0.23263383`
`0``-28.79160593``28.79160593`
`0``28.79160593``-28.79160593`


`R_2 larr R_2-:-28.79160593`

 = 
`1``-0.23263383``-0.23263383`
`0``1``-1`
`0``28.79160593``-28.79160593`


`R_1 larr R_1+0.23263383xx R_2`

 = 
`1``0``-0.46526765`
`0``1``-1`
`0``28.79160593``-28.79160593`


`R_3 larr R_3-28.79160593xx R_2`

 = 
`1``0``-0.46526765`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=57.58321186`

`(A' * A-57.58321186I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-0.46526765`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-0.46526765x_3=0,x_2-x_3=0`

`=>x_1=0.46526765x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=57.58321186` is

`v=`
`0.46526765x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_1=`
`0.46526765`
`1`
`1`
`v_1=`
`0.46526765`
`1`
`1`


2. Eigenvectors for `lamda=0.41678814`



2. Eigenvectors for `lamda=0.41678814`

`A' * A-lamdaI = `
61212
122626
122626
 - `0.41678814` 
100
010
001


 = 
61212
122626
122626
 - 
0.4167881400
00.416788140
000.41678814

 = 
`5.58321186``12``12`
`12``25.58321186``26`
`12``26``25.58321186`


Now, reduce this matrix
interchanging rows `R_1 harr R_2`

 = 
`12``25.58321186``26`
`5.58321186``12``12`
`12``26``25.58321186`


`R_1 larr R_1-:12`

 = 
`1``2.13193432``2.16666667`
`5.58321186``12``12`
`12``26``25.58321186`


`R_2 larr R_2-5.58321186xx R_1`

 = 
`1``2.13193432``2.16666667`
`0``0.09695902``-0.09695902`
`12``26``25.58321186`


`R_3 larr R_3-12xx R_1`

 = 
`1``2.13193432``2.16666667`
`0``0.09695902``-0.09695902`
`0``0.41678814``-0.41678814`


interchanging rows `R_2 harr R_3`

 = 
`1``2.13193432``2.16666667`
`0``0.41678814``-0.41678814`
`0``0.09695902``-0.09695902`


`R_2 larr R_2-:0.41678814`

 = 
`1``2.13193432``2.16666667`
`0``1``-1`
`0``0.09695902``-0.09695902`


`R_1 larr R_1-2.13193432xx R_2`

 = 
`1``0``4.29860099`
`0``1``-1`
`0``0.09695902``-0.09695902`


`R_3 larr R_3-0.09695902xx R_2`

 = 
`1``0``4.29860099`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=0.41678814`

`(A' * A-0.41678814I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``4.29860099`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1+4.29860099x_3=0,x_2-x_3=0`

`=>x_1=-4.29860099x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.41678814` is

`v=`
`-4.29860099x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_2=`
`-4.29860099`
`1`
`1`
`v_2=`
`-4.29860099`
`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
interchanging rows `R_1 harr R_2`

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


`R_1 larr R_1-:12`

 = 
`1``2.16666667``2.16666667`
`6``12``12`
`12``26``26`


`R_2 larr R_2-6xx R_1`

 = 
`1``2.16666667``2.16666667`
`0``-1``-1`
`12``26``26`


`R_3 larr R_3-12xx R_1`

 = 
`1``2.16666667``2.16666667`
`0``-1``-1`
`0``0``0`


`R_2 larr R_2-:-1`

 = 
`1``2.16666667``2.16666667`
`0``1``1`
`0``0``0`


`R_1 larr R_1-2.16666667xx 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 `(0.46526765,1,1)`, Length L = `sqrt(0.46526765^2+1^2+1^2)=1.48878272`

So, normalizing gives `v_1=(0.46526765/1.48878272,1/1.48878272,1/1.48878272)=(0.31251549,0.67168969,0.67168969)`

For Eigenvector-2 `(-4.29860099,1,1)`, Length L = `sqrt((-4.29860099)^2+1^2+1^2)=4.52525916`

So, normalizing gives `v_2=((-4.29860099)/4.52525916,1/4.52525916,1/4.52525916)=(-0.94991267,0.22098182,0.22098182)`

For Eigenvector-3 `(0,-1,1)`, Length L = `sqrt(0^2+(-1)^2+1^2)=1.41421356`

So, normalizing gives `v_3=(0/1.41421356,(-1)/1.41421356,1/1.41421356)=(0,-0.70710678,0.70710678)`

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

`:. Sigma = `
`sqrt(57.58321186)``0``0`
`0``sqrt(0.41678814)``0`
`0``0``sqrt(0)`
`=`
`7.58836029``0``0`
`0``0.64559131``0`
`0``0``0`


`:. U = ``[u_1,u_2,u_3]``=`
`0.21821511``0.78679656``-0.57735027`
`-0.57227826``0.58237811``0.57735027`
`0.79049337``0.20441846``0.57735027`


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

`:. V = `
`0.31251549``0.94991267``0`
`0.67168969``-0.2209818``-0.70710678`
`0.67168969``-0.2209818``0.70710678`


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

`:. Sigma = `
`sqrt(57.58321186)``0``0`
`0``sqrt(0.41678814)``0`
`0``0``sqrt(0)`
`=`
`7.58836029``0``0`
`0``0.64559131``0`
`0``0``0`


`:. V = ``[v_1,v_2,v_3]``=`
`0.31251549``-0.94991267``0`
`0.67168969``0.22098182``-0.70710678`
`0.67168969``0.22098182``0.70710678`


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

`:. U = `
`0.21821511``-0.78679657``-0.57735027`
`-0.57227826``-0.58237811``0.57735027`
`0.79049337``-0.20441846``0.57735027`



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


`U×Sigma`=
`0.218215108``0.786796564``-0.577350269`
`-0.572278258``0.582378109``0.577350269`
`0.790493366``0.204418455``0.577350269`
×
`7.588360288``0``0`
`0``0.645591314``0`
`0``0``0`


=
`0.218215108×7.588360288+0.786796564×0-0.577350269×0``0.218215108×0+0.786796564×0.645591314-0.577350269×0``0.218215108×0+0.786796564×0-0.577350269×0`
`-0.572278258×7.588360288+0.582378109×0+0.577350269×0``-0.572278258×0+0.582378109×0.645591314+0.577350269×0``-0.572278258×0+0.582378109×0+0.577350269×0`
`0.790493366×7.588360288+0.204418455×0+0.577350269×0``0.790493366×0+0.204418455×0.645591314+0.577350269×0``0.790493366×0+0.204418455×0+0.577350269×0`


=
`1.65589486+0+0``0+0.507949028+0``0+0+0`
`-4.342653607+0+0``0+0.375978249+0``0+0+0`
`5.998548466+0+0``0+0.131970779+0``0+0+0`


=
`1.65589486``0.507949028``0`
`-4.342653607``0.375978249``0`
`5.998548466``0.131970779``0`


`(U × Sigma)×(V^T)`=
`1.65589486``0.507949028``0`
`-4.342653607``0.375978249``0`
`5.998548466``0.131970779``0`
×
`0.312515487``0.67168969``0.67168969`
`0.949912673``-0.220981799``-0.220981799`
`0``-0.707106781``0.707106781`


=
`1.65589486×0.312515487+0.507949028×0.949912673+0×0``1.65589486×0.67168969+0.507949028×-0.220981799+0×-0.707106781``1.65589486×0.67168969+0.507949028×-0.220981799+0×0.707106781`
`-4.342653607×0.312515487+0.375978249×0.949912673+0×0``-4.342653607×0.67168969+0.375978249×-0.220981799+0×-0.707106781``-4.342653607×0.67168969+0.375978249×-0.220981799+0×0.707106781`
`5.998548466×0.312515487+0.131970779×0.949912673+0×0``5.998548466×0.67168969+0.131970779×-0.220981799+0×-0.707106781``5.998548466×0.67168969+0.131970779×-0.220981799+0×0.707106781`


=
`0.517492789+0.482507219+0``1.112247505-0.11224749+0``1.112247505-0.11224749+0`
`-1.357146507+0.357146503+0``-2.916915655-0.08308435+0``-2.916915655-0.08308435+0`
`1.874639295+0.125360715+0``4.02916316-0.02916314+0``4.02916316-0.02916314+0`


=
`1.000000007``1.000000015``1.000000015`
`-1.000000004``-3.000000005``-3.000000005`
`2.000000011``4.00000002``4.00000002`


`1^"st"` Solution is possible.

`1^"st"` Solution is possible.


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


`U×Sigma`=
`0.218215109``-0.786796568``-0.577350269`
`-0.572278261``-0.582378112``0.577350269`
`0.79049337``-0.204418457``0.577350269`
×
`7.588360288``0``0`
`0``0.645591314``0`
`0``0``0`


=
`0.218215109×7.588360288-0.786796568×0-0.577350269×0``0.218215109×0-0.786796568×0.645591314-0.577350269×0``0.218215109×0-0.786796568×0-0.577350269×0`
`-0.572278261×7.588360288-0.582378112×0+0.577350269×0``-0.572278261×0-0.582378112×0.645591314+0.577350269×0``-0.572278261×0-0.582378112×0+0.577350269×0`
`0.79049337×7.588360288-0.204418457×0+0.577350269×0``0.79049337×0-0.204418457×0.645591314+0.577350269×0``0.79049337×0-0.204418457×0+0.577350269×0`


=
`1.655894867+0+0``0-0.50794903+0``0+0+0`
`-4.342653629+0+0``0-0.375978251+0``0+0+0`
`5.998548497+0+0``0-0.13197078+0``0+0+0`


=
`1.655894867``-0.50794903``0`
`-4.342653629``-0.375978251``0`
`5.998548497``-0.13197078``0`


`(U × Sigma)×(V^T)`=
`1.655894867``-0.50794903``0`
`-4.342653629``-0.375978251``0`
`5.998548497``-0.13197078``0`
×
`0.312515485``0.671689687``0.671689687`
`-0.949912665``0.220981819``0.220981819`
`0``-0.707106781``0.707106781`


=
`1.655894867×0.312515485-0.50794903×-0.949912665+0×0``1.655894867×0.671689687-0.50794903×0.220981819+0×-0.707106781``1.655894867×0.671689687-0.50794903×0.220981819+0×0.707106781`
`-4.342653629×0.312515485-0.375978251×-0.949912665+0×0``-4.342653629×0.671689687-0.375978251×0.220981819+0×-0.707106781``-4.342653629×0.671689687-0.375978251×0.220981819+0×0.707106781`
`5.998548497×0.312515485-0.13197078×-0.949912665+0×0``5.998548497×0.671689687-0.13197078×0.220981819+0×-0.707106781``5.998548497×0.671689687-0.13197078×0.220981819+0×0.707106781`


=
`0.517492788+0.482507217+0``1.112247505-0.112247501+0``1.112247505-0.112247501+0`
`-1.357146505+0.357146502+0``-2.916915657-0.083084358+0``-2.916915657-0.083084358+0`
`1.874639293+0.125360716+0``4.029163162-0.029163143+0``4.029163162-0.029163143+0`


=
`1.000000005``1.000000005``1.000000005`
`-1.000000003``-3.000000015``-3.000000015`
`2.000000008``4.000000019``4.000000019`


`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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