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Home > Matrix & Vector > Matrix operation > Moore-Penrose Pseudoinverse of a Matrix example
18. Moore-Penrose Pseudoinverse of a Matrix example ( Enter your problem )
( Enter your problem )
  1. Example `[[4,0],[3,-5]]` `("Formula " A^(+)=V Sigma^(+) U^T)`
  2. Example `[[1,0,1,0],[0,1,0,1]]` `("Formula " A^(+)=V Sigma^(+) U^T)`
  3. Example `[[4,0],[3,-5]]` `("Formula " A^(+)=A^T * (A*A^T)^(-1))`
  4. Example `[[1,0,1,0],[0,1,0,1]]` `("Formula " A^(+)=A^T * (A*A^T)^(-1))`
  5. Example `[[1,-2,3],[5,8,-1],[2,1,1],[-1,4,-3]]` `("Formula " A^(+)=(A^T*A)^(-1) * A^T)`

1. Example `[[4,0],[3,-5]]` `("Formula " A^(+)=V Sigma^(+) U^T)`


1. Find Moore-Penrose Pseudoinverse ...
`[[4,0],[3,-5]]`

Solution:
The Moore-Penrose pseudoinverse `A^(+)` is calculated from SVD (Singular Value Decomposition) of a matrix A,
`A = U Sigma V^T`
then Moore-Penrose pseudoinverse `A^(+)` is given by
`A^(+) = V Sigma^(+) U^T`
where `Sigma^(+)` is obtained by taking the reciprocal of each non-zero data on the diagonal of `Sigma`, leaving all other zeros as it is, and then taking transpose of the resultant matrix.


`U, Sigma, V` using SVD : `A = U Sigma V^T`


`A = `
`4``0`
`3``-5`


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` = 
`4``0`
`3``-5`
T
 = 
`4``3`
`0``-5`


`A×(A^T)`=
`4``0`
`3``-5`
×
`4``3`
`0``-5`


=
`4×4+0×0``4×3+0×(-5)`
`3×4+(-5)×0``3×3+(-5)×(-5)`


=
`16+0``12+0`
`12+0``9+25`


=
`16``12`
`12``34`
`A * A' = `
`16``12`
`12``34`


Find Eigen vector for `A * A'`

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

 `(16-lamda)`  `12` 
 `12`  `(34-lamda)` 
 = 0


`:.(16-lamda) × (34-lamda) - 12 × 12=0`

`:.(544-50lamda+lamda^2)-144=0`

`:.(lamda^2-50lamda+400)=0`

`:.(lamda-10)(lamda-40)=0`

`:.(lamda-10)=0 or (lamda-40)=0`

`:.lamda=10 or lamda=40`

`:.` The eigenvalues of the matrix `A * A'` are given by `lamda=10,40`

1. Eigenvectors for `lamda=40`



1. Eigenvectors for `lamda=40`

`A * A'-lamdaI = `
1612
1234
 - `40` 
10
01


 = 
1612
1234
 - 
400
040

 = 
`-24``12`
`12``-6`


Now, reduce this matrix
`R_1 larr R_1-:(-24)`

 = 
`1``-0.5`
`12``-6`


`R_2 larr R_2-12xx R_1`

 = 
`1``-0.5`
`0``0`


The system associated with the eigenvalue `lamda=40`

`(A * A'-40I)`
`x_1`
`x_2`
 = 
`1``-0.5`
`0``0`
 
`x_1`
`x_2`
 = 
`0`
`0`


`=>x_1-0.5x_2=0`

`=>x_1=0.5x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=40` is

`v=`
`0.5x_2`
`x_2`


Let `x_2=1`

`v_1=`
`0.5`
`1`
`v_1=`
`0.5`
`1`


2. Eigenvectors for `lamda=10`



2. Eigenvectors for `lamda=10`

`A * A'-lamdaI = `
1612
1234
 - `10` 
10
01


 = 
1612
1234
 - 
100
010

 = 
`6``12`
`12``24`


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

 = 
`1``2`
`12``24`


`R_2 larr R_2-12xx R_1`

 = 
`1``2`
`0``0`


The system associated with the eigenvalue `lamda=10`

`(A * A'-10I)`
`x_1`
`x_2`
 = 
`1``2`
`0``0`
 
`x_1`
`x_2`
 = 
`0`
`0`


`=>x_1+2x_2=0`

`=>x_1=-2x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=10` is

`v=`
`-2x_2`
`x_2`


Let `x_2=1`

`v_2=`
`-2`
`1`
`v_2=`
`-2`
`1`


For Eigenvector-1 `(0.5,1)`, Length L = `sqrt(|0.5|^2+|1|^2)=1.118`

So, normalizing gives `u_1=((0.5)/(1.118),(1)/(1.118))=(0.4472,0.8944)`

For Eigenvector-2 `(-2,1)`, Length L = `sqrt(|-2|^2+|1|^2)=2.2361`

So, normalizing gives `u_2=((-2)/(2.2361),(1)/(2.2361))=(-0.8944,0.4472)`

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

`A' * A`
`A^T` = 
`4``0`
`3``-5`
T
 = 
`4``3`
`0``-5`


`(A^T)×A`=
`4``3`
`0``-5`
×
`4``0`
`3``-5`


=
`4×4+3×3``4×0+3×(-5)`
`0×4+(-5)×3``0×0+(-5)×(-5)`


=
`16+9``0+(-15)`
`0+(-15)``0+25`


=
`25``-15`
`-15``25`
`A' * A = `
`25``-15`
`-15``25`


Find Eigen vector for `A' * A`

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

 `(25-lamda)`  `-15` 
 `-15`  `(25-lamda)` 
 = 0


`:.(25-lamda) × (25-lamda) - (-15) × (-15)=0`

`:.(625-50lamda+lamda^2)-225=0`

`:.(lamda^2-50lamda+400)=0`

`:.(lamda-10)(lamda-40)=0`

`:.(lamda-10)=0 or (lamda-40)=0`

`:.lamda=10 or lamda=40`

`:.` The eigenvalues of the matrix `A' * A` are given by `lamda=10,40`

1. Eigenvectors for `lamda=40`



1. Eigenvectors for `lamda=40`

`A' * A-lamdaI = `
25-15
-1525
 - `40` 
10
01


 = 
25-15
-1525
 - 
400
040

 = 
`-15``-15`
`-15``-15`


Now, reduce this matrix
`R_1 larr R_1-:(-15)`

 = 
`1``1`
`-15``-15`


`R_2 larr R_2+15xx R_1`

 = 
`1``1`
`0``0`


The system associated with the eigenvalue `lamda=40`

`(A' * A-40I)`
`x_1`
`x_2`
 = 
`1``1`
`0``0`
 
`x_1`
`x_2`
 = 
`0`
`0`


`=>x_1+x_2=0`

`=>x_1=-x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=40` is

`v=`
`-x_2`
`x_2`


Let `x_2=1`

`v_1=`
`-1`
`1`
`v_1=`
`-1`
`1`


2. Eigenvectors for `lamda=10`



2. Eigenvectors for `lamda=10`

`A' * A-lamdaI = `
25-15
-1525
 - `10` 
10
01


 = 
25-15
-1525
 - 
100
010

 = 
`15``-15`
`-15``15`


Now, reduce this matrix
`R_1 larr R_1-:15`

 = 
`1``-1`
`-15``15`


`R_2 larr R_2+15xx R_1`

 = 
`1``-1`
`0``0`


The system associated with the eigenvalue `lamda=10`

`(A' * A-10I)`
`x_1`
`x_2`
 = 
`1``-1`
`0``0`
 
`x_1`
`x_2`
 = 
`0`
`0`


`=>x_1-x_2=0`

`=>x_1=x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=10` is

`v=`
`x_2`
`x_2`


Let `x_2=1`

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


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

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

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

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

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

`:. U = ``[u_1,u_2]``=`
`0.4472``-0.8944`
`0.8944``0.4472`


`:. Sigma = `
`sqrt(40)``0`
`0``sqrt(10)`
`=`
`6.3246``0`
`0``3.1623`


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

`:. V = `
`0.7071``-0.7071`
`-0.7071``-0.7071`



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

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

`:. U = `
`-0.4472``0.8944`
`-0.8944``-0.4472`


`:. Sigma = `
`sqrt(40)``0`
`0``sqrt(10)`
`=`
`6.3246``0`
`0``3.1623`


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



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


`U×Sigma`=
`0.44721``-0.89443`
`0.89443``0.44721`
×
`6.32456``0`
`0``3.16228`


=
`0.44721×6.32456+(-0.89443)×0``0.44721×0+(-0.89443)×3.16228`
`0.89443×6.32456+0.44721×0``0.89443×0+0.44721×3.16228`


=
`2.82841+0``0+(-2.82844)`
`5.65688+0``0+1.4142`


=
`2.82841``-2.82844`
`5.65688``1.4142`


`(U × Sigma)×(V^T)`=
`2.82841``-2.82844`
`5.65688``1.4142`
×
`0.70709``-0.70709`
`-0.70709``-0.70709`


=
`2.82841×0.70709+(-2.82844)×(-0.70709)``2.82841×(-0.70709)+(-2.82844)×(-0.70709)`
`5.65688×0.70709+1.4142×(-0.70709)``5.65688×(-0.70709)+1.4142×(-0.70709)`


=
`1.99994+1.99996``-1.99994+1.99996`
`3.99992+(-0.99997)``-3.99992+(-0.99997)`


=
`3.9999``0`
`2.99995``-4.99989`


`1^"st"` Solution is possible.

`1^"st"` Solution is possible.


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


`U×Sigma`=
`-0.44721``0.89442`
`-0.89442``-0.44721`
×
`6.32456``0`
`0``3.16228`


=
`-0.44721×6.32456+0.89442×0``-0.44721×0+0.89442×3.16228`
`-0.89442×6.32456+(-0.44721)×0``-0.89442×0+(-0.44721)×3.16228`


=
`-2.82841+0``0+2.82841`
`-5.65681+0``0+(-1.4142)`


=
`-2.82841``2.82841`
`-5.65681``-1.4142`


`(U × Sigma)×(V^T)`=
`-2.82841``2.82841`
`-5.65681``-1.4142`
×
`-0.70711``0.70711`
`0.70711``0.70711`


=
`-2.82841×(-0.70711)+2.82841×0.70711``-2.82841×0.70711+2.82841×0.70711`
`-5.65681×(-0.70711)+(-1.4142)×0.70711``-5.65681×0.70711+(-1.4142)×0.70711`


=
`1.99999+1.99999``-1.99999+1.99999`
`3.99999+(-1)``-3.99999+(-1)`


=
`3.99999``0`
`2.99999``-4.99999`


`2^"nd"` Solution is possible.

`2^"nd"` Solution is possible.


`:. U = `
`0.4472``-0.8944`
`0.8944``0.4472`
`, E = `
`6.3246``0`
`0``3.1623`
`, V = `
`0.7071``-0.7071`
`-0.7071``-0.7071`


Now, `Sigma^(+)` is obtained by taking the reciprocal of each non-zero data on the diagonal of `Sigma`, leaving all other zeros as it is, and then taking transpose of the resultant matrix.

`:. Sigma^(+) = `
`1/6.3246``0`
`0``1/3.1623`
T
=
`0.15811``0`
`0``0.31623`
T
=
`0.1581``0`
`0``0.3162`


Now, Moore-Penrose pseudoinverse `A^(+) = V Sigma^(+) U^T`

`V×(Sigma^(+))`=
`0.7071``-0.7071`
`-0.7071``-0.7071`
×
`0.1581``0`
`0``0.3162`


=
`0.7071×0.1581+(-0.7071)×0``0.7071×0+(-0.7071)×0.3162`
`-0.7071×0.1581+(-0.7071)×0``-0.7071×0+(-0.7071)×0.3162`


=
`0.1118+0``0+(-0.2236)`
`-0.1118+0``0+(-0.2236)`


=
`0.1118``-0.2236`
`-0.1118``-0.2236`


`(V × (Sigma^(+)))×(U^T)`=
`0.1118``-0.2236`
`-0.1118``-0.2236`
×
`0.4472``0.8944`
`-0.8944``0.4472`


=
`0.1118×0.4472+(-0.2236)×(-0.8944)``0.1118×0.8944+(-0.2236)×0.4472`
`-0.1118×0.4472+(-0.2236)×(-0.8944)``-0.1118×0.8944+(-0.2236)×0.4472`


=
`0.05+0.2``0.1+(-0.1)`
`-0.05+0.2``-0.1+(-0.1)`


=
`0.25``0`
`0.15``-0.2`


`:. A^(+) = `
`0.25``0`
`0.15``-0.2`



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