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Find Pseudoinverse [[4,0],[3,-5]]

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Your problem `->` Pseudoinverse [[4,0],[3,-5]]


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`


`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 `

`:.` 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` `1=-24-:-24`
`R_1 larr R_1-:-24`
 `-0.5` `-0.5=12-:-24`
`R_1 larr R_1-:-24`
`12``-6`


`R_2 larr R_2-12xx R_1`

 = 
`1``-0.5`
 `0` `0=12-12xx1`
`R_2 larr R_2-12xx R_1`
 `0` `0=-6-12xx-0.5`
`R_2 larr R_2-12xx R_1`


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
interchanging rows `R_1 harr R_2`

 = 
`12``24`
`6``12`


`R_1 larr R_1-:12`

 = 
 `1` `1=12-:12`
`R_1 larr R_1-:12`
 `2` `2=24-:12`
`R_1 larr R_1-:12`
`6``12`


`R_2 larr R_2-6xx R_1`

 = 
`1``2`
 `0` `0=6-6xx1`
`R_2 larr R_2-6xx R_1`
 `0` `0=12-6xx2`
`R_2 larr R_2-6xx R_1`


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)=1.11803`

So, normalizing gives `u_1=(0.5/1.11803,1/1.11803)=(0.44721,0.89443)`

For Eigenvector-2 `(-2,1)`, Length L = `sqrt((-2)^2+1^2)=2.23607`

So, normalizing gives `u_2=((-2)/2.23607,1/2.23607)=(-0.89443,0.44721)`



`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 `

`:.` 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_1 larr R_1-:-15`
 `1` `1=-15-:-15`
`R_1 larr R_1-:-15`
`-15``-15`


`R_2 larr R_2+15xx R_1`

 = 
`1``1`
 `0` `0=-15+15xx1`
`R_2 larr R_2+15xx R_1`
 `0` `0=-15+15xx1`
`R_2 larr R_2+15xx R_1`


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_1 larr R_1-:15`
 `-1` `-1=-15-:15`
`R_1 larr R_1-:15`
`-15``15`


`R_2 larr R_2+15xx R_1`

 = 
`1``-1`
 `0` `0=-15+15xx1`
`R_2 larr R_2+15xx R_1`
 `0` `0=15+15xx-1`
`R_2 larr R_2+15xx R_1`


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)=1.41421`

So, normalizing gives `v_1=((-1)/1.41421,1/1.41421)=(-0.70711,0.70711)`

For Eigenvector-2 `(1,1)`, Length L = `sqrt(1^2+1^2)=1.41421`

So, normalizing gives `v_2=(1/1.41421,1/1.41421)=(0.70711,0.70711)`

`1^"st"` Solution

`:. Sigma = `
`sqrt(40)``0`
`0``sqrt(10)`
`=`
`6.32456``0`
`0``3.16228`


`:. U = ``[u_1,u_2]``=`
`0.44721``-0.89443`
`0.89443``0.44721`


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

`:. V = `
`0.70711``-0.70711`
`-0.70711``-0.7071`


Or
`2^"nd"` Solution

`:. Sigma = `
`sqrt(40)``0`
`0``sqrt(10)`
`=`
`6.32456``0`
`0``3.16228`


`:. V = ``[v_1,v_2]``=`
`-0.70711``0.70711`
`0.70711``0.70711`


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

`:. U = `
`-0.44722``0.89443`
`-0.89443``-0.44722`


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


`U×Sigma`=
`0.4472``-0.8944`
`0.8944``0.4472`
×
`6.3246``0`
`0``3.1623`


=
`0.4472×6.3246-0.8944×0``0.4472×0-0.8944×3.1623`
`0.8944×6.3246+0.4472×0``0.8944×0+0.4472×3.1623`


=
`2.8284+0``0-2.8284`
`5.6569+0``0+1.4142`


=
`2.8284``-2.8284`
`5.6569``1.4142`


`(U × Sigma)×(V^T)`=
`2.8284``-2.8284`
`5.6569``1.4142`
×
`0.7071``-0.7071`
`-0.7071``-0.7071`


=
`2.8284×0.7071-2.8284×-0.7071``2.8284×-0.7071-2.8284×-0.7071`
`5.6569×0.7071+1.4142×-0.7071``5.6569×-0.7071+1.4142×-0.7071`


=
`2+2``-2+2`
`4-1``-4-1`


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


`1^"st"` Solution is possible.

`1^"st"` Solution is possible.


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


`U×Sigma`=
`-0.4472``0.8944`
`-0.8944``-0.4472`
×
`6.3246``0`
`0``3.1623`


=
`-0.4472×6.3246+0.8944×0``-0.4472×0+0.8944×3.1623`
`-0.8944×6.3246-0.4472×0``-0.8944×0-0.4472×3.1623`


=
`-2.8284+0``0+2.8284`
`-5.6569+0``0-1.4142`


=
`-2.8284``2.8284`
`-5.6569``-1.4142`


`(U × Sigma)×(V^T)`=
`-2.8284``2.8284`
`-5.6569``-1.4142`
×
`-0.7071``0.7071`
`0.7071``0.7071`


=
`-2.8284×-0.7071+2.8284×0.7071``-2.8284×0.7071+2.8284×0.7071`
`-5.6569×-0.7071-1.4142×0.7071``-5.6569×0.7071-1.4142×0.7071`


=
`2+2``-2+2`
`4-1``-4-1`


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


`2^"nd"` Solution is possible.

`2^"nd"` Solution is possible.


`:. U = `
`0.44721``-0.89443`
`0.89443``0.44721`
`, E = `
`6.32456``0`
`0``3.16228`
`, V = `
`0.70711``-0.70711`
`-0.70711``-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
=
`0.1581``0`
`0``0.3162`
T
=
`0.15811``0`
`0``0.31623`


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

`V×(Sigma^(+))`=
`0.70711``-0.70711`
`-0.70711``-0.7071`
×
`0.15811``0`
`0``0.31623`


=
`0.70711×0.15811-0.70711×0``0.70711×0-0.70711×0.31623`
`-0.70711×0.15811-0.7071×0``-0.70711×0-0.7071×0.31623`


=
`0.1118+0``0-0.22361`
`-0.1118+0``0-0.22361`


=
`0.1118``-0.22361`
`-0.1118``-0.22361`


`(V × (Sigma^(+)))×(U^T)`=
`0.1118``-0.22361`
`-0.1118``-0.22361`
×
`0.44721``0.89443`
`-0.89443``0.44721`


=
`0.1118×0.44721-0.22361×-0.89443``0.1118×0.89443-0.22361×0.44721`
`-0.1118×0.44721-0.22361×-0.89443``-0.1118×0.89443-0.22361×0.44721`


=
`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`







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