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Problem: Pseudoinverse [[4,0],[3,-5]] [ Calculator, Method and examples ]

Solution:
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 =
 16 12 12 34
- 40
 1 0 0 1

=
 16 12 12 34
-
 40 0 0 40

=
 -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 =
 16 12 12 34
- 10
 1 0 0 1

=
 16 12 12 34
-
 10 0 0 10

=
 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 2 6 12

R_2 larr R_2-6xx 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)=1.11803

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 -15 25
- 40
 1 0 0 1

=
 25 -15 -15 25
-
 40 0 0 40

=
 -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

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