SVD - Singular Value Decomposition Example [[4,0],[3,-5]]

1. Find SVD - Singular Value Decomposition ...
`[[4,0],[3,-5]]`

Solution:

`A = `

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

`A' * A`

`A^T`

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

	`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`

`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`

	`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`

`A' * A-lamdaI = `

	25	-15
	-15	25

- `10`

	1	0
	0	1

	25	-15
	-15	25

	10	0
	0	10

	`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)`

Solution
`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 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`

Solution is possible.

This material is intended as a summary. Use your textbook for detail explanation.
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1. Example `[[4,0],[3,-5]]`