SVD - Singular Value Decomposition Example [[2,3],[4,10]]

Find SVD - Singular Value Decomposition ...
`[[2,3],[4,10]]`

Solution:

`A = `

	`2`	`3`
	`4`	`10`

`A' * A`

`A^T`

	`2`	`3`
	`4`	`10`

	`2`	`4`
	`3`	`10`

`(A^T)×A`

	`2`	`4`
	`3`	`10`

	`2`	`3`
	`4`	`10`

	`2×2+4×4`	`2×3+4×10`
	`3×2+10×4`	`3×3+10×10`

	`4+16`	`6+40`
	`6+40`	`9+100`

	`20`	`46`
	`46`	`109`

`A' * A = `

	`20`	`46`
	`46`	`109`

Find Eigen vector for `A' * A`

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

	`(20-lamda)`	`46`
	`46`	`(109-lamda)`

= 0

`:.(20-lamda) × (109-lamda) - 46 × 46=0`

`:.(2180-129lamda+lamda^2)-2116=0`

`:.(lamda^2-129lamda+64)=0`

`:.(lamda-0.498)(lamda-128.502)=0`

`:.(lamda-0.498)=0 or (lamda-128.502)=0`

`:.lamda=0.498 or lamda=128.502`

`:.` The eigenvalues of the matrix `A' * A` are given by `lamda=0.498,128.502`

1. Eigenvectors for `lamda=128.502`

`A' * A-lamdaI = `

	20	46
	46	109

- `128.502`

	1	0
	0	1

	20	46
	46	109

	128.502	0
	0	128.502

	`-108.502`	`46`
	`46`	`-19.502`

Now, reduce this matrix
`R_1 larr R_1-:(-108.502)`

	`1`	`-0.424`
	`46`	`-19.502`

`R_2 larr R_2-46xx R_1`

	`1`	`-0.424`
	`0`	`0`

The system associated with the eigenvalue `lamda=128.502`

`(A' * A-128.502I)`

	`x_1`
	`x_2`

	`1`	`-0.424`
	`0`	`0`

	`x_1`
	`x_2`

	`0`
	`0`

`=>x_1-0.424x_2=0`

`=>x_1=0.424x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=128.502` is

`v=`

	`0.424x_2`
	`x_2`

Let `x_2=1`

`v_1=`

	`0.424`
	`1`

`v_1=`

	`0.424`
	`1`

2. Eigenvectors for `lamda=0.498`

`A' * A-lamdaI = `

	20	46
	46	109

- `0.498`

	1	0
	0	1

	20	46
	46	109

	0.498	0
	0	0.498

	`19.502`	`46`
	`46`	`108.502`

Now, reduce this matrix
`R_1 larr R_1-:19.502`

	`1`	`2.3587`
	`46`	`108.502`

`R_2 larr R_2-46xx R_1`

	`1`	`2.3587`
	`0`	`0`

The system associated with the eigenvalue `lamda=0.498`

`(A' * A-0.498I)`

	`x_1`
	`x_2`

	`1`	`2.3587`
	`0`	`0`

	`x_1`
	`x_2`

	`0`
	`0`

`=>x_1+2.3587x_2=0`

`=>x_1=-2.3587x_2`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.498` is

`v=`

	`-2.3587x_2`
	`x_2`

Let `x_2=1`

`v_2=`

	`-2.3587`
	`1`

`v_2=`

	`-2.3587`
	`1`

For Eigenvector-1 `(0.424,1)`, Length L = `sqrt(|0.42396|^2+|1|^2)=1.0862`

So, normalizing gives `v_1=((0.424)/(1.0862),(1)/(1.0862))=(0.3903,0.9207)`

For Eigenvector-2 `(-2.3587,1)`, Length L = `sqrt(|-2.35874|^2+|1|^2)=2.562`

So, normalizing gives `v_2=((-2.3587)/(2.562),(1)/(2.562))=(-0.9207,0.3903)`

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

`:. U = `

	`0.3125`	`-0.9501`
	`0.9499`	`0.312`

`:. Sigma = `

	`sqrt(128.502)`	`0`
	`0`	`sqrt(0.498)`

`=`

	`11.3359`	`0`
	`0`	`0.7057`

`:. V = `

`[v_1,v_2]`

`=`

	`0.3903`	`-0.9207`
	`0.9207`	`0.3903`

Verify Solution `A = U Sigma V^T`

`U×Sigma`

	`0.31252`	`-0.95009`
	`0.94992`	`0.31202`

	`11.33587`	`0`
	`0`	`0.70572`

	`0.31252×11.33587+(-0.95009)×0`	`0.31252×0+(-0.95009)×0.70572`
	`0.94992×11.33587+0.31202×0`	`0.94992×0+0.31202×0.70572`

	`3.54269+0`	`0+(-0.6705)`
	`10.76817+0`	`0+0.2202`

	`3.54269`	`-0.6705`
	`10.76817`	`0.2202`

`(U × Sigma)×(V^T)`

	`3.54269`	`-0.6705`
	`10.76817`	`0.2202`

	`0.39033`	`0.92068`
	`-0.92068`	`0.39033`

	`3.54269×0.39033+(-0.6705)×(-0.92068)`	`3.54269×0.92068+(-0.6705)×0.39033`
	`10.76817×0.39033+0.2202×(-0.92068)`	`10.76817×0.92068+0.2202×0.39033`

	`1.38282+0.61731`	`3.26168+(-0.26172)`
	`4.20314+(-0.20273)`	`9.91404+0.08595`

	`2.00013`	`2.99996`
	`4.00041`	`9.99999`

Solution is possible.

This material is intended as a summary. Use your textbook for detail explanation.
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4. Example `[[2,3],[4,10]]`