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Topic: PROOF? XS'*XS = eye ==> (XS\X0)' = X0'*XS
Replies: 1   Last Post: Mar 13, 2013 1:49 AM

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 Bruno Luong Posts: 9,822 Registered: 7/26/08
Re: PROOF? XS'*XS = eye ==> (XS\X0)' = X0'*XS
Posted: Mar 13, 2013 1:49 AM

"Greg Heath" <heath@alumni.brown.edu> wrote in message <khoe76\$bsd\$1@newscl01ah.mathworks.com>...
> help plsregress
>
> contains the line
>
> XL = (XS\X0)' = X0'*XS
>

If XS'*XS = eye(n), XS is full column-rank (the columns of XS is orthonormal)

(XS \ X0) is a not an overdetermined system, and there gives a unique solution Y of the least square solution:

Y = argmin |XS*Y - X0|^2

where |.| is the L2 norm. The Euler-Lagrange condition is:

XS'*XS*Y-XS'*X0 = 0.

So
Y = XS'*X0. (since XS'*XS = eye)

That means
Y = XS\X0 = XS'*X0

Transpose that you get your identity.

Bruno

Date Subject Author
3/12/13 Greg Heath
3/13/13 Bruno Luong