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From: "Bruno Luong" <b.luong@fogale.findmycountry>
Newsgroups: comp.soft-sys.matlab
Subject: Re: PROOF?  XS'*XS = eye ==> (XS\X0)' = X0'*XS
Date: Wed, 13 Mar 2013 05:49:15 +0000 (UTC)
Organization: FOGALE nanotech
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"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