Date: Mar 13, 2013 1:49 AM
Author: Bruno Luong
Subject: Re: PROOF? XS'*XS = eye ==> (XS\X0)' = X0'*XS
"Greg Heath" <firstname.lastname@example.org> wrote in message <email@example.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.
Y = XS'*X0. (since XS'*XS = eye)
Y = XS\X0 = XS'*X0
Transpose that you get your identity.