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Thread Subject:
PROOF? XS'*XS = eye ==> (XS\X0)' = X0'*XS

Subject: PROOF? XS'*XS = eye ==> (XS\X0)' = X0'*XS

From: Greg Heath

Date: 12 Mar, 2013 23:37:10

Message: 1 of 2

help plsregress

contains the line

XL = (XS\X0)' = X0'*XS

When I plugged in numbers from the problem it checks out. Since the second equality looked strange, I tried to prove it. I could not.

With my curiosity peaked, I put random numbers in and found that it is not true in general. More curious than ever, I went back to XS from my problem and played around.

AHA! I found that XS'*XS = the identity matrix but XS*XS' does not.

However, the proof still eludes me.

How 'bout it?

Greg

Subject: PROOF? XS'*XS = eye ==> (XS\X0)' = X0'*XS

From: Bruno Luong

Date: 13 Mar, 2013 05:49:15

Message: 2 of 2

"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

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