Date: Nov 13, 2012 12:13 PM
Author: Kaba
Subject: Re: Least-squares scaling
Ray Koopman wrote:

> On Nov 11, 10:55 am, Kaba <k...@nowhere.com> wrote:

>

> > Hi,

>

> >

>

> > Let

>

> >

>

> > R in R^{d times n}

>

> > P in R^{d times n}, and

>

> > S in R^{d times d}, S symmetric positive semi-definite.

>

> >

>

> > The problem is to find a matrix S such that the squared Frobenius norm

>

> >

>

> > E = |SP - R|^2

>

> >

>

> > is minimized. Geometrically, find a scaling which best relates the

>

> > paired vector sets P and Q. The E can be rewritten as

>

> >

>

> > E = tr((SP - R)^T (SP - R))

>

> > = tr(P^T S^2 P) - 2tr(P^T SR) + tr(R^T R)

>

> > = tr(S^2 PP^T) - 2tr(SRP^T) + tr(RR^T).

>

> >

>

> > Taking the first variation of E, with symmetric variations,

>

> > and setting it to zero gives that

>

> >

>

> > SPP^T + PP^T S = RP^T + PR^T

>

> >

>

> > holds in the minimum point. One can rearrange this to

>

> >

>

> > (SPP^T - RP^T)^T = -(SPP^T - RP^T),

>

> >

>

> > which says that SPP^T - RP^T is skew-symmetric.

>

> > But I have no idea how to make use of this fact. Anyone?

>

> >

>

> > --http://kaba.hilvi.org

>

>

>

> How do you intend to prevent S from having negative eigenvalues?

We can forget about S being positive-definite, and simply concentrate on symmetric S.

> What if R = -P ?

We get

PP^T (S + I) = -(S + I)PP^T.

<=>

PP^T S + SPP^T = -2PP^T

But I still don't know how to get forward.

Since I am away from home, it's Google Groups now... Hope this doesn't come out garbled.