```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^TBut 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.
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