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Topic: Least-squares scaling
Replies: 4   Last Post: Nov 13, 2012 6:57 PM

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Kaba

Posts: 3
Registered: 10/25/12
Re: Least-squares scaling
Posted: Nov 13, 2012 12:13 PM
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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.



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