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Kaba
Posts:
3
Registered:
10/25/12
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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
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> >
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> > R in R^{d times n}
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> > P in R^{d times n}, and
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> > S in R^{d times d}, S symmetric positive semi-definite.
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> >
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> > The problem is to find a matrix S such that the squared Frobenius norm
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> >
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> > E = |SP - R|^2
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> >
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> > is minimized. Geometrically, find a scaling which best relates the
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> > paired vector sets P and Q. The E can be rewritten as
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> >
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> > E = tr((SP - R)^T (SP - R))
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> > = tr(P^T S^2 P) - 2tr(P^T SR) + tr(R^T R)
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> > = tr(S^2 PP^T) - 2tr(SRP^T) + tr(RR^T).
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> >
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> > Taking the first variation of E, with symmetric variations,
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> > and setting it to zero gives that
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> >
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> > SPP^T + PP^T S = RP^T + PR^T
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> >
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> > holds in the minimum point. One can rearrange this to
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> >
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> > (SPP^T - RP^T)^T = -(SPP^T - RP^T),
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> >
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> > which says that SPP^T - RP^T is skew-symmetric.
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> > But I have no idea how to make use of this fact. Anyone?
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> >
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> > --http://kaba.hilvi.org
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>
>
> 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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