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Topic:
Leastsquares scaling
Replies:
4
Last Post:
Nov 13, 2012 6:57 PM



Kaba
Posts:
3
Registered:
10/25/12


Re: Leastsquares scaling
Posted:
Nov 13, 2012 12:13 PM


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 semidefinite. > > > > > > 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 skewsymmetric. > > > 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 positivedefinite, 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.



