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Topic:
Leastsquares using polar decomposition
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0



Kaba
Posts:
289
Registered:
5/23/11


Leastsquares using polar decomposition
Posted:
Nov 16, 2012 11:05 AM


(I'm sending this again, since the last post did not appear on my news server, nor in Google Groups)
Hi,
Let P, R in R^{d x n}, A, Q, S in R^{d x d} with Q^T Q = I and S^T = S.
The problem is to find A to minimize
AP  R,
where . is the Frobenius norm. It can be shown that the solution is
A = RP^T (PP^T)^{1}.
I want to derive the result differently. First, notice that every A can be decomposed as A = QS. For if A = UDV^T is the singular value decomposition of A, then Q = UV^T and S = VDV^T is such a decomposition. Now I want to solve a new problem, which is to find Q and S to minimize
QSP  R.
Clearly, I could just compute A and then decompose it as above. But I don't want to do that. Instead, I want to derive the result by optimizing Q and S.
If I vary Q (subject to it being orthogonal), I get
PP^T S + SPP^T = Q^T RP^T + PR^T Q.
If I vary S (subject to it being symmetric), I get
Q^T RP^T S = SPR^T Q.
But then I am stuck. In a previous post this week, concerning leastsquares scaling where Q = I, it was noticed that
PP^T S + SPP^T = RP^T + PR^T
is a Lyapunov equation, and can be solved as it is. Therefore, I would except that the solution of this more general problem yields something similar to a Lyapunov equation, perhaps a Sylvester equation (more general), or an algebraic Riccatti equation (even more general):
http://en.wikipedia.org/wiki/Sylvester_equation
http://en.wikipedia.org/wiki/Algebraic_Riccati_equation
It seems to me that first I should be able give an implicit equation for either S or Q alone. It probably is also the case that the QS decomposition (polar decomposition) is not unique, and therefore an additional condition would be needed to make the solution unique.
Ideas appreciated. The reason I am interested in this is that I suspect there is a new technique (to me) underlying here.
 http://kaba.hilvi.org



