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Topic: Least-squares using polar decomposition
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Posts: 289
Registered: 5/23/11
Least-squares using polar decomposition
Posted: Nov 16, 2012 11:05 AM
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(I'm sending this again, since the last post did not appear on my news
server, nor in Google Groups)


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


But then I am stuck. In a previous post this week, concerning
least-squares 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):

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.


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