# Talk:Polar decomposition

## Spectral decomposition when A is normal

It should surely be possible to say something about Λ and Q in terms of W, V and Σ. Can anybody give the relation? At the moment I can't quite see it. -- Jheald 02:49, 13 January 2006 (UTC).

Sorted. (I think). -- Jheald 11:50, 13 January 2006 (UTC).

## a comment explaining revert

by the expression that A = UP, A is implicitly assumed to be square, since unitarity and positive semidefiniteness make sense only for square matrices. if A is not square, then a similar decomposition can probably be obtained, but U would then be weakened to a partial isometry or isometry (in the sense of matrices). it seems to me that only in the latter, mor general, case the comment that anon was adding would become necessary. Mct mht 03:19, 2 October 2006 (UTC)

Don't think this is true. The implication surely for a non-square A is that you can either zero-pad it, adding null rows or null columns; or (alternatively/equivalently) that U may non-square and span less than a full basis.
that's what is meant by an isometry or partial isometry (in the sense of matrices). Mct mht 14:49, 2 October 2006 (UTC)
Either way, there should be no problem in practice obtaining a polar decomposition for a non-square matrix. Jheald 12:06, 2 October 2006 (UTC).
that's what i said. what exactly is not true? Mct mht 14:49, 2 October 2006 (UTC)

## Applications

It would be nice to read about applications. Anybody? CDaMama 16:56, 3 January 2007 (UTC)

## A invertible

The article says the following (which I agree with): "This decomposition always exists; and so long as A is invertible, it is unique, with P positive-definite." After this, the section implicitly assumes that A is invertible, and this assumption should be stated explicitly, or only when needed. For example, the formula P = sqrt(A*A) is always valid, even if A is singular, and P is always unique. But if A is singular, then so is P, and the formula U = A*inverse(P) is no longer valid; in this case, U is no longer unique, although it of course still has lots of meaningful structure. Lavaka (talk) 19:40, 25 September 2008 (UTC)

(brief addition to my previous comment) if A is not square, then all the "nonsingular" properties still hold as long as A is full-rank Lavaka (talk) 19:45, 25 September 2008 (UTC)

I agree with this comment. It seems to me that the only problematic place is the one flagged up by you, i.e. the formula U = A*inverse(P). The cure could be simply to add "If A is invertible..." to the beginning of this sentence containing this formula. Spectral Theorist (talk) 22:03, 25 November 2008 (UTC)

I edited the article accordingly. -- Jitse Niesen (talk) 11:56, 26 November 2008 (UTC)

## Proofs in the "Bounded operators on Hilbert space" section

The proofs in this section seem to break up the exposition somewhat. The presence of proofs here is also inconsistent with the style of the first section. Wouldn't it be better to stick to the statements without proofs? Or delegate the proofs to a separate subsection? Spectral Theorist (talk) 22:08, 25 November 2008 (UTC)

## Trivial fact about complex conjugate

Recently, this was added to the page:

"The decomposition of the complex conjugate of $A$ is given by $\overline{A} = \overline{U}$ $\overline{P}$."

I reverted this edit, since it's a trivial fact that has nothing to do with the rest of the article. Of course that's the polar decomposition of $\overline{A}$ -- you just take the complex conjugate of both sides. The person who originally added that to the article then undid my edit and put it back in without giving a reason. So I ask you, why include this fact in the article? We might as well add a line saying "The decomposition of $2A$ is given by $2A = U(2P)$". JokeySmurf (talk) 01:52, 19 November 2009 (UTC)

Consider "the polar decomposition... is... analogous to the polar form of a nonzero complex number z = r.e^ip". The complex conjugate of z is r.e^-ip, or r.conj(e^ip). By analogy, one might think the polar form of conj(A) is U conj(P) but it isn't. The edit was trivial, but I felt it's worth pointing out since the analogy with complex numbers fails here. I'd suggest the analogy may not be worth making if it fails on this point. BTW I won't revert the change if you decide to change it back :) —Preceding unsigned comment added by 132.239.142.130 (talk) 21:48, 25 November 2009 (UTC)
The analogy doesn't fail on that point at all. P (positive semidefinite) is analogous to r (positive real number), not p. U is analogous to e^ip. The complex conjugate gets applied to U just as it gets applied to e^ip. JokeySmurf (talk) 16:17, 26 November 2009 (UTC)
Yes I see - my error. But still, the analogy is broken since P (i.e. the real number r) must also be conjugated. —Preceding unsigned comment added by 132.239.142.130 (talk) 00:58, 1 December 2009 (UTC)
No it's not, because the operation on Mn that is analogous to the complex conjugate is the conjugate transpose (P^*, not conj(P)) -- this is related to the fact that Hermitian matrices have real eigenvalues and are thus analogous to real numbers (**not** matrices such that conj(A) = A). Since P^* = P, there's no problem. JokeySmurf (talk) 03:07, 1 December 2009 (UTC)

## Iterative solutions to the polar decomposition

I was wondering if there might be any benefit to adding information about some iterative solutions to this decomposition?

One method I know of for finding the "U" part of the decomposition involves:
U(0) = A
U(k+1) = (U(k) + pinv(U(k)')/2
XWolfRH (talk) 08:25, 1 April 2012 (UTC)

Hi, some time ago I added something like that to the german version, along with published articles for convergence proofs.--LutzL (talk) 11:29, 2 April 2012 (UTC)