Complex-Symmetric Matrix SVD with Matlab: Conjugate Relationship Not Seen

[Heavytortoise]

For a square, complex-symmetric matrix $A$, the columns of the right and left matrices $U$ and $V$ of the singular value decomposition should be complex conjugates, since for $$A=A^T, A\in{\mathbb C}^{N\times N}$$,
$$A = U\Sigma V^H, A^T=(U\Sigma V^H)^T$$
so that
$$U\Sigma V^H=(V^H)^T\Sigma U^T.$$
Then we have $$U=(V^H)^T$$, right? So why isn't this the case when I run a few experiments with Matlab? The magnitudes of the elements of $U$ and $V$ are equal, but they aren't conjugates. The expected relationship holds for real $A$, where $U$ and $V$ are real-valued, but not for complex symmetric matrices. Who's screwed up here, me or Matlab?

 

[scottdave]

This question has gone a long time without being answered.

I don't have a solution to your question but some insight.

Note that the documentation for Matlab's svd() command does not mention complex valued matrices.

The command will accept complex-valued input and produce results U, Sigma, and V. And you can use U*Sigma*V^T to get back the same A.

The functions like svd() in Matlab are numeric processes.

I can confirm that the open-source Octave produces similar results as Matlab.

It's likely that searching the Mathworks forum will get an answer from someone familiar with the inner workings of Matlab's svd() function.

https://www.mathworks.com/matlabcentral/content/communities.html

 

[Office_Shredder]

I don't have Matlab available right now, but the svd decomposition is not unique. In particular the U and V define the unit vectors that are being mapped to each other, and in complex land you can scale the unit vectors on each side by complex numbers of norm 1 in an appropriate way and get another svd representation. I'm guessing this is causing the problem, the algorithm doesn't guarantee if picks a specific choice of representation.