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The title might require a little explanation. Given a complex matrix $Q\in\mathbb{C}^{n\times n}$ it has a singular value decomposition $Q=UD_QV^T$ with unitary matrices $U$ and $V$ and real, positive definit diagonal $D_Q$.

Let's say we have the SVD another matrix $Q'=WD_{Q'}X^T$ and we know that for some unitary matrices $A$ and $B$ both $A^H Q B^*$ and $A^H Q' B^*$ (where $^H$ is hermitian conjugation and $^*$ is complex conjugation, so that $A^H Q B^*=D_Q$) are diagonal with only positive entries on the diagonal. That is, there exist matrices which simultaneously decompose $Q$ and $Q'$ into singular values.

We know, that the SVD is in general not unique. My question is, whether it is possible to make some statements about the product $U^H Q' V^*$. Will this necessarily also be a diagonal matrix?

HerpDerpington
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    Here's a relevant fact: given to matrices $P,Q$ of the same shape, there exist unitary matrices $A,B$ such that both $APB$ and $AQB$ are real and diagonal if and only if both $PQ^$ and $Q^P$ are Hermitian. I'm not sure about a reference and I don't have a proof off the top of my head, but the fact is used in this paper for instance. – Ben Grossmann May 09 '22 at 15:10
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    In general, we can't guarantee that $U^HQ'V^$ will have any special structure. A counterexample exists with $n = 2$ and $Q = I$ (the identity matrix). I suspect, however, that if $Q$ has no repeating singular values, then $U^HQ'V^$ can be guaranteed to be diagonal. – Ben Grossmann May 09 '22 at 15:15

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