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$\DeclareMathOperator\svd{svd}$Consider the SVD of rectangular matrices as operators

$$\svd:\mathbb C^{n\times m}\to \mathbb U_n\times \mathbb D_{n, m}\times\mathbb U_m$$

where $\mathbb U_n$ is the space of $n\times n$ unitary matrices and $\mathbb D_{n, m}$ is the space of real nonnegative diagonal rectangular matrices.

We know that this operator is not well defined, since there's more than one SVD for each matrix.

What I want to know is, there exists a specification of $\svd$ on $\mathbb C^{n\times m}$ that makes it a Borel-measurable function?

I already know that it can't be continuous, and that if I play enough with the signs of the entries, I guess I can make it not even Lebesgue measurable, but what I need here is for just one specification. If I had to try to do it, I guess I'd try to define it locally and then try to glue together the charts, but I feel like it might badly backfire.

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    $\begingroup$ There are wellknown algorithms to do the SVD decomposition, see f.i. Golub/van Loan, Matrix Computations (1996), Alg. 8.6.2. Is it possible that any of these algorithms lead to a non measurable decompoisition? $\endgroup$
    – Dieter Kadelka
    Commented Sep 21, 2021 at 9:39
  • $\begingroup$ You should consider the closed multimap taking M to the set of all triples (U,D,V) (unitary-diagonal-unitary) which are decompositions for M. Then apply the Kuratowski Ryll-Nardzewski measurable selection theorem. If I’m not wrong the map $(U,D,V)\mapsto UDV^*$ is a (smooth) submersion, hence open, hence measurable. en.wikipedia.org/wiki/… $\endgroup$
    – Pietro Majer
    Commented Sep 21, 2021 at 10:27
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    $\begingroup$ in fact if you have your continuous local specifications of SVD you don't have to bother to glue them: take them defined on members of a locally finite open cover $\{O_j\}_{j\in\mathbb N}$ of $\mathbb C^{n\times m}$, say $\text{sdv}_j:O_j\to \mathbb {U}_n\times \mathbb {D}_{n,m}\times \mathbb {U}_m$. Then you just map $M$ to $\text{sdv}_j(M)$ for the first index $j$ such that $M\in O_j$, and this defines a measurable map. $\endgroup$
    – Pietro Majer
    Commented Sep 21, 2021 at 11:26

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