TITLE: Does there exist a polar decomposition of matrices over finite fields? 
QUESTION [6 upvotes]: There exists a polar decomposition for matrices over the reals.
What I would like to know is if an analog has been shown for groups of matrices over finite fields. If not, it would be great to get some feedback as to how to proceed, or some reasons why it may not exist.
Specifically, I'm interested in decomposing invertible matrices over $\mathbb{F}_q$ as a product $x = s.o$ where $s$ is symmetric and $o$ is orthogonal.

REPLY [3 votes]: The answer is no for $\mathbb F_2$. A minimal invertible counterexample is $M = \left[\begin{matrix}1 & 1 & 0\\0 & 1 & 0\\0 & 1 & 1\end{matrix}\right]
$. This can be verified by observing that all orthogonal $3 \times 3$ matrices over $\mathbb F_2$ are permutation matrices. So if you multiply $M$ by each of the six permutation matrices, you never get a symmetric matrix.
The fact that $O_3(\mathbb F_2)$ only contains permutation matrices can be verified using the "inner-product preserving" property of orthogonal matrices.
Computational experiments show that a counterexample exists for $\mathbb F_3$ of size $3 \times 3$. For $\mathbb F_5$, there is a $2 \times 2$ counterexample.

Regarding alternatives to polar decomposition, in Gutin - Generalisations of singular value decomposition to dual-numbered matrices [1] I introduce an analogue of SVD for the ring of dual numbers equipped with the involution $a + b \varepsilon \mapsto a - b\varepsilon$. This analogue of SVD exists for every dual-numbered matrix, while the polar decomposition (its obvious generalisation) does not. What this means is that when generalising SVD it might be profitable to allow the matrix $S$ in $USV^*$ to not necessarily be Hermitian.
[1] - https://www.tandfonline.com/doi/full/10.1080/03081087.2021.1903830, but this has some formatting errors, so instead see https://arxiv.org/abs/2007.09693