# Does there exist a polar decomposition of matrices over finite fields?

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.

• since the singular value decomposition does not generalize to finite fields, doesn't the same apply to the polar decomposition? mathoverflow.net/questions/6987/… May 3, 2013 at 8:14
• Perhaps you should explain the motivation you have for such a decomposition. Sep 21, 2013 at 7:55

Orthogonal group consists of fixed-points of an involutary automorphism in $$\operatorname{GL}(n)$$. There is a general theory of involutions and symmetric varieties. For an involution $$\sigma$$ of a semi-simple algebraic group $$G$$, the subvariety $$\{x\sigma(x)\mid x\in G\}$$ is affine generalizes the notion set of symmetric matrices, viz. $$= \operatorname{GL}(n)/{\operatorname O(n)}$$.

See works of De Concini–Procesi or T.A. Springer for stratifying $$G/H$$, $$H$$ being the fixed-point subgroup of $$\sigma$$.

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

• When you are finding counterexamples, are you always computing the orthogonal group with respect to the quadratic form $x \mapsto x^{\mathsf T}x$, or do you consider other forms as well? Jul 5, 2021 at 17:41
• @LSpice I was considering those matrices which satisfy $AA^* = A^*A = I$ where $(-)^*$ is the "conjugate-transpose" operation. The conjugate-transpose operation generalises whatever involution you equip the ring of scalars with. For instance, over $\mathbb C$, you would usually equip the ring with the involution $a + bi \mapsto a - b i$. Over $\mathbb F_2$, $\mathbb F_3$ and $\mathbb F_5$ there is only one ring involution - so it's the only one I consider. I hope that makes sense.
• @LSpice "it is common to call such matrices unitary rather than orthogonal" - Yes, but if the involution is trivial, then both terms appear appropriate. "this is specialising to one choice of quadratic" - Interesting point. I only considered the sesquilinear form $u^* v$ when $(-)^*$ is derived from an arbitrary involution over the ring of scalars. "when you speak of ring involutions, in fact prime fields don't have any non-trivial ones" - And I'm working over prime fields, so indeed I only have the trivial ones.