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Out of curiosity, I came across the study of matrices of the form $U = B^\top B^{-1}$ for $B \in GL(n,\mathbb{F}_2)$. In essence, U represents how matrice $B$ is asymetric.

This raised the question: under what conditions is a given matrix $U$ representable as $B^\top B^{-1}$?

It is clear that $U \sim U^{-1}$, but I was unable to progress further. Perhaps there are certain conditions on the Jordan form of $U$?

I "tried" to find similar question on this site, but no luck.

Here some insight: take a substituion $U = J_k(\lambda)$ — Jordan block of size k for eigenvalue $\lambda$. For $\lambda = 1$:

  1. If $k = 2m + 1$, then $J_k(\lambda)$ can be a posible solution
  2. If $k = 2m$, then $J_k(\lambda)$ cannot be as U. Therefore, for such eigenvalue $\lambda$ we may have that U have even number of Jordan blocks.

This came from finding an inverse of $J_k(\lambda)$, which can be written down easily with respect to $\mathbb{F}_2$ addition.

Update: After quite some work, I finally made a sketch for an analysis:

  1. To reduce a problem, let $J_U = P U P^{-1}$ be a Jordan form of $U$. Assume that $J_U = C^\top C^{-1}$. Then $B = P C P^\top$. Now, this problem reduces to checking an assumption on $J_U$.
  2. First, let $\lambda \in \overline{\mathbb{F}}_2$ be the only eigenvalue of $U$. Keeping in mind that $U \sim U^{-1}$ and therefore $J_U \sim J_U^{-1}$, analyze $J_U = J_k(\lambda) \oplus J_k(\lambda^{-1})$.
  3. Second, let $\lambda = 1$. Take $J_U = J_k(1)$ and rewrite $J_U = C^\top C^{-1}$ as $$C + C^\top = J_k(0) C.$$ We will have a reccurence relation on $c_{i,j}$. Analyze those for even and odd $k$ separately.
  4. Lastly, check $J_U$ as direct sum of such Jordan blocks and analyze, how direct sum of matrices affects the representation of $J_U$ as $C^\top C^{-1}$.
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  • $\begingroup$ This post is relevant (but of limited applicability since it's about matrices over $\Bbb R$ instead of $\Bbb F_2$) $\endgroup$ Commented Nov 5 at 14:21
  • $\begingroup$ An interesting example of a $U$ not representable in the form $U = B^TB^{-1}$ (over any field): $$ U = \pmatrix{0&1\\1&0} $$ Note that $U = U^{-1}$, hence $U \sim U^{-1}$. However, $$ B^TB^{-1} = U \implies B^T = UB, $$ but it's straightforward to show that the only matrices $B$ satisfying $B^T = UB$ are those with a single repeated entry, and such a matrix $B$ cannot be invertible. $\endgroup$ Commented Nov 5 at 14:35
  • $\begingroup$ A probably useless alternative formulation: since every square matrix is the product of two symmetric matrices, if you express $B$ as such a product, you are essentially asking what $U$ is a commutator of invertible symmetric matrices. $\endgroup$ Commented Nov 5 at 16:41
  • $\begingroup$ @BenGrossmann , quite an interesting example indeed! But that post is quite too much too limited due to a structure for $\overline{\mathbb{F}}_2$ (not even $\mathbb{F}_2$) $\endgroup$ Commented Nov 5 at 17:58
  • $\begingroup$ @user1551 , it may be overkill, but nonetheless a valuable note! If we let $B = S_1 S_2$, then it might be reformulated as such: in which bases do $S_1$ and $S_2^{-1}$ commute? $\endgroup$ Commented Nov 5 at 18:10

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Finally finished this. Here's a

Theorem. A matrix $U \in GL(n,\mathbb{F}_2)$ is decomposable as $B^\top B^{-1}$ if and only if its Jordan normal form $J_U$ is a direct sum of the following three types of blocks:

  1. $J_k(\omega) \oplus J_k(\omega^{-1})$ for some $\omega$ in a finite extension field $E$ of $\mathbb{F}_2$ ($\omega \neq 1$);
  2. $J_{2k}(1) \oplus J_{2k}(1)$;
  3. $J_{2k-1}(1)$;

where $k \in \mathbb{N}$.

For clarity: each of these types, on its own, also has such a representation.

Sketch for a proof: First, note that $U$ is similar to $U^{-1}$. Therefore, $J_U$ is similar to $J_U^{-1}$. This simplifies the problem significantly.

Next, we prove the theorem for individual Jordan blocks. For the listed types, there is concrete way to describe such $B$ using the similarity between components $J_k(\lambda)$ and $J_k(\lambda^{-1})$.

After that, we need to check that for $J = H \oplus T$, where $H$ is representable but $T$ is not, $J$ itself cannot be represented as $B^\top B^{-1}$. This can be shown by examining the Schur complement.

Finally, for blocks of the first type ($J_k(\omega) \oplus J_k(\omega^{-1})$), we can find $\bar K \in GL(n, E)$, where $E$, once again, is (finite) extension of $\mathbb{F}_2$. Moreover, there exist a diagonal matrice $D \in GL(n, E)$ for which $D \bar K \in GL(n,\mathbb{F}_2)$.

It would be handy to rigorously prove that $J_{2k-1}(1)$ is indeed an allowed type; however, I have been unable to find a straightforward proof.

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