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.

While using a substitution above is quite well behaved, it is poorly behaving if we change basis. It insist that such factorization of $U$ also depends on eigenvectors.

However, with some algebraic manipulation, we can derive this criterion:

Let $B_0$ be a candidate for such representation of U. If there exists such $\mu : \chi_U(\mu) = 0$ and $\chi_{B^\top_0 B^{-1}_0}(\mu) \neq 0$, then $B^\top_0 B^{-1}_0 \neq U$.

It would be nice to find out criteria for (generalized) eigenvectors of $U$.

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 $C^\top C^{-1}$.

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

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.

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.

While using a substitution above is quite well behaved, it is poorly behaving if we change basis. It insist that such factorization of $U$ also depends on eigenvectors.

However, with some algebraic manipulation, we can derive this criterion:

Let $B_0$ be a candidate for such representation of U. If there exists such $\mu : \chi_U(\mu) = 0$ and $\chi_{B^\top_0 B^{-1}_0}(\mu) \neq 0$, then $B^\top_0 B^{-1}_0 \neq U$.

It would be nice to find out criteria for (generalized) eigenvectors of $U$.

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

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.

Update: So I tried a substituion $U = J_k(\lambda)$ — Jordan block of size k for eigenvalue $\lambda$ and it gave me some insights: For $\lambda = 1$

1. If $k = 2m + 1$, then $J_k(\lambda) \sim J_k(\lambda)^{-1}$
2. If $k = 2m$, then $J_k(\lambda)$ isn't similar to $J_k(\lambda)^{-1}$. Therefore, for such eigenvalue $\lambda$ we 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.

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

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.