Let $S=(s_{ij})$ be a skew-symmetric integral matrix of order $n$. We only consider the case that $n$ is even. Let $e$ be the all-one vector in $\mathbb{R}^n$. Define the walk matrix $$W(S)=[e,Se,\cdots,S^{n-1}e].$$ (the name "walk matrix" comes from graph theory, where $S$ is the adjacency matrix of an undirected graph. Of course, $S$ is not skew-symmetric in the setting of graphs.)

It is well-known that $\det(S)$ is always a square number. I find that the integer $\sqrt{\det(S)}$ is always a divisor of $\det W(S)$. But I cannot find any references on this relation.

For example, consider $$S=\left( \begin{array}{cccc} 0 & 4 & 0 & -3 \\ -4 & 0 & -2 & -1 \\ 0 & 2 & 0 & 3 \\ 3 & 1 & -3 & 0 \\ \end{array} \right).$$ Then, $$W(S)=\left( \begin{array}{cccc} 1 & 1 & -31 & -3 \\ 1 & -7 & -15 & 165 \\ 1 & 5 & -11 & -87 \\ 1 & 1 & -19 & -75 \\ \end{array} \right).$$ Using Mathematica, we find that $\det(S)=18^2$, $\det(W)=16128=18\times 896$ and $\sqrt{\det(S)}\mid \det(W)$.

It seems that the above relation $\sqrt{\det(S)}\mid \det(W)$ always hold for any skew-symmetric integral matrix of even orders. In particular, when $\det(S)=0$ then $\det(W)=0$. This particular case is not hard to show.