Determinant of walk matrix for a skew-symmetric matrix of even order

Question

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.

Answer 1

Surely, there is nothing special in the all-ones vector: the claim holds for any integer-valued $e$.

Notice that $$ \det W^TW =\det\bigl[e^T (-1)^iS^{i+j}e\bigr], $$ Since $S$ is skew-symmetric, we have $e^TS^ie=0$ for all odd $i$. Permuting now the rows and columns of $W^TW$ we get $$ \det W^TW=\det\begin{bmatrix} A_0&0\\ 0& -A_1\end{bmatrix}, $$ where $$ A_0= \begin{bmatrix} e^Te& e^TS^2e& e^TS^4e& \cdots& e^TS^{n-2}e\\ e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n-2}e& e^TS^ne& e^TS^{n+2}e& \cdots& e^TS^{2n-4}e \end{bmatrix}, \quad A_1= \begin{bmatrix} e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ e^TS^4e& e^TS^6e& e^TS^8e& \cdots& e^TS^{n+2}e\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n}e& e^TS^{n+2}e& e^TS^{n+4}e& \cdots& e^TS^{2n-2}e \end{bmatrix}. $$ We show that $$ \det A_1=\pm\det S\det A_0, \qquad(*) $$ so that $\det W=\pm\det A_0\sqrt{\det S}$, which yields what we want.

Denote $$ u_i= \begin{bmatrix} e^TS^ie& e^TS^{i+2}e& \cdots& e^TS^{i+n-2}e \end{bmatrix}. $$ The eigenvalues of $S$ are purely imaginary, so by Cayley—Hamilton we get $$ S^{n}=\sum_{j=0}^{n/2-1} \alpha_jS^{2j}, $$ where $\alpha_0=-\det S$. Therefore, $$ u_n= -\det S\, u_0+\sum_{j=1}^{n/2-1} \alpha_ju_{2j}. $$ Plug this into $\det A_1$, expand by linearity, and erase vanishing summands to get $(*)$.