Determinant of “skew-symmetric” matrices

For $$n\in\mathbb{N}$$ and $$m=\lfloor\frac{n}2\rfloor$$, consider the $$n\times n$$ skew-symmetric matrix $$A_n$$ where each entry in the first $$m$$ sub-diagonals below the main diagonal is $$1$$ and each of the remaining entries below the main diagonal is $$-1$$. Let $$I_n$$ be the $$n\times n$$ identity matrix.

Next, construct the matrix $$M_n=A_n+xI_n$$. For example, we have $$M_3=\begin{pmatrix} x&-1&1 \\ 1&x&-1 \\ -1&1&x \end{pmatrix} \qquad \text{and} \qquad M_4=\begin{pmatrix} x&-1&-1&1 \\ 1&x&-1&-1 \\ 1&1&x&-1 \\ -1&1&1&x \end{pmatrix}.$$

QUESTION. Is the following true? Experiments suggest to be so. $$\det(M_n)=\sum_{k=0}^m\binom{n}{2k}x^{n-2k}.$$

• Do you mean $A_n + x I_n$? – Gordon Royle Dec 12 '18 at 8:36
• wouldn't $$\det(M_n)=\sum_{k=0}^m e_{n-2k}(x_1,\dots,x_n)$$ be easier? – Martin Rubey Dec 12 '18 at 9:56
• The coefficient of $x^{n-m}$ is the sum of determinants of $m \times m$ principal submatrices of $A_n$. These are skew-symmetric, so have determinant zero when $m$ is odd. When $m=2k$ is even, such a determinant is a square of a Pfaffian, which (roughly speaking) counts weighted perfect matchings of, well, a size $2k$ subgraph of the oriented complete graph with oriented adjacency matrix $A_n$. I wonder if there is any combinatorial way to get $\binom{n}{2k}$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...). – Zach Teitler Dec 12 '18 at 23:15

For $$n$$ odd, $$M_n$$ is an $$n\times n$$ circulant matrix, and so Theorem 17 in Krattenthaler's marvellous text applies. Denoting by $$w$$ a primitive $$n$$th root of unity, it gives $$\det M_n=\prod_{i=0}^{n-1} (x-w^i-w^{2i}-\dots -w^{mi}+w^{(m+1)i}+\dots +w^{(n-1)i}),$$ something that should not be hard to simplify.
For $$n$$ even, you still have a special Hankel matrix, for which again there are general methods in [loc.cit.].
• For even $n$, there is an extension of circulant matrices sometimes called $\alpha$-circulant that should do the job; basically you can multiply the sub-diagonal entries by a certain complex number $\alpha$. In this case I guess that the same formula holds, but with $w$ a root of $-1$ instead of $1$. – Federico Poloni Dec 12 '18 at 10:58