Determinant of a $2k\times 2k$ matrix with zero diagonal and $+1$ and $-1$ off the diagonal entries

Question

Prove that the determinant of a $2k\times 2k$ matrix with zero diagonal and $+1$ and $−1$ off diagonal entries (with no particular order) is non-zero in $\mathbb{R}$ and $\mathbb{C}$. What can we say about such a matrix in other fields (particularly, is there any field in which the determinant is zero)?

Answer 1

For any even $n$, let $X$ be the collection of $n \times n$ matrices with off-diagonal entries $\pm 1$ and diagonal entries $0$. In particular, the matrix $B = (b_{ij})$ with off-diagonal entries $1$ and diagonal entries $0$ belongs to $X$. It is easy to show $\det(B) = (n-1)(-1)^{n-1}$.

For any $A = (a_{ij}) \in X$, we have $a_{ij} \equiv b_{ij} \pmod 2$ for $i, j$. This implies

$$\det(A) \equiv \det(B) = (n-1)(-1)^n \equiv 1 \pmod 2\quad\implies\quad\det(A) \ne 0$$

Not sure what happens in general. However, if $p$ is prime factor of $n-1$, then $\det(B) = 0$ on any field with characteristic $p$.