I'm looking at the following family of $n\times n$ matrices. The entries are 0 everywhere except above and below the diagonal. Above it takes values from $1 \to n-1$ and below from $ -n +1 \to -1$. Example when $n=4$:
$\left[ \begin{matrix} 0 & 1 & 0 & 0 \\ -3 & 0 & 2 & 0 \\ 0 & -2 & 0 & 3 \\ 0 & 0 & -1 & 0 \\ \end{matrix} \right] $
How can I find the characteristic polynomial of these matrices in general?
Let $D=\operatorname{diag}(1,i,i^2,\ldots,i^{n-1})=\operatorname{diag}(1,i,-1,-i,1,i,-1,-i,\ldots)$. Then $D^{-1}AD=iK$ where $$ K=\pmatrix{ 0&1\\ n-1&0&2\\ &n-2&\ddots&\ddots\\ & &\ddots&\ddots&\ddots\\ & & &\ddots&0 &n-1\\ & & & & 1 &0}. $$ is known as the Kac matrix. The spectrum of $K$ is given by $$ \{-n+1,\,-n+3,\,-n+5,\ldots,\,n-5,\,n-3,\,n-1\}. $$ Hence the eigenvalues of $A$ are $$ \{(-n+1)i,\,(-n+3)i,\,(-n+5)i,\ldots,\,(n-5)i,\,(n-3)i,\,(n-1)i\} $$ and the characteristic polynomial of $A$ is \begin{cases} x&\text{when $n=1$,}\\ [x^2+(n-1)^2][x^2+(n-3)^2]\cdots(x^2+4^2)(x^2+2^2)x&\text{when $n\ge3$ is odd,}\\ [x^2+(n-1)^2][x^2+(n-3)^2]\cdots(x^2+3^2)(x^2+1)&\text{when $n$ is even.}\\ \end{cases}