I came across this idea when studying perturbations in Q.M.
Is it possible to somehow show what the determinant would be of a matrix of this form:
\begin{matrix} 0 & a & 0 & \dots & 0 \\ b & 0 & a & \ddots & 0 \\ 0 & b & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & a \\ 0 & 0 & 0 & b & 0 \end{matrix}
It seems like the answer periodically switches between non-zero and zero depending on the number of dimensions taken for a matrix with the same entries.
Thanks
Make an induction proof, calling $M_n(a,b)$ a such matrix.Let $P_n="\det(M_{2n+1}(a,b))=0".$
For $n=1$ you get $\det(M_n(a,b))=\begin{vmatrix}0&a&0\\b&0&a\\0&b&0\end{vmatrix}=b(a\cdot 0-b\cdot 0)=0,$
For $n\in\mathbb{N^*},$ suppose $P_n$ is true. Then : \begin{align*}\det(M_{2n+3}(a,b))&=\begin{vmatrix} 0 & a & 0 & \dots & 0 \\ b & 0 & a & \ddots & 0 \\ 0 & b & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & a \\ 0 & 0 & 0 & b & 0 \end{vmatrix}\\ &=-b\cdot\begin{vmatrix} a & 0 & 0 & \dots & 0 \\ b & 0 & a & \ddots & 0 \\ 0 & b & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & a \\ 0 & 0 & 0 & b & 0 \end{vmatrix}\\ &=-ba\cdot\det(M_{2n+1}(a,b))+b^2\cdot\begin{vmatrix} 0 & 0 & \dots & 0 \\ b & \ddots & \ddots & \vdots \\ \ddots & \ddots & \ddots & a \\ 0 & 0 & b & 0 \end{vmatrix}=0\end{align*} by hypothesis. You can do something similar for the other case.