3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.