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This question is related to [Determinant of block tridiagonal Toeplitz matrices] (Determinant of block tridiagonal Toeplitz matrices).

$n\times n$ block tridiagonal matrix $A_{nm}$ constructed from $m\times m$ blocks $J_m,I_m$,

\begin{align} A_{nm} &= \begin{bmatrix} J_n & I_n & 0 & \cdots & \cdots & 0 \\ I_m & J_m & I_m & 0 & \cdots & 0 \\ 0 & I_m & J_m & I_m & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ 0 & \cdots & \cdots & I_m & J_m & I_m \\ 0 & \cdots & \cdots & \cdots & I_m & J_m \end{bmatrix} _{n\times n} , \end{align}

where $I_m$ is $m\times m$ identity matrix, and $J_m$ is a tridiagonal matrix with ones on the three diagonals.

\begin{align} \end{align}

The question: is this a known property or is there a simple way to prove that

\begin{align} \det(A_{nm})&=\det(A_{mn}) \tag{1}\label{1} \end{align}

for all $n,m\in\mathbb N$?


Statement \eqref{1} agrees for small $n,m$, for example this is a $9\times9$ matrix with elements $m_{ij}=\det(A_{ij}),\ i,j=1,\dots,9$:

\begin{align} M_{9} &= \begin{bmatrix} 1& 0& -1& -1& 0& 1& 1& 0& -1 \\ 0& -3& 0& 5& 0& -7& 0& 9& 0 \\ -1& 0& -7& -9& 0& -7& 119& 0& 369 \\ -1& 5& -9& 0& 55& 29& 279& -95& 0 \\ 0& 0& 0& 55& 0& -1183& 0& 0& 0 \\ 1& -7& -7& 29& -1183& 2197& -791& 28672& -165271 \\ 1& 0& 119& 279& 0& -791& -34391& 0& -2733921 \\ 0& 9& 0& -95& 0& 28672& 0& -4002939& 0 \\ -1& 0& 369& 0& 0& -165271& -2733921& 0& 0 \end{bmatrix} , \end{align}

where the first and second row/column shows an easy pattern, but the other (as well as a main diagonal) are not recognized as a known integer sequence.

For some reason, the choice of $n=5$ of $m=5$ result in seven zeros of nine.

Obviously, matrices $A_{nm}$ and $A_{mn}$ are of the same size, $mn\times mn$ elements and it's easy to show that they have the same number of $1$s.

As it is shown in this answer, $\det(A_{nm})$ boils down to the determinant of the $m\times m$ matrix, for example

\begin{align} \det(43)&= \left| \begin{matrix} 4&6&5 \\ 6&9&6\\ 5&6&4 \end{matrix} \right|=-9 ,\\ \det(34)&= \left| \begin{matrix} 2&3&3&1\\ 3&5&4&3\\ 3&4&5&3\\ 1&3&3&2 \end{matrix} \right|=-9 . \end{align}

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1 Answer 1

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Eigenvalues are given exactly by $f(\theta_1,\theta_2)=1+2\cos(\theta_1)+2\cos(\theta_2)$ where $\theta_1=i\pi/(m+1)$ $i=1,\ldots,m$ and $\theta_2=j\pi/(n+1)$ $j=1,\ldots,n$.

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