I want to ask a question regarding the invariance of determinants under permutation. The following matrix is the one I want to discuss here. (It's just a symmetric matrix with non-zero elements at the bands and two corners.)
\begin{bmatrix} \Gamma_{1,1} \cdot \mathcal{I} & \Gamma_{1,2} \cdot P_{1} & 0 & \underline{0} & \Gamma_{1,m} \cdot P_{m} \\ \hline \Gamma_{1,2} \cdot P_{1}^{T} & \Gamma_{2,2} \cdot \mathcal{I} & \Gamma_{2,3} \cdot P_{2} & \underline{0} & 0 \\ \hline 0 & \Gamma_{2,3} \cdot P_{2}^{T} & \Gamma_{3,3}\cdot \mathcal{I} & \Gamma_{3,4} \cdot P_{3} & \underline{0} \\ \hline \underline{0}^{T} & \ddots & \ddots & \ddots & \ddots \\ \hline \Gamma_{1,m} \cdot P_{m}^{T} & 0 & 0 & \cdots & \Gamma_{m,m}\cdot \mathcal{I} \end{bmatrix}
where $\{P_{i}\}_{i=1}^{m}$ are the permutation matrices with size $M \times M$.
I was trying to prove it by using "Newton's identities" that can connect the determinant with the traces of the n-th power of the matrix. Showing that the diagonal blocks of the n-the power of this matrix are always identity matrices up to some factors gives us the result.
I didn't find a systematic way to show that the n-th power of the above matrix has the same diagonal blocks. Or was my intuition wrong?
In this problem, we have two parameters the size of the base matrix m and the size of permutation matrices M. (Perhaps the invariance can be established under some region (m, M).）
Based on Lemma 1 in this paper, I can easily come up with this result by forcing $(\prod\limits_{i=1}^{m-1} P_{i}) \cdot P_{m}^{T} = \mathcal{I}$ (i.e., $P_{m} = \prod\limits_{i=1}^{m-1} P_{i}$). (Maybe it can also come up with the result without such a constraint.)
The determinant will be equivalent to the determinant of the matrix with the following form:
\begin{bmatrix} t_1 \cdot \mathcal{I}- h \cdot (\prod\limits_{i=1}^{m-1}P_{i})\cdot P_{m}^{T} & t_2 \cdot P_{m} \\ t_3 \cdot (\prod\limits_{i=1}^{m-1}P_{i})^{T} & t_4 \cdot (\prod\limits_{i=1}^{m-1}P_{i})^{T}\cdot P_{m} - \mathcal{I} \end{bmatrix}
Trivially, we can get its invariance under permutations with the constraint that $(\prod\limits_{i=1}^{m-1} P_{i}) \cdot P_{m}^{T} = \mathcal{I}$.
It seems that the determinant can be eventually reduced to be in the following form: $\det(\ell_{1}\cdot \mathcal{I} + \ell_{2}\cdot W + \ell_{3} \cdot W^{T})$, denoted by $D(W)$, where $W$ is an arbitrary permuation matrix with size $M \times M$. When $W = \mathcal{I}$, we denote the determinant as $D_{id}$. Based on this observation, we can reduce the number of possible determinants from $\left(M!\right)^{m}$ to $M!$ and we can conduct numerical simulations for $M < 10$. The simulation result shows that $<D(W)>$ are equal to $D_{id}$ and furthermore $\sqrt[M]{<D(W)>}$ are the same.
As $M \to \infty$, can we say something between the average of $D(W)$ with respect to $W$, denoted by $<D(W)>$, and $D_{id}$? (e.g. $\lim\limits_{M \to \infty} <D(W)> = D_{id}$.)