Determinant of a large symmetric block matrix

Question

Consider a given matrix $Q \in \text{Mat}_N(\mathbb{R})$, which is invertible, and $n \geq 1$. I am looking for the determinant of the symmetric block matrix $I_n(Q)$ of total size $nN \times nN$: $$I_n(Q) = \begin{pmatrix}Id_N & Q & \cdots & Q \\ Q^\intercal & Id_N & \cdots & Q \\ \vdots & \vdots & \ddots & \vdots \\ Q^\intercal & Q^\intercal & \cdots & Id_N \\ \end{pmatrix}$$

If $Q$ is symmetric, I know that $\det I_n(Q) = \det((Id_N-Q)^{n-1}) \det(Id_N+(n-1)Q)$, and I also have a formula for $I_n(Q)^{-1}$, but I couldn't derive a form for the non-symmetric case. Any help would be appreciated ! Thanks :)

Answer 1

Here is a partial answer. Let us consider the case where the number of block rows is $n=4$ first. The other cases are similar. We shall perform blockwise elementary row/column operations on the block matrix $$ \pmatrix{ I&Q&Q&Q\\ Q^T&I&Q&Q\\ Q^T&Q^T&I&Q\\ Q^T&Q^T&Q^T&I}. $$ First, for $i=n,n-1,\ldots$ down to $2$, subtract the $i$-th block row by the one above it. We get $$ \pmatrix{ I&Q&Q&Q\\ Q^T-I&I-Q&0&0\\ 0&Q^T-I&I-Q&0\\ 0&0&Q^T-I&I-Q}. $$ Suppose $1$ is not an eigenvalue of $Q$. Let $$ M=(I-Q)^{-1}(Q^T-I). $$ Then for $j=n-1,n-2,\ldots$ down to $1$, perform the block column operation $C_j\leftarrow C_j-C_{j+1}M$ successively. When $n=4$, we will obtain, in three steps, the following matrices: \begin{aligned} &\pmatrix{ I&Q&Q-QM&Q\\ Q^T-I&I-Q&0&0\\ 0&Q^T-I&I-Q&0\\ 0&0&0&I-Q},\\ &\pmatrix{ I&Q-QM+QM^2&Q-QM&Q\\ Q^T-I&I-Q&0&0\\ 0&0&I-Q&0\\ 0&0&0&I-Q},\\ &\pmatrix{ I-QM+QM^2-QM^3&Q-QM+QM^2&Q-QM&Q\\ 0&I-Q&0&0\\ 0&0&I-Q&0\\ 0&0&0&I-Q}. \end{aligned} Hence at the end we get a block upper triangular matrix. In general, if $1$ is not an eigenvalue of $Q$, the determinant of the original block matrix is given by $$ \det\left(I+Q\left[(-M)+(-M)^2+..+(-M)^{n-1}\right]\right)\det(I-Q)^{n-1}.\tag{1} $$ Here we do not replace the square bracket term by $[(-M)-(-M)^n](I+M)^{-1}$ because $-1$ may be an eigenvalue of $M$ (in particular, this always occurs if the size $N$ of the matrix $Q$ is odd).

Since the determinant of the original block matrix is a polynomial in the entries of $Q$, if a generic formula exists, we don't expect it to contain any matrix inverse term. Unfortunately I cannot manage to cancel out the inverse term $(I-Q)^{-1}$ inside the powers of $M$. Therefore formula $(1)$ cannot be further simplified at the moment and the requirement that $1$ is not an eigenvalue of $Q$ cannot be dropped yet.