# Determinant of a large symmetric block matrix

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 :)

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.