1
$\begingroup$

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 :)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

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.