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Is there any formula for determinant of a matrix $A$ that looks like:

$$A = \begin{pmatrix} \textbf{B}_{11} & \cdots & \textbf{B}_{1L} \\ \vdots & & \vdots \\ \textbf{B}_{L1} & \cdots & \textbf{B}_{LL} \end{pmatrix} $$

where each $\textbf{B}_{ij}$ is a $n \times n$ diagonal matrix ?

Edit: I have found something on a paper. If we write each $$\textbf{B}_{ij} = \cdot \begin{pmatrix} {b}^{(ij)}_1 & \ & 0 \\ \ & \ddots & \ \\ 0 & \ & {b}^{(ij)}_n \end{pmatrix}$$ it seems that $$\det(A) = \displaystyle\prod_{k=1}^{n}\det \begin{pmatrix} b^{(11)}_k & \cdots & b^{(1L)}_k \\ \vdots & & \vdots \\ b^{(L1)}_k & \cdots & b^{(LL)}_k \end{pmatrix}$$ Any hint on proving it?

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  • $\begingroup$ For what it's worth, I was writing my answer before I saw your edit :) $\endgroup$ – Omnomnomnom May 8 '17 at 16:03
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Yes. Let $\mathbf C_{ij}$ be the $L \times L$ matrix defined by $$ \mathbf C_{ij}(p,q) = \mathbf B_{pq}(i,j) $$ Notably, we find that $\mathbf C_{ij} = 0$ when $i \neq j$. By rearranging the rows and columns of $A$ appropriately, we see that there is a permutation matrix $P$ such that $$ PAP^{-1} = \pmatrix{\mathbf C_{11} & & \cdots & \mathbf C_{1n} \\ \\ \vdots && \ddots & \vdots\\ \mathbf C_{n1} & & \cdots & \mathbf C_{nn}} = \pmatrix{\mathbf C_{11} & 0& \cdots & 0 \\ 0 & \mathbf C_{22}& \ddots & \vdots \\ \vdots & \ddots & \ddots & 0\\ 0 & \cdots & 0 & \mathbf C_{nn}} $$ Conclude that $$ \det(A) = \det(\mathbf C_{11}) \det(\mathbf C_{22}) \cdots \det(\mathbf C_{nn}) $$


In particular, $P$ is the permutation matrix corresponding to $\tau:\{1,\dots,nL\} \to \{1,\dots, nL\}$ defined by $$ \tau(1 + (i-1) + n(j-1)) = 1 + (j-1) + L(i-1) \qquad 1 \leq i \leq n, \quad 1 \leq j \leq L $$ Notably: if $x \in \Bbb R^L, y \in \Bbb R^n$, and $\otimes$ denotes the Kronecker product, then $$ P(x \otimes y) = y \otimes x $$

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