Determinant diagonal blocks compound matrix Good afternoon,
I would like to prove the equation
\begin{equation}
\begin{vmatrix}
b_{1,1}I_d & b_{1,2}I_d & \cdots & b_{1,r}I_d \\
b_{2,1}I_d & b_{2,2}I_d & \cdots & b_{2,r}I_d \\
\vdots & \vdots & \ddots & \vdots \\
b_{r,1}I_d & b_{r,2}I_d & \cdots & b_{r,r}I_d \\
\end{vmatrix}
=
{\begin{vmatrix}
b_{1,1} & b_{1,2} & \cdots & b_{1,r} \\
b_{2,1} & b_{2,2} & \cdots & b_{2,r} \\
\vdots & \vdots & \ddots & \vdots \\
b_{r,1} & b_{r,2} & \cdots & b_{r,r} \\
\end{vmatrix}}^d
\end{equation}
with $b_{1,1}, b_{1,2}, \ldots, b_{r,r} \in \mathbb{C}$, $r, d \in \mathbb{N}^*$ and $I_d$ the order $d$ identity matrix. A recurrence argument should do but I cannot find a proper way of writing it. Could you help please?
Thanks!
 A: There is a general theorem, which can be proved by using Schur's complement formula: let a matrix $A\in M_{pq}(k)$ be written blockwise, with blocks $A_{ij}\in M_q(k)$ for $1\le i,j\le p$. Assume that the blocks $A_{ij}$ commutte pairwise, so that the determinantal expression
$$C=\sum_{\sigma\in\frak_p}\epsilon(\sigma)\prod_{i=1}^pA_{i\sigma(i)}$$
makes sense. Then
$$\det A=\det C.$$
In your case, $C=(\det B) I_d$ and one obtains readily $\det A=(\det B)^d$.
A: Suppose you have the relation for d (and indeed for d=1 it is obvious when written out); now do a block matrix with one block B in the upper left and a block for d in the lower right. You can do (r-1) column moves and (r-1) row moves to "move the upper left entries into their proper places", giving the desired matrix for d+1. Since the same permutation is done for rows as for columns, the sign of the determinant doesn't change. Use induction to add the additional (det B) factor.
Gerhard "Prefers Using Absolute Value Instead" Paseman, 2019.08.20.
