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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!

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2 Answers 2

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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 $\det A=(\det B)^d$.

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  • $\begingroup$ Surely this is far too complicated for something that can be proven by row and column manipulations …. $\endgroup$
    – LSpice
    Commented Aug 20, 2019 at 16:45
  • $\begingroup$ @LSpice. It isn't. Using a result valid for a more general situation is often a good way to understand what is going on. And the general result is interesting in its own. $\endgroup$
    – Denis Serre
    Commented Aug 20, 2019 at 18:53
  • $\begingroup$ @Denis Serre thank you for your answer, I was not aware of the theorem you mentioned and I am happy to discover it. The post indicated by Robert Israel is also a solution for a more general problem. $\endgroup$
    – Durzot
    Commented Aug 20, 2019 at 18:56
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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.

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  • $\begingroup$ Thank your for your answer. If I understand correctly you rewrite the matrix in a permutation of the canonical basis for $\mathbb{R}^{rd+r}$ so as to make $B$ appear in the upper left corner of size $r\times r$. If that is the case your idea is very similar to the idea in the post indicated by Robert Israel where no recurrence is needed. $\endgroup$
    – Durzot
    Commented Aug 20, 2019 at 19:07
  • $\begingroup$ The induction is even easier to start at $d = 0$. $\endgroup$
    – LSpice
    Commented Aug 20, 2019 at 19:07
  • $\begingroup$ @Durzot, indeed, but I wanted to paint the idea with broad strokes and almost no notation. Once you have the idea of the structure, you should have little problem writing it. You can go for a non inductive presentation as well, but this I find easier to communicate. Gerhard "It's All About The Understanding" Paseman, 2019.08.20. $\endgroup$
    – Gerhard Paseman
    Commented Aug 20, 2019 at 19:34

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