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Denis Serre
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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$$C=(\det B) I_d$ and one obtains readily $\det A=(\det B)^d$.

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

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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Denis Serre
  • 52.3k
  • 10
  • 146
  • 300

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