Determinante de una matriz de bloques anti-diagonal

Uno tiene:

$$J:=\begin{pmatrix}0 & A\\B & 0\end{pmatrix}=\begin{pmatrix}A& 0\\0 & B\end{pmatrix}\times\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}.$$ Solo tienes que calcular: $$\varepsilon:=\det\left(\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}\right).$$ De hecho, usando la primera igualdad, se tiene: $$\det(J)=\varepsilon\det(A)\det(B).$$ Si $n$es impar, el resultado parece ser falso, obtendrá: $$\det(J)=-\det(A)\det(B).$$

La definición de determinante por permutación dará la respuesta.

para una matriz$C=(c_{i,j})_{m \times m}$

$$\mathrm{det}(C)=\sum \limits_{\sigma \in S_{m}} \mathrm{sgn}~\sigma\cdot c_{1,\sigma(1)}c_{2,\sigma(2)} \cdots c_{m,\sigma(m)}$$

Si escribimos,$J=(x_{i,j})_{2n \times 2n}$entonces,

$$\mathrm{det}(J)=\sum \limits_{\sigma \in S_{2n}} \mathrm{sgn}~\sigma\cdot x_{1,\sigma(1)}x_{2,\sigma(2)} \cdots x_{n,\sigma(n)}x_{n+1,\sigma(n+1)} \cdots x_{2n,\sigma(2n)}$$

Para$\sigma \in S_{2n}$con$\sigma(i) \in \{1,2, \cdots,n\}~ \mathrm{for~some~} i \in \{1,2, \cdots,n\}$o,$\sigma(i) \in \{n+1,n+2, \cdots,2n\}~ \mathrm{for~some~} i \in \{n+1,n+2, \cdots,2n\}$, el término$x_{1,\sigma(1)}x_{2,\sigma(2)} \cdots x_{n,\sigma(n)}x_{n+1,\sigma(n+1)} \cdots x_{2n,\sigma(2n)} = 0$

Dejar,$H=\{\sigma \in S_{2n}~|~ \sigma(i) \in \{n+1,n+2, \cdots,2n\}~ \forall i \in \{1,2, \cdots,n\} ~\mathrm{and}~\sigma(i) \in \{1,2, \cdots,n\}~ ~ \forall i \in \{n+1,n+2, \cdots,2n\} \}$

Entonces,

$$\mathrm{det}(J)=\sum \limits_{\sigma \in H} \mathrm{sgn}~\sigma\cdot x_{1,\sigma(1)}x_{2,\sigma(2)} \cdots x_{n,\sigma(n)}x_{n+1,\sigma(n+1)} \cdots x_{2n,\sigma(2n)}$$

Para,$(\sigma_1,\sigma_2) \in S_n \times S_n$, definir$\sigma(i)=\sigma_1(i)+n~ \forall i \in \{1,2, \cdots,n\}$y$\sigma(i)=\sigma_2(i-n)~ \forall i \in \{n+1,n+2, \cdots,2n\}$.

Entonces,$\sigma \in H$y

$$S_n \times S_n \rightarrow H$$ $$(\sigma_1,\sigma_2) \mapsto \sigma$$ es una biyeccion y $\mathrm{sgn}~\sigma= (-1)^n \cdot \mathrm{sgn}~\sigma_1\cdot \mathrm{sgn}~\sigma_2$(verificar).

Entonces,

$$\begin{eqnarray*} \mathrm{det}(J) &=& (-1)^n \cdot\sum \limits_{(\sigma_1,\sigma_2) \in S_n \times S_n} \mathrm{sgn}~\sigma_1\cdot \mathrm{sgn}~\sigma_2 \cdot x_{1,\sigma_1(1)+n}x_{2,\sigma_1(2)+n} \cdots x_{n,\sigma_1(n)+n}x_{n+1,\sigma_2(1)} \cdots x_{2n,\sigma_2(n)}\\ &=& (-1)^n \cdot\sum \limits_{\sigma_1 \in S_n} \sum \limits_{\sigma_2 \in S_n} \mathrm{sgn}~\sigma_1\cdot \mathrm{sgn}~\sigma_2 \cdot x_{1,\sigma_1(1)+n}x_{2,\sigma_1(2)+n} \cdots x_{n,\sigma_1(n)+n}x_{n+1,\sigma_2(1)} \cdots x_{2n,\sigma_2(n)}\\ &=& (-1)^n \cdot\sum \limits_{\sigma_1 \in S_n} \left(\mathrm{sgn}~\sigma_1\cdot x_{1,\sigma_1(1)+n}x_{2,\sigma_1(2)+n} \cdots x_{n,\sigma_1(n)+n} \times \sum \limits_{\sigma_2 \in S_n} \mathrm{sgn}~\sigma_2\cdot x_{n+1,\sigma_2(1)} \cdots x_{2n,\sigma_2(n)} \right)\\ &=& (-1)^n \cdot\sum \limits_{\sigma_1 \in S_n} \left(\mathrm{sgn}~\sigma_1\cdot x_{1,\sigma_1(1)+n}x_{2,\sigma_1(2)+n} \cdots x_{n,\sigma_1(n)+n} \times \mathrm{det}(B) \right)\\ &=& (-1)^n \cdot\mathrm{det}(B)\times \left(\sum \limits_{\sigma_1 \in S_n} \mathrm{sgn}~\sigma_1\cdot x_{1,\sigma_1(1)+n}x_{2,\sigma_1(2)+n} \cdots x_{n,\sigma_1(n)+n}\right)\\ &=& (-1)^n \cdot\mathrm{det}(B)\mathrm{det}(A) \end{eqnarray*}$$