Is it true in general that if $A$ and $B$ are two $n \times n$ matrices, then the determinant of the anti-diagonal block matrix

$$ J = \left[\begin{array}{cc} 0 & A \\ B& 0 \end{array}\right] $$

is $\det(J)=\det(B)\det(A)$? It is simple to prove this if $n=2$, but I have no idea on how to generalize it.


One has: $$J:=\begin{pmatrix}0 & A\\B & 0\end{pmatrix}=\begin{pmatrix}A& 0\\0 & B\end{pmatrix}\times\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}.$$ You only have to compute: $$\varepsilon:=\det\left(\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}\right).$$ Indeed, using the first equality, one has: $$\det(J)=\varepsilon\det(A)\det(B).$$ If $n$ is odd, the result appears to be false, you will get: $$\det(J)=-\det(A)\det(B).$$


The definition of determinant through permutation will give the answer.

For a matrix $C=(c_{i,j})_{m \times m}$ $$\mathrm{det}(C)=\sum \limits_{\sigma \in S_{m}} \mathrm{sgn}~\sigma\cdot c_{1,\sigma(1)}c_{2,\sigma(2)} \cdots c_{m,\sigma(m)}$$

If we write, $J=(x_{i,j})_{2n \times 2n}$ then, $$\mathrm{det}(J)=\sum \limits_{\sigma \in S_{2n}} \mathrm{sgn}~\sigma\cdot x_{1,\sigma(1)}x_{2,\sigma(2)} \cdots x_{n,\sigma(n)}x_{n+1,\sigma(n+1)} \cdots x_{2n,\sigma(2n)}$$

For $\sigma \in S_{2n}$ with $\sigma(i) \in \{1,2, \cdots,n\}~ \mathrm{for~some~} i \in \{1,2, \cdots,n\}$ or, $\sigma(i) \in \{n+1,n+2, \cdots,2n\}~ \mathrm{for~some~} i \in \{n+1,n+2, \cdots,2n\}$, the term $x_{1,\sigma(1)}x_{2,\sigma(2)} \cdots x_{n,\sigma(n)}x_{n+1,\sigma(n+1)} \cdots x_{2n,\sigma(2n)} = 0$

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

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

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

Then, $\sigma \in H$ and $$S_n \times S_n \rightarrow H$$ $$(\sigma_1,\sigma_2) \mapsto \sigma$$ is a bijection and $\mathrm{sgn}~\sigma= (-1)^n \cdot \mathrm{sgn}~\sigma_1\cdot \mathrm{sgn}~\sigma_2$ (verify).

So, \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*}


There are block matrix determinant identities that will give you this, but if you want to solve using basic determinant formulas consider the full expansion formula for the determinant using permutations, giving a sum of $n!$ terms for an $n \times n$ matrix, where each term is either $1$ or $-1$ multiplied by $n$ entries in the matrix. If a term includes an entry in either of the zero blocks, then the term is $0$. So every non-zero term is a product of entries selected from $A$ and entries selected from $B$. If you consider all terms you can get by selecting entries just from $A$ and $B$, and group them together according to entries selected in $A$, you will see that for each choice of entries in $A$, when you consider the sum of terms, you can factor out $\det B$ from the sum and get just the product of terms selected in $A$. So then when you add those up (after factoring out $\det B$) you get $\det A$. There is a small detail in handling the signs of the permutations, but this is pretty straightforward.


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