Determinant of $2 \times 2$ block matrix whose diagonal blocks are zero $\Bbb A$ is an $n\times n$ matrix, and $\Bbb B$ is an $m × m$ matrix.$\space$What is the determinant of matrix $\Bbb C$?
\begin{equation*}
\mathbb{C}=
\begin{pmatrix}
\begin{array}{@{}c|c@{}}
  \begin{matrix}
  0
  \end{matrix}
  & \mathbb{A} \\
\hline
  \mathbb{B} &
  \begin{matrix}
 0
  \end{matrix}
\end{array}
\end{pmatrix}
\end{equation*}
I thought it could be just
($\det\Bbb A\cdot\det\Bbb B$)
or ($-\det\Bbb A\cdot\det\Bbb B$),
but I'm not sure, as it seems too easy.
 A: This determinant can be computed by the Laplace expansion theorem (the generalized form). Fix the first $1, 2, \ldots, n$ rows, and let columns range over $(j_1, j_2, \ldots, j_n) \in \{1, 2, \ldots, n + m\}$. Since the square minor is non-zero only if $(j_1, j_2, \ldots, j_n) = (m + 1, m + 2, \ldots, m + n)$,  it follows that
\begin{align*}
\det C = & C\begin{pmatrix}1 & 2 & \cdots & n \\
m + 1 & m + 2 & \cdots & m + n\end{pmatrix}(-1)^{1 + \cdots + n + m + 1 + \cdots + m + n}C\begin{pmatrix}n + 1 & n + 2 & \cdots & n + m \\
1 & 2 & \cdots & m \end{pmatrix} \\
= & (-1)^{n(n + 1) + mn}\det(A)\det(B).
\end{align*}
A: As Robert Israel noted,
$${\rm C} = \underbrace{\begin{bmatrix} {\rm O} & {\rm I}_n \\ {\rm I}_m & {\rm O} \end{bmatrix}}_{=: \rm P} \begin{bmatrix} {\rm B} & {\rm O} \\ {\rm O} & {\rm A} \end{bmatrix}$$
Let $\sigma := \det ({\rm P})$. Hence,
$$ \det({\rm C}) = \det\begin{bmatrix} {\rm O} & {\rm I}_n \\ {\rm I}_m & {\rm O} \end{bmatrix} \cdot \det\begin{bmatrix} {\rm B} & {\rm O} \\ {\rm O} & {\rm A} \end{bmatrix} = \sigma \cdot \det({\rm A}) \cdot \det({\rm B})$$
Note that $\rm P$ is an $(m+n) \times (m+n)$ permutation matrix. It is also circulant. Hence, its $k$-th eigenvalue is
$$\lambda_k = \exp \left( -i 2\pi \left( \frac{m}{m+n} \right) k \right)$$
where $k \in \{0,1,\dots,m+n-1\}$. Since the determinant is the product of the eigenvalues,
$$\sigma = \prod_{k=0}^{m+n-1} \exp \left( -i 2\pi \left( \frac{m}{m+n} \right) k \right) = \exp \left( -i 2\pi \left( \frac{m}{m+n} \right) \sum_{k=0}^{m+n-1} k \right)$$
Since $\displaystyle\sum_{k=0}^{m+n-1} k = \frac{(m+n)(m+n-1)}{2}$,
$$\sigma = \exp \left( -i \pi m (m+n-1) \right) = (-1)^{m (m+n-1)} = \begin{cases} +1 & \text{ if } m (m+n-1) \text{ is even}\\ -1 & \text{ if } m (m+n-1) \text{ is odd}\end{cases}$$
Note that $m (m+n-1)$ is odd only when both $m$ and $n$ are odd. Thus,
$$\sigma = \begin{cases} -1 & \text{ if } m \text{ and } n \text{ are odd}\\ +1 & \text{ otherwise}\end{cases}$$


*

*Robert M. Gray, Toeplitz and Circulant Matrices: a Review.


linear-algebra matrices block-matrices circulant-matrices permutation-matrices determinant eigenvalues-eigenvectors permutations
