Determinant of $2 \times 2$ block matrix whose diagonal blocks are zero

Question

$\Bbb A$ 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.

Answer 1

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,

$$\sigma = \begin{cases} -1 & \text{ if } m \text{ and } n \text{ are odd}\\ +1 & \text{ otherwise}\end{cases}$$

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

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

Answer 2

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*}