# 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.

• Well... what methods do you know to compute the determinant? – Alex Kruckman Jun 23 '20 at 23:37
• You might note that $$\mathbb C = \pmatrix{ 0 & I_n\cr I_m & 0\cr} \pmatrix{ \mathbb B & 0\cr 0 & \mathbb A}$$ – Robert Israel Jun 23 '20 at 23:53

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*}

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,

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

• I believe $\det P$ should be derivable through a certain permutation of rows/columns, without such heavy machinery. – lisyarus Jun 24 '20 at 18:08
• @lisyarus That is almost certainly the case. However, coming from signal processing, I am more familiar with circulant matrices than with permutations. – Rodrigo de Azevedo Jun 24 '20 at 18:16
• For a permutation matrix, the easiest way to compute the determinant is via the Leibniz formula: $\det(A) = \sum_{\sigma\in S_n} \operatorname{sign}(\sigma)\prod_{i} A_{i, \sigma(i)}$. Since $P$ is a permutation matrix, all terms in the sum are zero, except for the one corresponding to the permutation represented by $P$ itself. The parity of the permutation here should be $(-1)^{m+n+1}$ if I'm not mistaken. – Hyperplane Jun 24 '20 at 23:30