Analogue of the Cauchy-Schwarz inequality for determinants: a simple proof

In this note, we present a simple proof of an analogue of the Cauchy-Schwarz inequality relevant to products of determinants. Specifically, we show that $$\mathrm{\mid }\det (A^{\ast }MB){\mathrm{\mid }}^2\le \det (A^{\ast }MA)\cdot \det (B^{\ast }MB),A,B\in {\mathbb{C}}^{m\times n},$$ where $M\in {\mathbb{C}}^{m\times m}$ is hermitian positive definite. Here $m$ and $n$ are arbitrary. In case $m\le n$, equality holds trivially. Equality holds when $m>n$ and $\text{rank}(A)=\text{rank}(B)=n$ if and only if the columns of $A$ and the columns of $B$ span the same subspace of ${\mathbb{C}}^m$.