Determinantal inequality for block matrices: if $A=(B,C)$ is a square matrix, then $|A|^2\le |B^TB|\cdot |C^TC|$

Question

Suppose $A=(B,C)$ is a $n\times n$ matrix, $B$ is a $n\times s$ matrix, $C$ is a $n\times (n-s)$ matrix. Show that $|A|^2\leq |B^TB|\cdot |C^TC|$.

If $A$ is singular, then it is obvious. If $A$ is not singular, then $$A^TA=\begin{pmatrix} B^TB&B^TC\\ C^TB&C^TC \end{pmatrix}$$ and thus by Schur's lemma, $|A|^2=|A^TA|=|B^TB|\cdot |C^TC-C^TB(B^TB)^{-1}B^TC|$. Then what should we do?

Answer 1

A general linear algebra result says that for $S\succeq 0$, $R\succeq 0$ such that $R-S\succeq 0$, we may infer that $|S|\leq |R|$. Apply this with $$ S\equiv C^TC-C^TB(B^TB)^{-1}B^TC\succeq 0,\quad R\equiv C^TC\succeq0,\quad R-S\succeq0. $$