Proving positive semidefiniteness using the Schur complement

Question

Let $\mathbb{R}^{n \times n} \ni C = C^\top \succ 0$. Let $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = m$, where $m \leq n$. How do I show that

\begin{equation} C - CA^\top(ACA^\top)^{-1}AC \succeq 0 \end{equation}

holds?

Answer 1

If $G$ is nonsingular, we have that $X\geq 0$ if and only if $G^TXG\geq 0$. Hence $$ X:=C-CA^T(ACA^T)^{-1}AC\geq 0 $$ if and only if $C^{-1/2}XC^{-1/2}\geq 0$ ($G:=C^{-1/2}$), that is, $$ I-B^T(BB^T)^{-1}B\geq 0, $$ where $B:=AC^{1/2}$. Now note that $P:=I-B^T(BB^T)^{-1}B$ is an orthogonal projector ($P=P^T$, $P^2=P$) and all orthogonal projectors are positive semidefinite: $$x^TPx=x^TP^2x=x^TP^TPx=(Px)^T(Px)\geq 0$$ for any vector $x$.