Matrix Identity Let $A$ be a  $ n \times n $ positive definite matrix, $P$ be a $n\times m$ real matrix with full column rank, I'm wondering whether we have the following inequalty
$$ P( P^{T} A P) ^{-1} P^{T} \preceq A ^{-1} ,$$
And in which sufficient conditions the the equality holds.
 A: First we consider a easier version when $A=I$. We have to prove the matrix below is semi-positive defined:
$$I-P(P^TP)^{-1}P^T=I-PP^+$$
Here $P^+$ refers to Moore–Penrose pseudoinverse. Because $P$ is column full rank, we have $P^+=(P^TP)^{-1}P^T$.
A wellknown result for $PP^+$ is that it's an orthogonal projection onto $R(P)$, the range of $P$. We can prove it within three steps:


*

*$PP^+$ is idempotent.

*$PP^+$ is Hermite.

*$R(PP^+)=R(P)$.


And then $I-PP^+$ is also an orthogonal projection, onto the $N(P^+)=\mathrm{Ker}P^+$. It is a simple deduction from those information that $I-PP^+$ is semi-positive defined, and it's $\mathbf{0}$ iff $\mathrm{rank}(P)=n$.

Now let's get to the general case. Because $A$ is positive defined, we have a invertible $n\times n$ matrix $Q$ so that $A=Q^TQ$. Notice
$$P(P^TQ^TQP)^{-1}P^T\preceq (Q^TQ)^{-1}\Leftrightarrow S(S^TS)^{-1}S^T\preceq I$$
where $S=QP$ is also full column rank. That's because pre-multiply by $Q$ and post-multiply by $Q^T$ cannot change the property of semi-positive defined. And now we can apply the result we've already obtained above.

The conclusion is: the inequality holds, and equality holds iff $\mathrm{rank}(P)=n$.
