Matrix Identity II I asked a question on an matrix inequality several days ago, see the link 
Matrix Identity.
Is this inequality still true for this strengthened version?
Let $A$ be a  $ n \times n $ positive definite matrix, $P, Q $  be  $n\times m$ and $n\times l$ real matrix with full-rank orthongonal column vector($m+l \leq n$), which is to say
$$P^{T}P=I_{m \times m}, \quad Q^{T}Q = I_{l \times l}, \quad P^{T}Q=O_{m \times l}.
 $$ I'm wondering whether we have the following inequalty
$$ P( P^{T} A P) ^{-1} P^{T} + Q( Q^{T} A Q) ^{-1} Q^{T} \preceq A ^{-1}.$$
And the equality holds when $ m+l =n$?
 A: With the same argument that Wild Chan gave in the other answer, we can at least show that if $A=I$ , we would want to show that $I - PP^\dagger - QQ^\dagger$ is positive semidefinite. But $PP^\dagger + QQ^\dagger$ is just an orthogonal projection onto the direct sum of the column spaces of $P,Q$ as they have orthogonal column spaces, and thus it is true.
So let us transfer to the other problem. Letting $A = LL^T$, we have (after multipling on the left by $L^T$ and on the right by $L$,
$L^TP(P^T L L^TP)^{-1}P^TL + L^TQ(Q^T L L^T Q)^{-1} Q^TL = R(R^TR)^{-1}R^T + S(S^T S)^{-1}S^T \preceq I$ where $R = L^TP$ and $S = L^TQ$.
As per comments, the same argument would hold if $P^T A Q = R^T S = 0$ but it may not be true.
We have a negative result if $R$ and $S$ are not orthogonal: For a counterexample we would need to find an $x$ for which $|x_R|^2 + |x_S|^2 > |x|^2$ where $x_R$ is the orthogonal projection to the column space of $R$ and likewise for $x_S$. If we take $x$ to be a column of $L^TP=R$ which is not orthogonal to  $L^TQ=S$, then $x_R = x$ but $x_S \neq 0$, and thus we have strict inequality which violates the claim. 
I suppose it remains to determine when $P^T A Q = 0$. One such condition is if the column space of $Q$ is $A$-invariant or the column space of $P$ is $A^T$-invariant.
