I have trouble with proving the following.
P is $m \times m$ positive definite and diagonal matrix. Q and R are $ n \times m $ matrices that $n < m$. Q and R have full rank, but $ Q \neq cR $ for any real number $c$. The elements of Q and R are not 0.
Let, $ S = QPQ^T - QPR^T (RPR^T)^{-1}RPQ^T $
I found out that S is always positive semi-definite with Matlab simulation, but I couldn't prove it analytically. Moreover, there are interesting properties which are,
If $ 2n \leq m$, S is always positive definite. (Not semi-definite)
These are all from numerical results, so I don't have confidence that these are true. How can I prove it mathemetically?
Notice that $S$ is of the form $A - X B^{-1} X^T$, i.e., it is the Schur complement of the symmetric block matrix $$ Z=\begin{pmatrix} Q P Q^T & Q P R^T \\ R P Q^T & R P R^T \end{pmatrix}. $$
Now we know that as $P$ is positive definite and $R$ has full rank that $RPR^T$ is positive definite. Under this condition there's a result that says that $S\geq 0$ iff the block matrix $Z \geq 0$. (Note that there are several other possibilities -- see the bottom of the linked wiki page). So our problem reduces to showing $Z \geq 0$. However, note that $$ Z = \begin{pmatrix} Q P Q^T & Q P R^T \\ R P Q^T & R P R^T \end{pmatrix} = \begin{pmatrix} Q \\ R \end{pmatrix} P \begin{pmatrix} Q^T & R^T \end{pmatrix} = M P M^T $$ where $M = \begin{pmatrix} Q \\ R\end{pmatrix}$. But as $P$ is positive definite we have $MPM^T$ is positive semidefinite and the result holds.