Let $Q$ be a real-valued, symmetric, and positive definite (PD) matrix, which is partitioned as
$$ Q=\begin{bmatrix} Q_{xx} & Q_{xy} \\ Q_{yx} & Q_{yy} \end{bmatrix} $$ where diagonal blocks are square. I know, for such a matrix [1], $Q_{xx}$, $Q_{yy}$ and its Schur complement $S=Q_{yy} - Q_{yx} Q_{xx}^{-1} Q_{xy}$ are PD as well. What can I say about $M = A S A^T$ ($A=Q_{xx}^{-1} Q_{xy}$)? My numerical studies shows that it is positive semi-definite, but how can I prove that?
Edit: $Q$ is sparse.
We know that $S$ is positive definite and $Q_{xx}$ has full rank and is symmetric. To check definiteness of M lets check if $v^T M v \geq 0 \ \forall v$. From definition of $M$ $$ v^T M v = v^T Q_{xx}^{-1} Q_{xy} S Q_{xy}^T Q_{xx}^{-1}v. $$ Now let $w = Q_{xx}^{-1}v$. Since $Q_{xx}$ has full rank it can be interpreted as some change of basis so we can conclude that $v^T M v \geq 0 \ \forall v$ if and only if $w^T Q_{xy} S Q_{xy}^T w \geq 0\ \forall w$.
Now to consider the effect of $Q_{xy}$. In a similar fashion, we can set $z = Q_{xy}^T w$ and substitute. This will produce $z^T S z$. From positive definiteness follows that $z^T S z \geq 0$ and $z^T S z = 0$ only if $z=0$. From this can be concluded that $v^T M v \geq 0 \ \forall v$ and $v^T M v = 0$ if and only if $Q_{xy}Q_{xx}^{-1}v = 0$.