TITLE: If a matrix $Q$ is symmetric and positie definite, is it possible to show that the matrix $Q-A^T(AQ^{-1}A^T)^{-1}A$ is also positive definite?
QUESTION [7 upvotes]: If I have a symmetric and positive definite $n\times n$ matrix $Q$ and a full row-rank totally unimodular $m\times n$, where $m<n$, matrix  $A$, is it posible to show that the matrix
 $$Q-A^T(AQ^{-1}A^T)^{-1}A$$
is also positive definite?
I have show that it is true when the matrix $Q$ has dimention $2\times 2$, by doing all the posible situations. Also, If $m=n$, I have that the matrix $Q-A^T(AQ^{-1}A^T)^{-1}A$ can be zero.

REPLY [4 votes]: The matrix $Q - A^T(AQ^{-1}A^T)^{-1}A$ will generally be positive semidefinite. Let 
$$M/Q : = Q - A^T(AQ^{-1}A^T)^{-1}A$$
and note that $M/Q$ is the lower Schur complement of the block matrix
$$
M= 
\begin{bmatrix}
AQ^{-1}A^T & A \\
A^T & Q\\
\end{bmatrix}.
$$
Since $Q$ is positive definite, it follows that $Q^{-1}$ is positive definite and that $Q^{1/2}$ exists. Furthermore $(Q^{-1})^{1/2}$ exists, hence let $Q^{-1/2} := (Q^{-1})^{1/2}.$ Note that
$$
\begin{bmatrix}
AQ^{-1}A^T & A \\
A^T & Q\\
\end{bmatrix}
=
\begin{bmatrix}
AQ^{-1/2} \\
Q^{1/2}\\
\end{bmatrix}
\begin{bmatrix}
Q^{-1/2}A^T & Q^{1/2} \\
\end{bmatrix}.
$$
Therefore $M$ is positive semidefinite. Since $M$ is positive semidefinite, so too is its lower Schur complement.