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.
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.