If I have an $n \times n$ real-valued non-symmetric matrix $\mathbf{M}$, which has determinant $|\mathbf{M}| > 0$, what can I say about the determinant of the matrix $\mathbf{Q}^T \mathbf{M}^{-1} \mathbf{Q}$, where $\mathbf{Q}$ is a real-valued $n \times m$ matrix?
In particular, can I say that $|\mathbf{Q}^T \mathbf{M}^{-1} \mathbf{Q}| > 0$?
For background, the matrix $\mathbf{Q}^T \mathbf{M}^{-1} \mathbf{Q}$ is the Schur complement of the block matrix in the following linear system:
$$
\left( \begin{array}{cc}
\mathbf{M} & \mathbf{Q} \\
\mathbf{Q}^T & \mathbf{0} \\
\end{array} \right)
\left( \begin{array}{c}
\beta \\
\gamma \\
\end{array} \right) =
\left( \begin{array}{c}
\mathbf{f} \\
\mathbf{0} \\
\end{array} \right).
$$
In the case that $\mathbf{M}$ is symmetric positive definite (i.e. symmetric, and $|\mathbf{M}| > 0$,), I believe I'm right in saying that $|\mathbf{Q}^T \mathbf{M}^{-1} \mathbf{Q}| > 0$.
Best Answer
Obviously, no. For example $Q = 0$. In fact, whenever $m > n$, you'll get a zero determinant.
We observe the case when $m \le n$. Let $Q = U \Sigma V^T$ be an SVD of $Q$. Then $$Q^T M^{-1} Q = V \Sigma^T U^T M^{-1} U \Sigma V^T.$$ So, $$\mathop{\rm det}(Q^T M^{-1} Q) = \mathop{\rm det}(V) \mathop{\rm det}(\Sigma^T U^T M^{-1} U \Sigma) \mathop{\rm det}(V^T) = \mathop{\rm det}(\Sigma^T U^T M^{-1} U \Sigma).$$
Denoting $\Sigma = \mathop{\rm diag}(\sigma_1,\dots,\sigma_m) \in \mathbb{R}^{n \times m}$, we see that $\Sigma^T U^T M^{-1} U \Sigma$ is $U^T M^{-1} U$ with $i$-th row multiplied by $\sigma_i$ and $j$-th row multiplied by $\sigma_j$, so $$\mathop{\rm det}(\Sigma^T U^T M^{-1} U \Sigma) = \left( \prod_{k=1}^m \sigma_k \right)^2 \mathop{\rm det}(U^T M^{-1} U) = \left( \prod_{k=1}^m \sigma_k \right)^2 \mathop{\rm det} M^{-1} = \frac{\left( \prod_{k=1}^m \sigma_k \right)^2}{\mathop{\rm det} M}.$$ So, $\mathop{\rm det}(Q^T M^{-1} Q) > 0$ if and only if $\mathop{\rm rank} Q = m$; otherwise $\mathop{\rm det}(Q^T M^{-1} Q) = 0$.