# Determinant of Schur Complement

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 $|\mathbf{Q}^T \mathbf{M}^{-1} \mathbf{Q}| > 0$.

## 1 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 $\mathop{\rm det}(Q^T M^{-1} Q) = 0$.

• Thanks very much for your detailed reply. I should have said that one of the conditions of the linear system in my application is that $rank(Q) = m <= n$. From the last statement of your answer I guess that does confirm it then, $det(Q^T M^{-1} Q) > 0$ . Brilliant, thanks again! – Tom Jul 14 '13 at 3:01