# Proof that the following composite matrix is invertible

Let $$B\in \mathbb{R}^{n\times n}$$ and $$C\in\mathbb{R}^{m\times n}$$ (where $$m\leq n$$) and let $$v^TBv > 0$$ $$\forall v \neq 0$$ with $$Cv = 0$$, where $$C$$ has full rank.

Proof that $$A = \begin{pmatrix} B & C^T \\ C & 0 \end{pmatrix}$$ is an invertible Matrix.

So far, I've concluded that $$B$$ is positive definite and therefore also $$det(B)>0$$, which could help prove that $$det(A) \neq 0$$. But other than that I had no idea for a solution.

Any help is appreciated, thanks in advance!

Take $$B = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$ and $$C = \begin{bmatrix} 1 & 0 \end{bmatrix}$$ to see that $$B$$ does not have to be positive definite (replace the $$-1$$ with $$0$$ in $$B$$ to see non-invertibility of $$B$$).

To show that $$A$$ is invertible we will show $$Av = 0 \implies v = 0$$. We have

$$\begin{bmatrix} B & C^T \\ C & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0$$

gives $$Cv_1 = 0$$ and $$Bv_1 + C^Tv_2 = 0$$. Left multiplying the second equation by $$v_1^T$$ yields

$$v_1^TBv_1 + v_1^TC^Tv_2 = 0 \iff v_1^TBv_1 + (Cv_1)^Tv_2 = 0$$

which gives $$v_1^TBv_1 = 0$$ since $$Cv_1 = 0$$. Thus $$v_1 = 0$$ (otherwise $$v_1^TBv_1 > 0$$). Plugging this back in gives $$C^Tv_2 = 0$$ which implies $$v_2 = 0$$ since $$C^T$$ is injective (full column rank).

$$\square$$

This amounts to simply using the Schur complement.

In detail, note that $$A$$ is invertible if and only if the matrix $$\pmatrix{I & 0\\-CB^{-1} & I} \pmatrix{B & C^T\\ C & 0} \pmatrix{I & 0\\-CB^{-1} & I}^T = \pmatrix{B&0\\0 & - CB^{-1}C^T}$$ is invertible. Thus, $$A$$ is invertible if and only if both $$B$$ is invertible and $$CB^{-1}C^T$$ is invertible. Using the fact that $$C$$ has full row-rank and $$B^{-1}$$ is positive definite, conclude that $$CB^{-1}C^T$$ is positive definite and therefore invertible.

• Hi Ben, I don't think $B$ has to be positive definite or even invertible (see counterexample in my answer). Commented Jan 29, 2021 at 16:01