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!
Best Answer
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).
$\square$