Let $A$ be an $m \times n$ matrix of rank $m$ and let $L$ be an $n \times n$ matrix that is symmetric and positive definite on the subspace $M = \{x : Ax = 0\}$.
Show that the $(n + m) \times (n + m)$ matrix $\left(\begin{matrix} L&A^T\\ A&0\end{matrix}\right)$ is nonsingular.
Set $N =\left(\begin{matrix} L&A^T\\ A&0\end{matrix}\right)$.
Suppose that $Nu = 0$. We aim to show that $u = 0$.
Splitting $u = [v,w]^T$ where $v$ is $n$-dimensional and $w$ is $n$-dimensional,
we have $$Av = 0$$ and $$Lv + A^Tw =0$$
Multiply the second equation on the left by $v^T$
to find $v^TLv + v^T A^T w = 0$. Hence $v^T L v = 0$ and since $v \in M$ and $L$ is positive definite on $M$, we have $v = 0$.
Hence we have established $A^T w = 0$. Since $A^T$ has full column rank, this establishes $w =0$, and in fact $u=0$.