Let matrix $A\in\mathbb{R}^{n\times n}$ be symmetric positive definite and let fat matrix $B\in\mathbb{R}^{m\times n}$, where $m<n$, have full row rank. Let \begin{equation}\mathcal{C}=\begin{bmatrix}A&B^T\\B&0\end{bmatrix}.\end{equation} Prove that block matrix $\mathcal{C}$ is symmetric indefinite.
I have been able to show the symmetry of the given matrix as follows:
When taking a tranpose of a block matrix we must also tranpose each individual block, but note that $A$ is SPD, hence $A^T=A$;
\begin{equation}\mathcal{C}^T=\begin{bmatrix}A^T&B^T\\(B^T)^T&0\end{bmatrix}=\begin{bmatrix}A&B^T\\B&0\end{bmatrix}\end{equation}
and hence $\mathcal{C}=\mathcal{C}^T$ and $\mathcal{A}$ is symmetric.
However, I do not know how to show the indefiniteness of this matrix.
$$ \begin{bmatrix} x^T & y^T \end{bmatrix}\begin{bmatrix}A&B^T\\B&0\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = x^TAx + x^TB^Ty +y^TBx = x^TAx + 2y^TBx. $$
We know that when $y = 0$ and $x \neq 0$, we have $x^TAx > 0$. So all we need to do is argue that this expression can also be negative.
Note that $B$ has full row rank means that $y^TB = 0$ if and only if $y = 0$.
So pick an $x$ not in the kernel of $B$ and then look at the function $$f(y) = y^TBx.$$ Argue that you can make this sufficiently negative to make the whole expression at the top negative.