TITLE: is this kind of symmetric matrix invertible?
QUESTION [5 upvotes]: Give a matrix $A=\begin{bmatrix}M&B\\ B^T&0\end{bmatrix}$, where $M\in\mathbb{R}^{n\times n}$, $B\in \mathbb{R}^{n\times m}, (m<n)$. If we know that $rank(B)=m$ and  for any $v\neq 0$ and $v\in Null(B^T)$, i.e $B^T v=0$, we have $v^T Mv >0$. Then is the matrix $A$ invertible?

REPLY [3 votes]: Suppose there are vectors $w$ and $v$ such that
\begin{equation}
\begin{bmatrix}
M&B\\
B^\top{}&0
\end{bmatrix}
\begin{bmatrix} 
w\\
v
\end{bmatrix}
=0
\end{equation}
Since $B^\top{}w = 0$, we have from above,
\begin{equation}
0=
\begin{bmatrix} 
w^\top{}&v^\top{}
\end{bmatrix}
\begin{bmatrix}
M&B\\
B^\top{}&0
\end{bmatrix}
\begin{bmatrix} 
w\\
v
\end{bmatrix}
=w^\top{}Mw.
\end{equation}
Since $w$ lies in $\mathrm{Null}(B^\top{})$ and $w^\top{}Mw=0$, we get $w = 0$. 
$Bv=0$ and $B$ has the full column rank, therefore $v=0$. We could get $\mathrm{Null}(A)=\{0\}$, so $A$ is invertible.