$\left[\begin{array}{cc} P & A^T\\ A & 0\end{array}\right]$ is non-singular if and only if $\mathcal N(P) \cap \mathcal N(A)=\{0\}?$ Suppose $P\succeq 0$ and $A$ is of full row rank.  I want to show that $\left[\begin{array}{cc} P & A^T\\ A & 0\end{array}\right]$ is nonsingular if and only if $\mathcal N(P) \cap \mathcal N(A)=\{0\},$ where $\mathcal N(\bullet)$ denotes the null space of a matrix.
The only-if part is straightforward, because $\left[\begin{array}{c} P \\ A \end{array}\right]$ would not even be full-rank otherwise.  
I think I managed to show the if part, but would appreciate it if someone can confirm it or point out where I'm mistaken.  I'd also appreciate an alternative, simpler or more intuitive proof/comments.  Here's my attempted proof: ($0$ denotes a scalar or a matrix below, depending on the context.)
Suppose $\left[\begin{array}{cc} P & A^T \\ A & 0\end{array}\right] \left[\begin{array}{c} x \\ y\end{array}\right]=0.$  Then we have $Ax=0$ and $Px+A^Ty=0.$  Therefore, $x^TPx+x^TA^Ty=0$. Since $Ax=0$, this implies that $x^TPx=0$, hence $P^{1/2}x=0$ and $Px=0.$  It then follows that $x=0$, since $\mathcal N(P) \cap \mathcal N(A)=\{0\}.$ As a result, $y=0$ since $A^T$ is of full column rank. Q.E.D.
 A: Let $A$ be an $n \times n$ matrix and the rank of $A$ is $r$ where $r < n$.
There is an $n \times n$ matrix $U=[U_1|U_2]$ with orthogonal columns such that
 $$
AU = [AU_1| AU_2] = [AU_1 | 0]
$$
where $AU_1$ and $U_1$ are $n \times r$ matrices and $AU_1$ is full rank.. The matrix $U$ may be computed using the singular value decomposition (SVD) or the QR factorisation.
We now consider the orthogonal transformation
$$
\left[ \begin{array}{cc}
U^t & 0 \\
0    & I 
\end{array} \right]
\left[ \begin{array}{cc}
P & A^t \\
A & 0 
\end{array} \right]
\left[ \begin{array}{cc}
U & 0 \\
0 & I
\end{array} \right]
= \left[ \begin{array}{cc}
R & U^tA^t \\
AU  & 0 
\end{array} \right]
=: M
$$
which does not change the rank of the composite matrix. Also the rank of $R$ is the same as the rank of $P$. We partition the matrix $R$,
$$
R = \left[ \begin{array}{cc}
R_{11} & R_{12} \\
R_{21} & R_{22}
\end{array} \right]
= \left[ \begin{array}{cc}
U_1^t P U_1 & U_1^tPU_2 \\
U_2^t P U_1 & U_2^t P U_2 
\end{array} \right]
$$
where $R_{11}$ is an $r \times r$ matrix.
The first $n$ columns of the matrix $M$ is then given by
$$
\left[ \begin{array}{cc}
R_{11} & R_{12} \\
R_{21} & R_{22} \\
AU_1    & 0 
\end{array}
\right]
$$
If the null space of $P$ is defined by a column of $U_2$ then the matrix
$$
\left[ \begin{array}{c}
R_{12} \\
R_{22}
\end{array}
\right]
$$
will have zero column which makes the $M$ a non-singular matrix.
This completes the missing part of the proof.
