Sorry that the title is a bit ambiguous, I can't think of anything better.
Given a matrix
$ C= \begin{bmatrix} A & B^T \\ B & 0 \end{bmatrix} $
where $C$ is not singular (so rank(C) = n+m), $0\in\mathbb{R}^{m\times m}$ is the zero matrix, $A\in \mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{m\times n}$, $m \leq n$. Assume that $A$ is not full rank.
If $rank(B) = m$ and $null(B) \cap null(A) = \{0\}$, then $A$ has rank at least $n - m$.
Why is the above statement true?
There's some intuition about "$null(B) \cap null(A) = \{0\}$" which I just don't see.
Edit: If it helps anymore, I've included the snippet of where this came from: snippet
This has nothing to see with all the hypotheses, except what's in the statement: the rank-nullity theorem says that $$\DeclareMathOperator\rk{rank}\rk A+\dim\ker A=n,$$ hence if $\rk A \ge r$, then $\;\dim\ker A\le n-r$.
Added:
Note the hypothesis $C$ has full rank implies $B$ has full rank, so $\ker B=0$, and the hypothesis $\ker A\cap\ker B=0$ is redundant.
Without the hypothesis on $C$, $\ker A\cap\ker B=0$ implies $\ker C=0$