We have a $2 \times 2$ block matrix M, in which the diagonal blocks are $0$. The sizes of the upper rigt block is $N_B \times N_A$ and the size of the lower left block is $N_A \times N_B$, where we know $N_B \geq N_A$. We also know that rank(M) = $2 N_A$.
An example of such matrix M with $N_A = 2$ and $N_B = 3$ is $ \begin{pmatrix} 0& 0 & 1 & 2 & 3\\ 0& 0 & 2 & 3 & 4 \\ 1& 1 & 0 & 0 & 0 \\ 1& 4 & 0 & 0 & 0 \\ 5& 7 & 0 & 0 & 0 \\ \end{pmatrix} .$
QUESTION: We need to show that for every vector $v \in$ ker($M$), it holds that the first $N_A$ elements of $v$ are $0$ (or equivalently we need to show that $v \in \{0\}^{N_A} \times R^{N_B}.$.) (For the above example, vector $\begin{pmatrix} 0\\ 0 \\ 1\\ 1 \\ 5\\ \end{pmatrix}$ does have the first $N_A = 2$ elements equal to $0$ and the vector $\begin{pmatrix} 0\\ 2 \\ 1\\ 1 \\ 0\\ \end{pmatrix}$doesn't. )
I tried to prove it, but I'm not sure if it's correct:
$M$ maps $V = R ^{N_A + N_B}$ to $V$. We define $A = R^{N_A} \times \{0\}^{N_B}$ and $B = \{0\}^{N_A} \times R^{N_B}$, which means $V = A \bigoplus B , dim(A) = N_A $ and $dim(B) = N_B.$ We can define two maps $$f: A \to B, f(x) = Mx$$ and $$g: B \to A, g(x) = Mx.$$ Since $dim(B) = N_B \geq dim(A) = N_A$ and $g$ is a linear map, we know $dim(ker(g))\geq N_B - N_A$.
Since $rank(M) = 2_$, we know $dim(ker(M)) = N_B - N_A$.
We see that $ker(g) \subseteq ker(M)$, since for every $v \in B \subseteq V$ it holds $v \in ker(g) \implies g(v) = 0 \implies Mv = 0 $$\implies v \in ker(M)$.
Since $ker(g) \subseteq ker(M)$ and $dim(ker(g)) \geq dim(ker(M))$, it follows that $ker(g) = ker(M)$. Therefore for every vector $v \in$ ker(M), it holds that the first $N_A$ elements of $v$ are $0$ (otherwise $v$ would not be in $B$).