Why is the rank of an element of a null space less than the dimension of that null space

Question

I'm trying to understand a statement I came across in a paper. Say $A$ and $B$ are $n \times n $-dimensional real matrices, and $B$ is in the null subspace of $A$ i.e. $AB=0$ (in fact $BA=0$ is also true in this case but I´m not sure if this is relevant here). The paper states that, since $B$ is in the null subspace of $A$,
$$ \textrm{rank}(B) \leq \textrm{dim}\big(\mathcal{N}(A)\big)\,, $$
where $\mathcal{N(.)}$ returns the null subspace of the argument. How can one prove this inequality? I tried to show it with the rank-nullity theorem, to no success. I suspect the proof may be even simpler than this.

Answer 1

$AB = 0$ if and only if all columns of $B$ are in the null space of $A$. This is in turn equivalent to the span of the columns of $B$ being contained in the null space of $A$. So the rank of $B$, which is the dimension of the span of the columns of $B$, is at most the dimension of the null space of $A$.