Let $A\in\mathbb R^{n\times c}, n>c$. Let $\operatorname{rank}(A)\leq c$. Hence the dimensions of the left null space of $A$ should be at least $n-c$. Given $B\in\mathbb R^{n\times d}$, if $d\leq n-c$, then $\operatorname{rank}(B)\leq\operatorname{N}(A^\top)$. Is this argument sufficient to say that there will exist $\bar A\neq 0$ that satisfies $\bar A^\top B=0$?
It also makes intuitive sense that the non-trivial solution space of $A^\top B=0$ given $B$ can be of any rank, i.e., $1\leq \operatorname{rank}(\bar A)\leq c$ and is independent of the magnitude of vectors of $\bar A$ in the solution space, since, we can arbitrarily fix the norm of each vector in $A$ and "rotate" them until they are perpendicular to the column space of $B$.
Does this make sense and what would be the correct way to prove this?