This is a simple observation and it is similar to Robert's answer, but perhaps it is worth making explicit anyway: Let $\mathcal A$ be the set of rows of $A$. Then your condition will hold precisely if it is not possible to partition $\mathcal A=\mathcal A_1 \cup \ldots \cup\mathcal A_p$ into (at most $p$) subsets with $\dim L(\mathcal A_j)<m$ for $j=1,\ldots ,p$.

Indeed, if we had such a partition, we could take $b_j\perp \mathcal A_j$ as the columns of $B$, and the converse is also obvious, by defining $\mathcal A_j =\{b_j\}^{\perp}\cap \mathcal A$ for given $b_j$'s.