Let $A\in \mathbb R^{m\times n}$ and $B\in \mathbb R^{n\times l}$. Assume that $$\begin{gathered} \operatorname{rank}(A) = m \hfill \\ \operatorname{rank}(B) = n \hfill \\ \end{gathered} $$ and also that $l\gg n$ and $m\leq n$. Could I conclude that $\operatorname{rank}(AB)=m$.
Attempt: I know how to upper bound the rank of the multiplication:
$\operatorname{rank}\left( {AB} \right) \leqslant \min \left( {\operatorname{rank}(A),\operatorname{rank}(B)} \right) = m$ but how to get the equality instead of the inequality!!
$\mathrm{rank}(AB)=\mathrm{rank}(A)$ if the column space of $B$ contains the entire row space of $A$. In your situation that is guaranteed, since the column space of $B$ contains all of $\mathbb{R}^n$. The shapes themselves don't guarantee this alone, you needed that $B$ has full row rank.