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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!!

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  • $\begingroup$ Are you sure about that $l>>n$. And what has this to do with matrix-calculus? $\endgroup$ Commented Mar 1, 2022 at 14:37
  • $\begingroup$ @JoséCarlosSantos Yes I am sure. it is a fat matrix with full row rank. $\endgroup$
    – Mark
    Commented Mar 1, 2022 at 14:38
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    $\begingroup$ More generally, the composition of surjective functions is surjective. Your particular question corresponds to the case when said functions are linear. $\endgroup$ Commented Mar 1, 2022 at 14:52

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$\mathrm{rank}(AB)=\mathrm{rank}(A)$ 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.

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  • $\begingroup$ you mind giving a proof why that hold? $rank(AB)=rank(A)$ iff the column space of B contains the entire row space of A. $\endgroup$
    – Mark
    Commented Mar 1, 2022 at 14:47
  • $\begingroup$ @Mark For your specific case it comes down to what Brian Moehring said in the comments, that the composition of surjective functions is surjective. The more general statement about linear functions specifically takes a fair number of steps to prove from scratch. $\endgroup$
    – Ian
    Commented Mar 1, 2022 at 15:10
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    $\begingroup$ Thanks, I already know how to navigate my way!!! $\endgroup$
    – Mark
    Commented Mar 1, 2022 at 15:24
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    $\begingroup$ @Ian $\text{row}A \subseteq \text{col}B$ implies $\text{rank}AB = \text{rank}A$ but the converse is false. E.g., $A = \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix}1 & 0 \\ 1 & 0 \end{bmatrix}$. $\endgroup$
    – Coriolanus
    Commented Mar 1, 2022 at 21:22
  • $\begingroup$ @Coriolanus Thanks. $\endgroup$
    – Ian
    Commented Mar 1, 2022 at 21:37

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