$A\in M_{m*n}, B\in M_{n*m}$ and $AB=I$ then $m\le n$

Question

I'm trying to prove the proposition in the title. Unlike similar claims for which the intuition was clear to me, even if a rigorous proof was sometimes not, in this case I'm not sure I understand why this is true. Any help would be appreciated.

Answer 1

$\DeclareMathOperator{\rank}{rank}$You may use the fact that $$ \rank (A B) \le \min \{ \rank(A), \rank(B) \}. $$ Here $\rank (A B) = \rank(I) = m$ and $\rank(A) = \rank(B) = \min \{m, n \} \le n$.

One way of seeing the above fact is to see the rank as the dimension of the image, i.e. of the column space, of the matrix.

Clearly the column space of $A B$ is a subspace of the column space of $A$, so $\rank(A B) \le \rank(A)$. And the column space is the image under $A$ of the column space of $B$, so that $\rank(A B) \le \rank(B)$.