matrices multiplication $AB=I$ imply that span of cols A is $F^m$

Question

$A\in M_{m,n} $ and $B\in M_{n,m}$, suppose that $AB=I_m$ does it imply that $ \operatorname{span}( \operatorname{col}(A))$ = $F^m$?

Answer 1

Let $f:F^n\to F^m$ the linear map associated to the matrix A in the canonical bases of $F^n$ and $F^m$. Similarly,let $g:F^m\to F^n$ be associated to $B$.

The span of the columns of $A$ is simply the image of the linear map $f$; and it is asserted this map is surjective.

This is pbvious from the hypothesis that $AB=I_m$, since $AB$ corresponds to the composition $f\circ g$ which is $\operatorname{id}_{F^m}$, and if $f\circ g$ is surjective, it is well known that $f$ is.