a question about rank of a matrix

Question

Suppose $A$ is a $m\times n$ matrix. Show that $\mbox{rank}\,A=m$ if and only if there exists a $n\times m$ matrix $B$ such that $AB=I_m$.
I have proved the case $AB=I_m$ eventuates $\mbox{rank}\,A=m$, but the main part, the inverse, I couldn't establish.
Would be grateful for your help.

Answer 1

HINT: Because $A$ has rank $m$, you can solve $Ax=e_j$, $j=1,\dots,m$. Here, by $e_j$ I mean the $j$th standard basis vector.

Answer 2

Without loss of generality, suppose $A = (A_1, A_2)$, where $A_1$ is an $m\times m$ invertible matrix, then we can find an $m\times m$ matrix $B_1$ such that $A_1B_1 = I_m$. Then $B = \left( \begin{array}{c} B_1 \\ O \end{array} \right)$ where $O$ is a $(n-m) \times m$ zero matrix is what you want.

If we have to permute the columns of $A$ to make it in the above form, make the corrsponding permutation of rows of $B$ at the end