Rank, and existence of Matrices Let $A$ be an $m\times n$ matrix.  If the rank of $A$ is $m$, then prove there exists a matrix $B$, wich is $n \times m$, such that $AB=\text{I}_m$

Suppose $A$ is an $(mxn)$ matrix with $rank(A) = m$
then $n \ge m $ and the column space of $A$ spans all of $\mathbb{R}^m$.
So, $\forall b\in \mathbb{R}^m, \exists x\in \mathbb{R}^n$ s.t. $Ax = b$.
Choose $b_i$ to be the $i^{th}$ column of $I_m$. The the columns of $B (nxm)$ are the vectors $x_i \in \mathbb{R}^n$ s.t. $Ax_i = b_i$.
