# Elem. proof help?

Consider the linear transormation $R_A : \mathbb{R}^m \rightarrow \mathbb{R}^n : v \mapsto vA$. By definition of the rank, $R_A$ is an injection, hence is left-invertible; hence there exists a transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $T\circ R_A=1$. Writing $T=R_B$ we have $R_B\circ R_A=1$; but $R_B\circ R_A=R_{AB}$ so $AB=I_m$.