
Elem. proof help?
Hi, just reviewing some linear algebra this summer. Came across this in one of my exercise books, and was looking for a proof.
Let A be an m x n matrix with rank m. Prove that there exists an n x m matrix B such that AB = I(m).
[That last part reads "AB equals I sub m."]

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