Proving there exists an $n \times m$ matrix $S$ such that $ST=I_n$, where $I_n$ is the $n×n$ identity matrix I'm learning linear algebra and came across this problem which I am having a difficult time trying to solve: The problem is:
Suppose $T$ is a $m\times n$ matrix with linearly independent columns. Prove there exists an $n \times m$ matrix $S$ such that $ST=I_n$, where $I_n$ is the $n×n$ identity matrix.
I was thinking of starting the proof with something like supposing that $T{\bf x}={\bf z}$ and if $S$ were such that $ST=I_n$, then finding $S{\bf z}$ but I'm not sure if that helps to prove the question. Do I need to find the number of solutions to $T{\bf x}={\bf z}$?
Any help would be appreciated!

Hint:
Interpret $T$ as the matrix of a linear map $f:K^n\longrightarrow K^m$ ($K$ is the  base field). The hypothesis implies $f$ is injective. Prove an injective linear map has a linear retraction, i.e. a linear map $r:K^m\longrightarrow K^n$ such that $r\circ f=\operatorname{id}_{K^n}.$
