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!
 A: 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 $r\circ f=\operatorname{id}_{K^n}.$
A: Using RREF: if $T$ has linearly independent columns, then its RREF is of the form
$$
R = \pmatrix{I_n \\ 0_{m \times (m-n)}}.
$$
By the nature of RREF, there exists an invertible matrix $P$ such that $PT = R$.
Now, note that by multiplying $R$ from the left by the matrix $Q = [I_n \ \ 0_{(n-m) \times m}]$, we end up with the matrix $I_n$ (the first $n$ rows of $R$). Thus, we have
$$
I_n = QR = Q(PT) = (QP)T.
$$
So, $S = QP$ satisfies our requirement.
A: There are multiple ways to see it. One matrix approach goes as follows:
Since $T$ is of full column rank, we have $n \leq m$, and there exists an order $m$ invertible matrix $P$ such that
\begin{align*}
T = P\begin{pmatrix} I_{(n)} \\ 0 \end{pmatrix}. \tag{1}
\end{align*}
Take $S = \begin{pmatrix} I_{(n)} & 0 \end{pmatrix}P^{-1}$, it then follows that
\begin{align*}
ST = \begin{pmatrix} I_{(n)} & 0 \end{pmatrix}P^{-1}
P\begin{pmatrix} I_{(n)} \\ 0 \end{pmatrix} = I_{(n)}.
\end{align*}
Remark: $(1)$ is known as a full rank factorization of $T$, which can be verified by using the equivalence normal form of $T$.
