Prove the column span of $A$, an $m\times n$ matrix, is $F^m$ iff $\exists$ an $n\times m$ matrix $B$ such that $AB = I_m$ This problem has a second part, about the columns of $A$ being linearly independent in $F^m$ iff $\exists n \times m$ matrix $B$ such that $BA = I_n$ but I think if I understand the first part, I'll get the second.
I've tried making the matrix into a linear transformation $T \in \mathcal{L}(F^n,F^m)$ such that $A = M_T(B_{F^n},B_{F^m})$, and then $B$ would give us the right inverse of $A$, i.e. $\exists S \in \mathcal{L}(F^m,F^n)$ such that $T\circ S = I_{F^m}$, the identity transformation in $F^m$, but I don't know how to use the column span property or how to finish the proof.
Any advice would be greatly appreciated! Thanks!
 A: If the column span of $A$ is $F^m$, then every vector $F^m$ can be expressed as a linear combination of the columns of $A$, which means that the linear system
$$
Ax=c
$$
is solvable for every $c\in F^m$. In particular, the systems $Ax=e_i$ $(i=1,2,\dots,m)$ are solvable, where $\{e_1,e_2,\dots,e_m\}$ is the standard basis of $F^m$. Then, let $Ab_i=e_i$ $(i=1,2,\dots,m)$ and consider the matrix
$$
B=\begin{bmatrix} b_1 & b_2 & \dots & b_m\end{bmatrix}
$$
having those solutions as columns. By the definition of matrix product, you have $AB=I_m$.
Conversely, if $AB=I_m$ and $c\in F^m$, we can write
$$
A(Bc)=I_mc=c
$$
so $Bc$ is a solution of $Ax=c$, which means that every vector in $F^m$ is a linear combination of the columns of $A$.
A: If $A$ is of a full column rank, then there is a permutation such that
$$AP = \begin{bmatrix} A_1 & A_2 \end{bmatrix}, \quad A_1 \in \mathbb{F}^{m \times m}, \quad A_2 \in \mathbb{F}^{m \times (n-m)},$$
where $A_1$ is invertible. In other words, $P$ moves $m$ linearly independent columns to the front.
Let $B := PB'$, where
$$B' := \begin{bmatrix} A_1^{-1} \\ 0 \end{bmatrix}.$$
Then
$$AB = A(PB') = (AP)B = \begin{bmatrix} A_1 & A_2 \end{bmatrix} \begin{bmatrix} A_1^{-1} \\ 0 \end{bmatrix} = A_1A_1^{-1} + A_20 = I.$$
For a converse, just use the fact that $\operatorname{rank}(XY) \le \min \{ \operatorname{rank} X, \operatorname{rank} Y \}$.
