Prove that $A\in R^{m\times n}$ has a left inverse if and only if $Rank(A)=n$

Question

By definition, A has a left inverse implies that $LA=I_n$.

Clearly, $Rank(A)=n$ if A has a left inverse since $Rank(A)=min(m,n), Rank(LA)=min(Rank(L),Rank(A))=min(m,n)=n$

How can I prove that if $Rank(A)=n$ than A has a left inverse?

Answer 1

Think in terms of linear transformations.

If $\operatorname{rank}A=n$, then $A$ is injective. Let $\{e_1,\ldots,e_n\}$ be the canonical basis of $\mathbb{R}^n$. For each $j\in\{1,2,\ldots,n\}$, let $f_j=A.e_j$. Then the set $\{f_1,f_2,\ldots,f_n\}$ is linearly independent and therefore the are vectors $f_{n+1},\ldots,f_m$ such that $\{f_1,f_2,\ldots,f_m\}$ is a basis of $\mathbb{R}^m$. Define $B$ such that $B.f_j=e_j$ for each $j\in\{1,2,\ldots,n\}$ (you can take $B.f_j=0$ when $j>n$, if you want). Then $B.A=\operatorname{Id}_n$.