Left inverse of matrix exists if $A\mathbf x=\mathbf b$ has a unique solution? If the equation $A\mathbf{x} = \mathbf{b}$ has a unique solution for some $\mathbf{b}$ is it true that $A$ has a left inverse? $A$ is an $m\times n$ matrix.
 A: You already reasoned (in a comment) that the null space of $A$ must be $\{0\}$, which means that the equation $Ax=b'$ has a unique solution $x$ for every $b'$ for which it has a solution at all (having at least two different solutions would by subtraction give a nonzero element of the null space, which does not exist). [Stated differently $x\mapsto Ax$ is an injective map.]
The set of such $b'$ is precisely $\def\R{\Bbb R}C=\{\,Ax\mid x\in\R^n\,\}$, which is a subspace of $\R^m$ (the column space of $A$). The above means there is a well-defined map $L:C\to\R^n$ that sends $b'\in C$ to the corresponding unique solution$~x\in\R^n$. In formula, for every $x\in\R^n$, putting $b=Ax$, one has $L(b)=x$, that is $L(Ax)=x$ for all $x\in\R$. Now show that


*

*$L$ is a linear map

*$L$ can be extended (non uniquely) to a linear map $L':\R^m\to\R^n$

*The matrix of any such $L'$ is a left inverse of $A$.

A: What have you tried? When does a matrix have a left inverse?
If $Ax = b$ has a unique solution for some $b$, what can you say about the nullspace of $A$? The rank of $A$?
A: Hint : Compute the relation between $\textrm{rank}(A^T A)$ and $\textrm{rank}(A)$. 
Use the fact that $\textrm{rank}(A) = \textrm{dim}(R(A))$.
Is $\textrm{rank}(A^T A)= \textrm{rank}(A) = n$? 
A: $A x= b$ can be written as a system of equations:
$$
\begin{cases}
b_1 = \sum_{i=1}^n a_{1i} x_i \\
\vdots \\
b_m = \sum_{i=1}^n a_{mi} x_i
\end{cases}
$$
If $m<n$ you have $m$ equations and $n$ variables. More variables than equations. So is the system underspecified, overspecified or fully specified ?
What when $m>n$ and $m=n$ ?
