If a matrix has linearly independent columns, does it automatically have a left inverse? 
If a matrix has linearly independent columns, does it automatically have a left inverse?

So I know the opposite is true.  That is, if a matrix has a left inverse, that means that the columns of the matrix are linearly independent.  Was wondering if a matrix has linearly independent columns, does that automatically mean it has a left inverse?
Thanks!
 A: Suppose $A$ is an $m \times n$ matrix with linearly independent columns. Let $L_A$ be the linear transformation defined by $L_A(x) = Ax$. Because the columns of $A$ are linearly independent, the null space of $L_A$ is trivial. Hence, $L_A$ is one-to-one. It follows that $L_A$ has a left inverse $K:R(A) \to \mathbb R^n$. Although the linear transformation $K$ is only defined on a subspace of $\mathbb R^m$, it can be extended to a linear transformation $T:\mathbb R^m \to \mathbb R^n$. This linear transformation $T$ also satisfies
$$
T \circ L_A = I
$$
where $I$ is the identity transformation on $\mathbb R^n$. Finally, if $M$ is the matrix representation of $T$ (with respect to the standard bases of $\mathbb R^m$ and $\mathbb R^n$) then
$$
M A = I.
$$
A: Yes, it does mean that. There are several ways to see this, but here is one:
If the matrix is $m\times n$, then the columns being linearly independent means the matrix has rank $n$. Thus the $m$ rows span an $n$-dimensional subspace of $\Bbb R^n$, which must be $\Bbb R^n$ itself. In particular, that means that there are linear combinations of the rows that make up each of the basis vectors.
The $k$th row of any left inverse will be the coefficients of such a linear combination for the $k$th basis vector, and any matrix consisting of such rows will be a left inverse. (In general, in a matrix product $AB=C$, the $k$th row in $C$ is a linear combination of the rows in $B$ given by the coefficients in the $k$th row of $A$. Also, more commonly, the $k$th column in $C$ will be a linear combination of the columns of $A$ given by the coefficients in the $k$th column of $B$.)
