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!

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 $$B$$.)
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 $$M A = I.$$