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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!

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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 $B$.)

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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 $$ M A = I. $$

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  • $\begingroup$ I think this argument could use some elaboration about how to choose a suitable left inverse. Left inverses need not be linear maps! $\endgroup$ – user804886 Jul 14 at 7:51
  • $\begingroup$ @user804886 I think I fixed it. Think it's ok now? $\endgroup$ – littleO Jul 14 at 8:26
  • $\begingroup$ Agreed! I'd give you a +1 if I could. $\endgroup$ – user804886 Jul 14 at 8:27

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