If $AB=C$ and $C_{n \times m}$ has $n$ linearly independent columns, then $A$ is invertible

Question

From Hubbard & Hubbard:

Let $A$ be an $n \times n$ matrix, let $B$ be an $n \times m$ matrix, and let $C$ be an $n \times m$ matrix. The matrices satisfy the relation $AB=C$. $C$ has $n$ linearly independent columns. Prove that $A$ is invertible.

Here is my solution for the special case when $m=n$: Since $C$ is square and its columns are linearly independent, $C$ is invertible. So we can write $ABC^{-1}=I$. So $A$ is invertible.

Unfortunately, this method clearly does not generalize at all to the case when $m \neq n$! How do you solve the problem in general? Any hints or solutions would be appreciated!

Answer 1

Just pick $n$ linearly independent columns in $C$ and delete the other columns to get $C'$. Similarly delete the corresponding columns of $B$ to get $B'$. Then $AB'=C'$ and you have already shown this implies $A$ invertible.