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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!

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    $\begingroup$ What about $A=\pmatrix{1&0\\1&0}$, $B=\pmatrix{1\\1}=C$? $\endgroup$ – Jens Schwaiger Jun 2 '20 at 4:01
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    $\begingroup$ @PrudiiArca: Not anymore! 😂 $\endgroup$ – Jens Schwaiger Jun 2 '20 at 4:05
  • $\begingroup$ @JensSchwaiger whoops! The problem was supposed to require $C$ to have $n$ linearly independent columns. I have edited my post. Nice catch $\endgroup$ – Math2718 Jun 2 '20 at 4:08
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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.

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If you know about the rank of a matrix you can use that $C$ has rank $\geq n$. Hence $$n \leq \operatorname{rk}C = \operatorname{rk} AB \leq \min\{\operatorname{rk}A,\operatorname{rk}B\}$$ shows that $A$ has rank $n$ and thus is invertible.

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  • $\begingroup$ $ \operatorname{rk} AB =\leq\min\{\operatorname{rk}A,\operatorname{rk}B\}$? $\endgroup$ – Jens Schwaiger Jun 2 '20 at 4:19
  • $\begingroup$ @JensSchwaiger Woopsie. Thanks! $\endgroup$ – PrudiiArca Jun 2 '20 at 4:24
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Since, $C$ has $n$ linearly independent columns and has $m$ columns in total, $m\ge n$. Now, let $B=[b_1,b_2,....,b_m]$(columnwise). Then, the set $S=\{Ab_1,Ab_2,....Ab_m\}$ (distinct ones among them) has $n$ linearly independent vectors. But, since $S\subseteq \text{image}(A)$, we conclude that $\text{image}(A)$ has $n$ linearly independent vectors. Thus, $\text{rank}(A)=n$. Hence, $A$ also has $n$ linearly independent vectors. Thus, $A$ is invertible.

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