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Suppose $C$$A$ = $I$. ($C$ and $A$ may not be square.) Show that the equation Ax = 0 has only the trivial solution. Why can $A$ not have more columns than rows?

I understand having only trivial solutions implies that there's linear independence, and the second part of the question most likely implies there can't be any free variables, but I can't pinpoint exactly the answer they are looking for.

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Assume that $A$ has $n$ columns and $x = (x_1, x_2, \ldots,x_n)$ is a nontrivial solution to the equation $Ax = 0$. Then:

$$ Ax=0 \implies CAx = 0 \implies x = 0 $$

a contradiction, since we assumed $x$ was nontrivial. Thus there are no nontrivial solutions.

If $A$ has more columns than rows, then $A$ must have less than $n$ pivots. Since there is a unique solution only when the number of pivots is equal to $n$ this is not possible.

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