Linear algebra proof that AB = On with A invertible only if B = On

Question

$A,B \in Mn(R)$ so that $AB=0n$ and $A$ is an invertible matrix. Proof that $B=0n$

by definition $A$ is invertible so: $\exists C \in Mn : AC=CA=In$ so $A \ne 0n$ Then $AB=0n$ if $B=0n$ Here I can only say if and not if and only if because the product can be 0 even though both matrices are not 0. I would like to know if I had it all wrong.

I also tryed this way:

$AB = 0_n$ so if I multiply each side by $C$ I get $ CA B = C 0_n $ which is $ I_nB = 0_n$ which can be true only if $B = 0_n$

Answer 1

You are correct that the first argument is flowed for the reason stated, and that the second argument is correct.