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$A,B \in Mn(R)$ 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 $B = 0_n$

user144037
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    Correct.${}{}{}$ –  Apr 18 '14 at 14:01
  • Your proof looks correct, but I'd state it a bit more concisely as $AB = 0_n$ implies $A^{-1}AB = I_nB = B = 0_n$. That way it also clear that it suffices for $A$ to have a left-inverse for this to work, meaning $A$ must be injective but not necessarily surjective. For square matrices, there's no material difference, but if $A$ has more rows than columns, i.e. maps to a space with a higher dimension, then it'll never be surjective yet may very well be injective. – fgp Apr 18 '14 at 14:12

1 Answers1

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

Belgi
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