Aside from field and vector space definitions, the only two other propositions I should limit myself to using are $0v =0$ for all $v\in V$ and $a0 =0$ for all $a \in \mathbb{F}$.

I feel like I'm missing something obvious. One implication was straightforward: $$av=a0 \implies a^{-1}av = a^{-1}a0 \implies v =0.$$ As for the other (showing $a=0$ is a possibility) I haven't been able to make much progress. At most I showed that $$(\underbrace{a+a+\cdots +a}_\text{$n$-many})v=0v$$ for positive integers $n \geq 1$, but since $V$ lacks multaplicative inverses it feels like I hit a dead end and can't isolate $a$ somehow.

Could someone at most provide a nudge in the right direction?


1 Answer 1


Let $av = 0$ where $a \ne 0$ and $v \ne 0$.

Then, in fact we have $a^{-1} av = a^{-1} 0 = 0$, but also $a^{-1} av = (a^{-1}a)v = 1v = v \ne 0$, contradiction.

  • $\begingroup$ @AndrewTawfeek the complement of "$a = 0$ or $v =0$" is "$a \ne 0$ and $v \ne 0$". Other than that, I've no idea what you are talking about. $\endgroup$
    – Kenny Lau
    Jan 7, 2018 at 11:42
  • $\begingroup$ That's exactly what I meant :) I thought there'd be a seperate method for showing $a=0$ is a possibility and hadn't thought to just take your approach and prove both cases together in one go $\endgroup$ Jan 7, 2018 at 11:49

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