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I understand it is similar to question ($ax=0$ if and only if $a=0$ or $x=0$) but i haven't yet defined vector devision so is there an alternative way to solve that if $ax=0$,a=0 or x=0_v$ (I can prove alternative). I can prove x=0 but can't see how a=0 is possible?

($a$ is scalar, $v$ is vector)

$ax = 0_v$

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    If $a=0$, done. If $a\neq 0$ then.... – 2017-02-23
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    Those other questions don't use vector division. I'm not sure vector division is allowed unless your structure is something more, like an algebra over a field – 2017-02-23
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    It is identical to the question you linked. Also, division by a vector is undefined. – 2017-02-23
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    It's not *vector* division you need; it's *scalar* division. And since the scalars form a field where multiplicative inverses exist, you have scalar division. – 2017-02-23
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    I can't see how obtaining a=0 is trivial. As the we're proving from left to right? – 2017-02-23
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    A couple things: first remember to surround your mathematical expressions (and variables) with \$ signs when you post so that it renders with Mathjax. You can trigger subscript with underscores, e.g. 0_v becomes $0_v$. – 2017-02-23
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    Second, it's not that $a= 0$ is 'trivial', exactly... The logic is as follows: we want to show $ax = 0$ implies $a = 0$ or $x = 0$. We divide the proof into cases. $a$ can be only one of two things, $0$ or nonzero. In the first case $a = 0$, so we have nothing to prove. In the second case $a$ is not $0$ so we can divide the equation $ax= 0$ by $a$ on both sides to see $x = \frac{0}{a} = 0$. – 2017-02-23
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    I can see the logic and it feels very intuitive, however as we're proving from left to right, in the case a=0 we've gone essentially from right to left which doesn't seem like the proof holds? As in we've just said a=0? – 2017-02-23

2 Answers 2

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If $a=0$, then you have nothing to do. If $a\neq 0$, then you have $a^{-1}$, so: $$ax=0 \Rightarrow a^{-1}ax=a^{-1}\cdot0=0 \Rightarrow x=0$$

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There's no "vector division" in a vector space. The matter is not to show directly that $x \neq 0_{v} \implies a=0$, but rather to show the contraposate $a \neq 0 \implies x=0_{v}$, which is easy since you can use the scalar inverse $a^{-1}$ of $a$.