How it follows that $a=0$ If $a \in \mathbb F$ is a scalar and $v \in V$ is a vector and I prove that if $av = 0$ then $a=0$ or $v=0$ I argue like this: Assume $a \neq 0$. Then multiplyaing both sides with its inverse implies $v=0$. Assume $v \neq 0$. Then how to deduce that $a=0$ from $av=0$? Is it possible?
 A: You have proven the statement:
$a \ne 0 \implies v = 0$
The contrapositive of this statement is
$v \ne 0 \implies a = 0$
The two statements are logically equivalent, so you really have already proven what you asked for in your question.
A: If you want to prove (under certain hypotheses) that either condition $A$ or $B$ holds (and possibly both), then one possible approach is to assume that $A$ does not hold, and show that this implies that $B$ holds, in other words establish that $\lnot A\Rightarrow B$. This suffices to prove $A\lor B$, and indeed is logically equivalent to it. So you don't need to separately prove $\lnot B\Rightarrow A$, that is just another logically equivalent way to say $A\lor B$.
This applies to your question with $A\equiv(a=0)$ and $B\equiv(v=0)$. In other words after showing $(a\neq0)\Rightarrow(v=0)$ you should just stop the proof; you're done.
A: You asked about proving that $(av = 0, v \ne 0) \Rightarrow a = 0$ using only linear algebra, which I assume means "simpler than shown above". I hope you find this helpful.
Let
$$v = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} \ne \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = 0, \quad v_i \in \mathbb{F}.$$
Since $v \ne 0$, there exists $k$ such that $v_k \ne 0$. Since
$$av = \begin{bmatrix} av_1 \\ av_2 \\ \vdots \\ av_n \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = 0\!,$$
for such $k$ we have
$$a v_k = 0.$$
Since $\mathbb{F}$ is a field, it has no zero divisors, so you may now conclude that $a = 0$ (the same way you concluded that $a \ne 0$ means that $v = 0$).
