TITLE: Proof if $a \vec v = 0$ then $a = 0$ or $\vec v = 0$
QUESTION [10 upvotes]: I'm kicking myself over this one, but I just can't seem to make the argument rigorous. From Axler's Linear Algebra Done Right:
for a vector space $V$ with an underlying field $F$:
Take an element $a$ from $F$ and $\vec{v}$ from $V$. $a\vec{v}=\vec{0}\implies a=0 $ or $ \vec{v}=\vec{0}$
After only being able to come up with half of a direct proof, I tried doing this by proving the contrapositive $a\neq 0 \wedge \vec{v} \neq \vec{0} \implies a\vec{v}\neq \vec{0}$
Say $a\vec{v}=\vec{u}$.
Since $a$ is non-zero, we can divide both sides by $a$.
$$\vec{v}=\frac 1 a \vec{u}$$
If $\vec{u}$ were $\vec{0}$ then by
$$\frac 1 a \vec{0}=\frac 1 a (\vec{0}+\vec{0})\implies\frac 1 a \vec{0}=\vec{0}$$
$v$ would be $0$ as well. Since it isn't by assumption, $\frac 1 a \vec{u}$ cannot be zero and so $\vec{u}$ cannot be as well.

Is this fully rigorous? It seems like a very simple question, but I'm not sure about it. Namely, the last step of $\frac 1 a \vec{u}\neq 0 \implies \vec{u}\neq 0$ doesn't seem obvious. I think I need to use the $1\vec{v}=\vec{v}$ axiom, but I'm not sure how. 
Is there a more direct proof? This whole contrapositive business seems a bit clunky for something so simple.

REPLY [2 votes]: We can also show the contrapositive: If $a \neq 0$ and $v \neq 0$ implies $av \neq   0.$
Assume $a\neq 0$ and $v \neq 0.$ Let $V$ has dimension $n,$ and express $v$ in terms of the standard basis vectors: $$v = v_1 e_1 + \ldots + v_n e_n.$$ Then by our assumption that $v \neq 0$ and by the definition of linear independence, at least one of the $v_i$'s is non-zero. Now, multiply $v$ by $a$ to get $$av = (av_1) e_1 + \ldots + (av_n) e_n.$$ But since $a \neq 0,$ at least one of the $av_i$'s will be non-zero, and hence the non-zero linear combination $$(av_1) e_1 + \ldots + (av_n) e_n$$ cannot equal to the $0$ vector. So, $av \neq 0.$