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.