# Show that if some nontrivial linear combination of vectors $\vec{u}$ and $\vec{v}$ is $\vec{0}$, then $\vec{u}$ and $\vec{v}$ are parallel.

I've never been that great at writing proofs, but I'm getting a bit better. I think I have the answer correct, but I don't know if I'm missing anything. My logic seems right but there may be some minute detail that I'm leaving out. Can anybody give any feedback on this? Thanks.

$\vec{0}$ being a nontrivial linear combination of $\vec{u}$ and $\vec{v}$ implies that there exists a non-zero $a$ or $b$ such that $a\vec{u}=-b\vec{v}$. Without loss of generality, assume $a\neq 0$. Then divide by $a$ and the equality holds: $\vec{u}=-\frac{b}{a}\vec{v}$. And since $-\frac{b}{a}\vec{v}$ is a scalar multiple of $\vec{u}$, it remains that $\vec{u}$ and $\vec{v}$ are parallel.

More rigorous proof: \begin{align*} \vec{0}=a\vec{u}+b\vec{v}&\Longrightarrow a\neq 0\vee b\neq 0&&\text{Given}\\ &\Longrightarrow a\vec{u}=-b\vec{v}\\ &\Longrightarrow \vec{u}=-\frac{b}{a}\vec{v}&&\text{WLOG assume $a\neq 0$}\\ &\Longrightarrow \vec{u}\text{ and }\vec{v}\text{ are parallel.} \end{align*}

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By hypothesis, at least one scalar is non zero. Suppose $a\neq 0$. So you can divide by $a$ and then $a\vec{u}+b\vec{v}=\vec{0}$ implies that $\vec{u}=-b/a\vec{v}$. – Sigur May 15 '13 at 0:03
This should be fine: it shows that the relationship between the two vectors is that one is a scalar multiple of the other, that scalar being the proportion between $a$ and $b$, and also indicating why $a$ and $b$ must be non-zero. The one "correction" I would suggest is this also shows that $\overrightarrow{u}$ and $\overrightarrow{v}$ are parallel if $a$ and $b$ have opposite signs and anti-parallel if they have the same sign. – RecklessReckoner May 15 '13 at 0:06

But, we don't have $a\ne 0$ and $b\ne 0$, only $a\ne 0$ or $b\ne 0$. So, one side might be $0$, but then the other is again parallel to it.
But if one is zero, how can it be that $a\vec{u}=-b\vec{v}$? – agent154 May 15 '13 at 0:05
@agent154, perhaps $a=0$ and $\vec{v}=0$, but the other two are nonzero. – vadim123 May 15 '13 at 0:06
The magic phrase is WLOG or Without Loss of Generality. One has either $a\neq 0$ or $b\neq 0$ and then by possibly relabeling $u,v$ and $a,b$ we assume $b\neq 0$. – Sharkos May 15 '13 at 0:21