Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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*}

share|cite|improve this question
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
up vote 3 down vote accepted

Almost perfect.

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.

share|cite|improve this answer
But if one is zero, how can it be that $a\vec{u}=-b\vec{v}$? – agent154 May 15 '13 at 0:05
Although in fairness you could dismiss the case where one of them is zero since then one concludes one of the vectors is zero and it's trivial. – Sharkos 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
@agent154, please, see my comment above. – Sigur May 15 '13 at 0:18
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.