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This question already has an answer here:

Suppose $V$ is a vector space over a field $F$. Let $v \in V\setminus \{0\}$ and $\lambda \in F$. Does $\lambda v= 0$ imply $\lambda = 0$?

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marked as duplicate by Namaste, Brian Borchers, Micah, Community Mar 31 '18 at 2:13

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    $\begingroup$ Yes, indeed. Let $v=(\mu_1,\ldots ,\mu_n)$. Then $\lambda v=0$ means $(0,\ldots ,0)=(\lambda\mu_1,\ldots ,\lambda\mu_n)$. Now you have the property in the field of real numbers. Say, $\mu_1\neq 0$, then $\lambda\mu_1=0$ forces $\lambda=0$. $\endgroup$ – Dietrich Burde Mar 30 '18 at 18:23
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    $\begingroup$ @DietrichBurde What if $V$ has infinite dimension? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 30 '18 at 18:25
  • $\begingroup$ @DietrichBurde While this can help OP, I think it's worthwhile to raise up the infinite dimensional case (especially if $V$ is a function space) so that OP can better appreciate the proof and the need of such abstraction. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 30 '18 at 18:31
  • $\begingroup$ Yes, no problem with infinite dimension. $\endgroup$ – Dietrich Burde Mar 30 '18 at 18:47
  • $\begingroup$ Though it does hold in any vector space over a field, it does not hold for instance in a free module over a finite ring (which is in computer science often treated as it were a vector space). $\endgroup$ – leftaroundabout Mar 30 '18 at 23:55
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Of course, if $\lambda \ne 0$ then exist $\lambda ^{-1}$ so $$v = \lambda ^{-1}\cdot (\lambda v) = \lambda ^{-1}(0) =0$$ A contradiction, so $\lambda =0$.

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    $\begingroup$ @downvoters Why downvote? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 30 '18 at 18:25
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    $\begingroup$ I did not downvote, but needing to resort to proof by contradiction feels slightly clumsy to me. It seems like this should have an intuitively "obvious" proof based on the fundamental properties of vector spaces. Scalars only scale vectors up or down, so of course you can't scale a vector into the zero vector unless you multiply it by zero. Yours is a perfectly valid proof, of course, it just feels a bit algebraic (i.e. "just blindly shuffle the symbols around until you get the result you want") for such a basic identity. $\endgroup$ – Kevin Mar 30 '18 at 23:12
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    $\begingroup$ @Kevin, I'm not sure a constructive proof is possible, and even if it is, I suspect it would be far more complicated and hard to follow. I also note that your verbal explanation has the same form as a proof by contradiction - the giveaway being the word "unless". :-) $\endgroup$ – Harry Johnston Mar 31 '18 at 1:43
  • $\begingroup$ If anyone has a constructive proof, please post it! $\endgroup$ – yammatack Mar 31 '18 at 2:03

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