# Does the zero product property hold in vector spaces? [duplicate]

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 $\lambda = 0$?

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

• 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 $\lambda=0$. – Dietrich Burde Mar 30 '18 at 18:23
• @DietrichBurde What if $V$ has infinite dimension? – GNUSupporter 8964民主女神 地下教會 Mar 30 '18 at 18:25
• @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. – GNUSupporter 8964民主女神 地下教會 Mar 30 '18 at 18:31
• Yes, no problem with infinite dimension. – Dietrich Burde Mar 30 '18 at 18:47
• 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). – leftaroundabout Mar 30 '18 at 23:55

Of course, if $\lambda \ne 0$ then exist $\lambda ^{-1}$ so $$v = \lambda ^{-1}\cdot (\lambda v) = \lambda ^{-1}(0) =0$$ A contradiction, so $\lambda =0$.