Let $V=\{(x,y): x,y \in \mathbb R\}$ with the operations:

  • $(x_1 , y_1) + (x_2 , y_2) = (x_1 + x_2 , 0)$
  • $c(x , y) = (cx , 0)$; $c\in \mathbb R$

I know this set doesn't have the zero vector nor the scalar identity, hence $V $ is not a vector space. Another way I was trying to prove this is by arguing that for the vector $(1,1)$ there doesn't exist $(x_1,y_1)$ , $(x_2,y_2)$ such that $(x_1,y_1) + (x_2,y_2) = (1,1)$ but it seems this doesn't violate any vector space property directly, is this a dead end? Any help is appreciated.

  • $\begingroup$ According to the vector space axioms listen here, $V$ breaks the axiom for having zero vector, and scalar identity, as you pointed out. It satisfies everything else. $\endgroup$
    – Guy
    May 29 '17 at 9:05
  • 1
    $\begingroup$ @Guy I just found it curious that some vectors couldn't be expressed as a sum, I thought maybe that could've led to something. Thanks! $\endgroup$
    – taue2pi
    May 29 '17 at 9:09
  • $\begingroup$ Note that for a vector space (or any abelian group) $V$ the addition map $+ : V \times V \to V$ is surjective (because $0 + v = v$ for all $v \in V$). So the fact that the map $+$ defined in your question is not surjective implies immediately that the given set is not a vector space with this $+$ as addition. $\endgroup$ May 29 '17 at 10:03
  • $\begingroup$ @MatthiasKlupsch could you expand on how this immediately proves $V$ is not a vector space? I only have the basic properties of a vector space. $\endgroup$
    – taue2pi
    May 29 '17 at 10:08
  • $\begingroup$ As I said you have $0 + v = v$ for all $v \in V$, so $v$ is the image of $(0,v)$ under $+$. Thus every element of $V$ is in the image of $+ : V \times V \to V$, that is, this map is surjective. However, you proved that your map $+$ is not surjective (since $(1,1)$ is not in the image). This implies that your $+$ cannot be the addition of a vector space structure. $\endgroup$ May 29 '17 at 10:21

It's like finding a needle in a hay stack . Well for vector space operations 1(x,y)=(x,y) |=(x,0) when y|=o so your scalar multiplication fails .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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