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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 .

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