Is the following in $\mathbb{R}^2$ a vector space?


Addition defined as follows: $u+v=[x_1+x_2+1,y_1+y_2]$

Multiplication defined as $ru=[rx+r-1,ry], r\in\mathbb{R}$

One of the axioms say that there exists an element in V, denote by $0$ such that v + $0$=v

Does this $0$ have to be the $0$ vector $[0,0]$?

Because in this case, $v+\text{ 0 vector}=[x_2+1,y_2]\neq v$

So does this make is not a vector space? Or can $0=[-1, 0]$, in which case it would work?

  • 2
    $\begingroup$ No, $0$ doesn't have to be $(0,0)$. $\endgroup$
    – Kenny Lau
    Sep 10, 2017 at 17:44
  • $\begingroup$ There is something up with your indices. $y_2$ is a component in both $u$ and $v$, and are you certain that your addition works the way it does? Because as it stands, the first component of $u+v$ only cares about $u$ and the second component only about $v$. $\endgroup$
    – Arthur
    Sep 10, 2017 at 17:52
  • $\begingroup$ Sorry it is a mistake typing I will fix $\endgroup$
    – K Split X
    Sep 10, 2017 at 18:09

1 Answer 1


$0$ has to be the vector satisfying $v+0=v$ for all vectors $v$ in your space. If your vector addition is the standard vector addition, then $0$ is $(0,0)$. If your addition is not the standard addition, then $0$ might be something else.

  • $\begingroup$ Yeah I was confused on the wording, thanks $\endgroup$
    – K Split X
    Sep 10, 2017 at 17:51

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