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Let $V = \{(x,y)|x,y \in \mathbb{R}\}$ and let $+$ be defined pointwise

a) If we define the scalar multiplication as $r(x,y) = (rx,y)$ for all $r \in \mathbb{R} $ and $(x, y) \in V$ , determine if $V$ with the above addition and this scalar multiplication is a vector space on $\mathbb{R} $ (justify your answer).

b) If we define the scalar multiplication as $r(x,y) = (rx,0)$ for all $r \in \mathbb{R} $ and $(x, y)\in V$ , determine if $V$ with the above addition and this scalar multiplication is a vector space on $\mathbb{R}$ (justify your answer).

My attempt:

a) not a vector space, because:

$(\alpha+\beta) u= (\alpha+\beta) (x,y)=((\alpha+\beta)x,y)=(\alpha x,{{1}\over{2}} y)+(\beta x,{{1}\over{2}} y)$

And that is not equal to $\alpha u+\beta u=(\alpha x,y)+(\beta x, y)= ((\alpha+\beta)x,2y) $

That’s mean, $(\alpha+\beta) u $ is equal to $\alpha u+\beta u$, $\iff$ $y=2y$ $\iff$ $y=0$

b) All of the first $7$ conditions is fine, but for $8-th$ condition $1.u=u$, I got that:

$1.u=1.(x,y)=(1.x,0)=(x,0)$ not equal to $(x,y)=u$

So this not a Vector space.

Is that true? Thanks.

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    $\begingroup$ Yep, your reasoning is correct $\endgroup$ – freakish Sep 24 '18 at 12:29
  • $\begingroup$ @freakish Thank you. $\endgroup$ – Dima Sep 24 '18 at 12:40
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Both proofs are OK (you'd get all points if I was grading a test and you wrote them down as a solution), but they could be shortened and made more clear (I'd maybe deduct a tiny point if this was homework and I was grading it).


for (a), there is no real need for the $\frac12$. You said that $(\alpha+\beta)(x,y) = ((\alpha+\beta)x, y)$, while $\alpha(x,y)+\beta(x,y) = ((\alpha+\beta), 2y)$, and the fact that these two expressions are not always equal is enough.

To make it even simpler, you really only need one particular setting of $\alpha, beta, x,y$ for which the equality $$(\alpha+\beta)(x,y)=\alpha(x,y)+\beta(x,y)$$ is anot true, so you could also just say $\alpha=beta=x=0, y=1$ and see that it is not true for this particular setting (and many others, of course).


For (b), technically, you need to write down that $(x,0)\neq (x,y)$ if $y\neq 0$, but again, you could just say that you take $x=0,y=1$ and note that for this particular $(x,y)$, we have $(x,y)\neq 1\cdot (x,y)$.

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    $\begingroup$ Thank you so much,your answer was so helpful. $\endgroup$ – Dima Sep 24 '18 at 12:41

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