# Is that $V = \{(x,y)|x,y \in \mathbb{R}\}$ vector space?

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.

• Yep, your reasoning is correct – freakish Sep 24 '18 at 12:29
• @freakish Thank you. – Dima Sep 24 '18 at 12:40

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 $$(x,y)\neq 1\cdot (x,y)$$.