Answer
This set does not have a vector space structure, because sums of vectors in the set do not remain in the set.
Work Step by Step
The first axiom doesn't hold. Let $u=(1,0)$ and let $v=(0,-1)$. $u$ and $v$ are in the aforementioned set, but $u+v = (1,-1)$ is not since $(-1)(1)=-1 < 0$. Therefore, the set doesn't have a vector space structure.
