Chapter 6 - Vector Spaces - 6.1 Vector Spaces and Subspaces - Exercises for 6.1 - Page 441: 2

Not a vector space, third axiom does not hold.

This set does not have a vector space structure. The third axiom doesn't hold. Let $k=-1$ and let $u = (1,1)$. Since $u$ is in the aforementioned set but $ku = (-1,-1)$ is not in the set; the third axiom doesn't hold so the set doesn't have a vector space structure.