Answer
The set is a vector space.
Work Step by Step
The vector $(0,0)$ is in the set of of the vectors, so the first axiom holds.
Let $u, v$ be in the aforementioned set. So, $u = (a,a), v=(b,b)$ for some $a,b \in \mathbb{R}$. Therefore, $u+v = (a+b, a+b)$. $a+b\in \mathbb{R}$ and the first and second components match, so $u+v$ in the set.
Finally, let $u$ be in the aforementioned set and let $k \in \mathbb{R}$. So, $u=(a,a)$ for some $a \in \mathbb{R}$. Therefore, $ku = (ka,ka)$. Since the first and second component match and $ka \in \mathbb{R}$, $ku$ is in the aforementioned set.
Since all three axioms hold, the set has a vector space structure.
