How to determine whether or not a set in $\Bbb R^3$ is a vector space? I have to determine whether or not the following is a vector space:
$$ S=\{ u ∈ \mathbb R^3 |  u_1 ≥ 0\}$$

I have determined the following (please let me know if any of these assumptions are wrong!):


*

*u is a set of vectors in $ \mathbb R^3 $ (i.e. the vectors have three components, $u_1, u_2$, and $u_3$)

*Since $u_1 ≥ 0,u_1$ can equal zero

*Since there are no restrictions placed on $u_2$ or $u_3$ these could both also equal 0, therefore the set contains the zero vector.



Now I'm stuck on how to proceed. I know that in order to show that something is a vector space, I have to prove that it is closed under both addition and scalar multiplication. However, I'm not sure how I would go about doing that since I'm not given specific vectors.

Thanks!
 A: If $u\in S$, you should have, in particular, that $-u\in S$. 
A: You first show that this set is nonempty, which you have done (the zero vector is in here). Now that you have this, you are now allowed to write the following:

Let $x, y \in \{ u \in \mathbb{R}^3 | u_1 \geq 0 \} \ldots$

You now have two objects in this set. Will their sum be in the set? Most certainly, since the sum of two non-negative numbers is itself non-negative (takes care of the first component) while you don't care about the second and third.
But here's the kick - scalar multiplication. Surely if you multiply any vector in this set by a positive scalar you're okay. But what of negative scalars? No. You violate the restriction that the first component be non-negative. Hence this thing is not a vector space.
A: $S$ is not closed by scalar multiplication. For instance $(1,0,0)\in S$, but $$ -1\cdot (1,0,0)=(-1,0,0)\notin S.$$
A: Hint:
if you define addition and scalar product in the usual way, what is the opposite of a vector in the set $S$.?
