Shortcut to proving that the set forms a vector space To prove that a set forms a vector space, we of course must prove that it satisfies the vector space axioms.
However, I vaguely remember that when I used to do these, there was a specific expression that I tested and if it worked, it ticked off a good number of these axioms automatically. But alas, I no longer remember.
Do you recall any such technique? 
 A: I believe you are talking about proving that a set $S \subset V$ is a vector space when $V$ is a vector space, i.e. showing that $S$ is a subspace of $V$.
In that case, you need to show that $\vec{0} \in S$ and that $S$ is closed under addition and constant multiplication. In other words, if $\vec{x}, \vec{y} \in S$ and $a,b \in \mathbb{R}$, then $a\vec{x} + b\vec{y} \in S$.
The reason why these two conditions are sufficient is because the other vector space axioms carry over from the fact that $V$ itself is a vector space.
UPDATE
As pointed out in the comments by @celtschk, setting $a = b = 0$ forces to conclude that $0 \in S$ provided the second condition holds. So there really is only one condition, namely that if $a,b \in \mathbb{R}$ and $\vec{x}, \vec{y} \in S$, then $a\vec{x} + b \vec{y} \in S$.
A: A brief way to define a group is a non-empty set $G$ with an associative binary operation such that for all $x,y\in G$ there are unique $z_1, z_2\in G$ with $xz_1=z_2x=y.$
Vector addition in a vector space forms a commutative group. So if $U$ is a non-empty subset of a vector-space $V,$ then U is a vector subspace iff (1) $ au\in U$ for all scalar $a$ and all $u\in U,$ and (2) for all $x,y\in U$ there exists $z\in U$ with $x+z=y$ (equivalently $y-x\in U$ for all $x,y\in U.$)
