Why is the following set not a vector space? I have the following set and I want to know whether it's a vector space or not:

$W = \{(x, y, z) ∈ \Bbb R^3 : (x + y)(2y − z) = 0\}$

Now, I understand that if I have a set W and it's a vector space then $0\in W$ and it should have vector addition and scalar multiplication for some $w_1, w_2 \in W$. 
But how to I use the definition to check if the vector space is not linear in that case? 
I know that $0\in W$. What about the other checks? How is it done?
Thanks.
 A: It is not closed under addition. For example, $(0,0,z)$, where $z$ is an arbitrary number, is in $W$. 
Now find another vector in $W$, for example $(1,1,2)$. Add these two together. You get $(1,1,z+2)$, again, where $z$ is arbitrary. This is clearly not in $W$, except for $z=0$.
A: $\newcommand{\Reals}{\mathbf{R}}$In case some theory helps clarify your strategy: Your set $W$ is the union of two planes through the origin:
$$
W_{1} = \{(x, y, z) \in \Reals^{3} : x + y = 0\},\qquad
W_{2} = \{(x, y, z) \in \Reals^{3} : 2y - z = 0\},
$$
since $(x + y)(2y - z) = 0$ if and only if $x + y = 0$ or $2y - z = 0$.
In an arbitrary vector space, a union of subspaces is non-empty and closed under scalar multiplication. In this example, you should therefore focus immediately on closure under addition.
A union of two subspaces, neither contained in the other, is never closed under addition: If $w_{1} \in W_{1} \setminus W_{2}$ and $w_{2} \in W_{2} \setminus W_{1}$, then $w_{1} + w_{2} \not\in W_{1} \cup W_{2}$. This is an easy, worthwhile exercise. (Hint: Assume contrapositively that $w_{1}$ and $w_{1} + w_{2}$ are elements of $W$. What can you say about $w_{2}$?)
To check that your $W$ is not closed under addition, just pick a vector $w_{1}$ in $W_{1}$ that is not an element of $W_{2}$, and similarly pick $w_{2}$ in $W_{2}$ not not in $W_{1}$, and verify $w_{1} + w_{2} \not\in W$.
