Showing that linear subset is not a subspace of the Vector space $V$ I am given the following
$V = \mathbb R^4$
$W = \{(w,x,y,z)\in \mathbb R^4|w+2x-4y+2 = 0\}$
I have to prove or disprove that $W$ is a subspace of $V$.
Now, my linear algebra is fairly weak as I haven't taken it in almost 4 years but for a subspace to exist I believe that:
1) The $0$ vector must exist under $W$
2) Scalar addition must be closed under $W$
3) Scalar multiplication must be closed under $W$
I don't think the first condition is true because if I were to take the vector, there is no way I can get the zero vector back. Is that correct or am I doing something very wrong?
 A: You're right in what you think, but as how you argue that I am not sure what you mean by "getting the zero vector back"? Back...from where?
Anyway, if we take $\;(w,x,y,z)=(0,0,0,0)\;$ , this vector belongs to $\;W\;$ iff
$$0+2\cdot0-4\cdot0+2=0\iff 2=0$$
and since the last equality is false we get that $\;(0,0,0,0)\notin W\implies W\;$ is not a subspace and you were right.
A: No, it is not a subspace.
It is because you have to verify your first point: "$0\in W$".
But $(0,0,0,0)\notin W$ because 
$$0+2\times 0-4\times 0+2\ne 0.$$
A: Let me give you a more general result: Consider a linear system of equations 
$$\begin{cases}
a_{11}x_1 + a_{12}x_2 + \ldots + a_{1n}x_n &= b_1\\
a_{21}x_1 + a_{22}x_2 + \ldots + a_{2n}x_n &= b_2\\
&\vdots&\\
a_{m1}x_1 + a_{m2}x_2 + \ldots + a_{mn}x_n &= b_m
\end{cases},$$
with $m \leq n$.
Suppose one of the $b_i \neq 0$, then clearly this system of equations does not form a subspace, since in the $i$th equation, we would find that (filling in the zero-vector $(0, 0, \ldots, 0)$ that $0 = b_i \neq 0$, which is not possible.
However, if all of the $b_i$ are zero, then this forms a subspace: clearly the zerovector is a solution to all equations and hence to the system. 
If both $(y_1, \ldots, y_n), (z_1, \ldots, z_n)$ are solutions, then so is $(y_1 + z_1, \ldots y_n + z_n)$ (in order to see this: we fill in this vector and use distributivity of the product over the summation). Analogously, $(\lambda y_1, \ldots, \lambda y_n)$ for some $\lambda \in \mathbb{R}$ will form a solution (note that $(\lambda y_1, \ldots, \lambda y_n) = \lambda (y_1, \ldots, y_n)$). 
A: I feel I should indicate that condition 2 and condition 3 fails as well. The vector $v = (2, 0 , 0 , 0)$ belongs in the subset, but neither $2v$ nor $v+v$ belongs in the subset.
It is thus not a subspace.
