Proving that the following ensemble is not vector space if $\mathbb{K}=\mathbb{C}$ I would like to check if my method to show that $F$ is not a vector space with $\mathbb{K}=\mathbb{C}$ is true.
$$ F = \{(x,y,z) / x^2+y^2+2z^2-2xy-2yz+2xz=0 \}$$
$x^2+y^2+2z^2-2xy-2yz+2xz=0 \Leftrightarrow (x+z-y)^2+z^2=0 \Leftrightarrow (x+z-y)^2=-z^2$
This equation admit solutions only if $z=i \alpha$.
We have :
$(x+i\alpha-y)^2=\alpha^2 \Leftrightarrow x=\pm \alpha +y -i \alpha$
Finally my space of solutions is :
$F=Span((\pm 1-i;0;i),(1;1;0))$
And because of the $\pm$, I can say that I have a union of two vectorial spaces :
$F_1=Span((+1-i;0;i),(1;1;0))$ and $F_2=Span((-1-i;0;i),(1;1;0))$ 
Thus $F=F_1 \cup F_2$
And as $F_1$ and $F_2$ are disjoint (the first is not included and the second and same for the second is not in the first), the global ensemble is not a vector space.
Am I right ?
 A: Based on this:
\begin{align*}
(x,y,z) \in F \iff (x+z-y)^2+z^2 &= 0 \\\iff (x+z-y+iz)(x+z-y-iz) &= 0
\end{align*}
I agree with that $F$ is the union of two subspaces $F_1$ and $F_2$:
\begin{align*}
    F_1 &= \left\{(x,y,z)\mid x - y + (1+i) z = 0\right\}\\
    F_2 &= \left\{(x,y,z)\mid x - y + (1-i) z = 0\right\}
\end{align*}
You're right that neither $F_1$ nor $F_2$ contains the other (I don't think it's conventional to use disjoint for this condition; I think of disjoint as meaning “having empty intersection”, and subspaces always share at least the zero vector).  
I'm not sure exactly how best to get from here to the conclusion that $F$ is not a subspace.  My preference would be to show it violates one of the subspace conditions.  For instance, let $v_1 = (-1-i, 0, 1)$ and $v_2 = (-1+i,0,1)$.  Then $v_1 \in F_1$ and $v_2 \in F_2$, so they are both in $F$.  Their sum is
$$
    v_1 + v_2 = (-2,0,2)
$$
This point does not satisfy the equation defining $F$.  Since $F$ is not closed under addition, $F$ cannot be a subspace.
