For what $\alpha$ is $\{(x,y,z) \in \mathbb{R}^3| x^2+y^2=z^2, z-y+\alpha=0\}$ a subspace For each $\alpha \in \mathbb{R}$, let $S_{\alpha}=\{(x,y,z) \in \mathbb{R}^3| x^2+y^2=z^2, z-y+\alpha=0\}$. For which $\alpha$ (if any) 
is $S_{\alpha}$ a subspace of $\mathbb{R}^3$?
We know that $S_{\alpha}$ if a subspace if:
1) $\vec{0} \in S_{\alpha}$ 
2) if $\vec{a}$ and $\vec{b}$ are in $S_{\alpha}$, then $\vec{a}+\vec{b} \in S_{\alpha}$
3) For any $c\in\mathbb{R}$ and $\vec{a} \in S_{\alpha}$, $c\vec{a} \in S_{\alpha}$.
I am not sure if I understand what this statement is saying, but would it be correct to say that for $\alpha=0$ $S_{\alpha}$ is a subspace because then $z=y \rightarrow z^2=y^2 \rightarrow x^2=0$ and then only $(0,0,0) \in S_{\alpha}$, and this satisfies the three axioms above? 
How could I find a more general solution?
 A: Since $0$ must be inside $S_\alpha$, we need $0-0+\alpha=0$, that is $\alpha =0$.
Hence $S_\alpha$ is a subspace implies that $\alpha = 0$.
If $\alpha = 0$, then $y = z$ and $x^2+y^2=z^2$ which reduces to $x=0$.
That is $$S_0 = \{ (x,y,z) \in \mathbb{R}^3|x=0, y=z\}$$
I will leave the verification of closure of addition and multiplication as exercise.
A: Let's assume that $S_{\alpha}$ is a subspace, and then check what restrictions this assumption imposes on $\alpha.$
We can check the properties one at a time:


*

*We need the origin to be in $S_{\alpha}$. That means that we need


*

*$0^2 + 0^2 = 0^2$ (good so far)

*$0-0+\alpha=0$.



So already we have that $S_{\alpha}$ can't be a subspace unless $\alpha=0$. We now need to check that the rest of the properties are satisfied for $S_{\alpha}$. First, as you point out in your question, if $\alpha=0$ then $y=z$ and $x^2=0$. Therefore
$$S_{\alpha} = \{(0,y,y)\,\vert\,y\in\mathbb{R}\}.$$
We check.


*If $(0,y_1,y_1)$ and $(0,y_2,y_2)$ are two arbitrary elements of $S_{\alpha}$, then
$$(0,y_1,y_1)+(0,y_2,y_2) = (0,y_1+y_2, y_1+y_2)\in S_{\alpha}$$
so this property checks out;

*If $(0,y,y)$ is an element of $S_{\alpha}$ then
$$c(0,y,y) = (0,cy,cy)\in S_{\alpha}.$$
Therefore we have shown that $\alpha=0$ is the only possible candidate, and that when $\alpha=0$, $S_{\alpha}$ is indeed a subspace.
