Part of proving a set is a subspace of $\mathbb{R}^3$

Question

Say I have the set $S=\{\bar{x}\in\mathbb{R}^3:x_1+x_2=0\}$

One of the things I have to prove is that every two vectors in $S$, their sum is also in $S$

Does the following prove that?:

Let $\bar{a},\bar{b} \in S$

Assume $a_1+a_2=0$ and that $b_1+b_2=0$

$\bar{a} + \bar{b}=(a_1+b_1,a_2+b_2,a_3+b_3)$

I want to prove that $(a_1+b_1) + (a_2+b_2)=0$ and so

$(a_1+b_1) + (a_2+b_2)=(a_1+a_2)+(b_1+b_2)=0+0=0$

Does this prove that the sum of any two vectors in $S$ is also in $S$? Or is there any information that I'm missing?...or is this just outright wrong..

Answer 1

You're proof is correct. wording could be better, I'd write it like this:

suppose $\overline{a},\overline{b} \in S$. This implies that $a_1 + a_2 = b_1 + b_2 =0$.

Now, $(\overline{a} + \overline{b})_1 + (\overline{a} + \overline{b})_2$ equals $(a_1 +b_1) + (a_2 + b_2)$, by the definition of vector addition, and this we can rewrite as $(a_1 + a_2) +(b_1+ b_2) = 0 + 0 = 0$. So $\overline{a},\overline{b} \in S$ by the definition of $S$.