is this subset a subspace - redux OK, I have been bothering people here with this for days and with luck I finally have this. People have helped a lot here so far. (Doing these examples is I hope helping me learn the proofs, but I want to know that I am doing this right). 
Let W be a subset of vector space V. Is it s subspace as well?
W = {($a_1$, $a_2$, $a_3$) ∈ $ℝ^3$ : $2a_1 - 7a_2 + a_3=0$}
So, to check if this is a subspace I need to satisfy the following:


*

*That 0 is in the set. Plugging (0,0,0) into the equation $2a_1 - 7a_2 + a_3=0$ yields 0=0 so yes, it is.

*That it is closed under addition.
Let ($b_1, b_2, b_3$) be an arbitrary vector in W.
For this to be closed under addition ($b_1, b_2, b_3$)+($a_1, a_2, a_3$) ∈ W.
$2(a_1+b_1) - 7(a_2+b_2) + (a_3+b_3) = 0$
can also be written as $(a_3+b_3) = -2(a_1+b_1) + 7(a_2+b_2)$
There are real-valued solutions to this, whenever $b_i = -a_i$ is one, so the answer is yes, it is closed under addition. 


*

*Is it closed under multiplication?


Any arbitrary λ($2a_1 - 7a_2 + a_3)=0 =(λ)0$
So since that's still part of the set, it is closed under multiplication.
So, did I do this one correctly? God I hope so. 
 A: Excellent work. It will get easier! Trust me. 
Very nicely done. You "covered all the bases", a bit awkwardly but you got the job done!
I'd just add: "Therefore, (since ${\bf 0} \in W$, and $W$ is closed under vector addition and scalar multiplication,) $\;W\,$ is a subspace of $\,V$.

This is how I'd approach the "closed under addition" component of the proof.
Let $w_1 = (a_1, a_2, a_3), w_2 = (b_1, b_2, b_3) \in W$. 
So $2a_1 - 7 a_2 + a_3 = 0, \text{ and}\; 2b_1 - 7b_2 + b_3 = 0.$
Now, it follows that $w_1 + w_2 = (a_1 + b_1, a_2 + b_2, a_3 + b_3).$ 
And since we have $$2(a_1+b_1) - 7(a_2+b_2) + (a_3+b_3)$$ 
$$ = 2a_1 + 2b_1 -7a_2 -7b_2 + a_3 + b_3 $$ 
$$ = (2a_1-7a_2+a_3)+(2b_1-7b_2+b_3) $$
$$ ={\bf 0 + 0} = {\bf 0}\in W$$
...$W$ is closed under vector addition.
A: I'm not sure I understand your proof that it is closed under addition (specifically, the $b_i=-a_i$ part seems like a special case, and not a proof). I recommend instead that you note: $$\begin{align}2(a_1+b_1)-7(a_2+b_2)+(a_3+b_3) &= 2a_1+2b_1-7a_2-7b_2+a_3+b_3\\ &= (2a_1-7a_2+a_3)+(2b_1-7b_2+b_3)\\ &= \textbf{0}+\textbf{0}\\ &= \textbf{0}.\end{align}$$
