Prove S is a subspace of $R^3$ Let S={(a,b,c)$\in$$R^3$|2a+b-3c=0}. Prove S is a subspace of $R^3$
I just asked another question like this earlier and this is just to confirm im on the right track when approaching problems like these. After starting this problem, i am finding myself confused when it comes to making sure its closed under addition(so far). 
so far i think i have proved the set is not empty with: 
let 0=(0,0,0),
2(0)+0-3(c)=0 therefore, set is not empty. 
let u=($a_1,b_1,c_1$) and w=($a_2,b_2,c_2$). 
u+w= $(a_1+a_2),(b_1+b_2),(c_1+c_2) $
At this point, where does the definition "2a+b-3c=0" come into play? How do i proceed from here to show closure under addition?
 A: Just to be pedantic, you are trying to show that $S$ is a linear subspace (a.k.a. vector subspace) of $\Bbb R^3$. The context is important here because, for example, any subset of $\Bbb R^3$ is a topological subspace.
There are two conditions to be satisfied in order to be a vector subspace:
$(1)$ we need $v+w\in S$ for all $v,w\in S$
$(2)$ we $rv\in S$ for any $r\in\Bbb R$ and $v\in S$.
We set $S=\{(a,b,c)\in \Bbb R^3|2a+b-3c=0\}$. Two elements of $S$ can be written $v=(a,b,c)$ and $w=(d,e,f)$. Then $v+w=(a+d,b+e,c+f)$. To check if this is in $S$, we need $2(a+d)+b+d-3(c+f)=0$. Just notice
$$2(a+d)+b+e-3(c+f)=(2a+b-3c)+(2d+e-3f)=0+0=0,$$
and thus $v+w\in S$.
Next, take $r\in \Bbb R$ and any $v=(a,b,c)\in S$. Then $rv=(ra,rb,rc)$. Now
$$2ra+rb-3rc=r(2a+b-3c)=0,$$
which means $rv\in S$.
A: $$2a_1+b_1-3c_1=0$$ $$2a_2+b_2-3c_2=0$$
Thus $2(a_1+a_2)+(b_1+b_2)-3(c_1+c_2)=0+0=0$
So $u+v \in S$
Similarly prove that $au \in S ,\forall a \in \Bbb{R}$
A: If we suppose $u=(u_1,u_2,u_3)$ and $w=(w_1w_2,w_3)$ are arbitrary vectors in $S$, so what we are saying is that
$2u_1+u_2-3u_3=0$ and that $2w_1+w_2-3w_3=0$. Now, we see that the vector
$$u+w=(u_1+w_1,u_2+w_2,u_3+w_3)$$
satisfy 
$$2(u_1+w_1)+(u_2+w_2)-3(u_3+w_3)$$
$$\begin{align}  
&= (2u_1+u_2-3u_3)+(2w_1+w_2-3w_3) \\
&= 0+0 \\
&= 0
\end{align}$$
that is, the vector $u+w$ also belongs in $S$. Do the same thing with the vector $cu$, where $c$ is an arbitrary scalar.
