Vector space and vector subspace We know that a vector space V is a set of vectors on which two operations are defined, sum, which satisfies the following properties:
\begin{equation}
u+v=v+u
\end{equation}
\begin{equation}
u+(v+w)=(v+u)+w
\end{equation}
\begin{equation}
u+0=0+u=0
\end{equation}
\begin{equation}
u+(-u)=(-u)+u=0
\end{equation}
(for each element u,v,w in V, -u opposite element, 0 neutral element) and product by scalar, which satisfies the following properties:
\begin{equation}
a(u+v)=av+au
\end{equation}
\begin{equation}
(a+b)u=au+bu
\end{equation}
\begin{equation}
(ab)u=a(bu)
\end{equation}
\begin{equation}
1 u=u
\end{equation}
(for each element u,v in V; a,b real scalar; 1 neutral element of product). Not all sets of vectors are vector spaces. For example, space $R^2$ with the following operations:
\begin{equation}
(x,y)+(x',y')=(x+x',y+y')
\end{equation}
\begin{equation}
a(x,y)=(ax,0)
\end{equation}
is not a vector space because $1(x,y)=(x,0)\neq (x,y)$.
A vector subspace, instead, is a set S of vectors included in the vector space V, which in turn is a vector space. In other words, S is a vector subspace if
\begin{equation}
0\in S
\end{equation}
\begin{equation}
u+v\in S, \ \ \ u,v\in S
\end{equation}
\begin{equation}
av\in S, \ \ \ \ v\in S, a \in R
\end{equation}
For example:
\begin{equation}
S=\left\{(x,y)\in R^2 | y=2x \right\}
\end{equation}
is a vector subspace. In fact, the generic element of S is thus formed: $(x,2x)$. If $x=0$, we have the neutral element $(0,0)\in S$. My doubt is involved in product by scalar, because: $av=a(x,2x)$ and the next step should be, of course, $(ax,2ax)$. Namely: $a(x,2x)=(ax,2ax)$. But this step is legitimate? We are trying to prove that S is a subspace, then we do not know if $a(x,2x)=(ax,2ax)$. For example, it may happen that $a(x,2x)=(ax,0)$ (see example above on vector space). Perhaps, after considering: 1) S included in V, 2) V vector space; we are assuming that: V vector space with the following operations:
\begin{equation}
(x,y)+(x',y')=(x+x',y+y')
\end{equation}
\begin{equation}
a(x,y)=(ax,ay)
\end{equation}?
I hope to be able to explain. Thank you very much.
 A: You are correct; $S\subset\mathbb{R}^2$, as given, may not be a subspace of $\mathbb{R}^2$ if the action of $\mathbb{R}$ is not compatible.  And similarly with addition—how do we know that $(x,2x) + (x',2x') = (x+x',2x+2x')$, unless we have specified the addition law on $S$?
The answer is that these kinds of questions and statements usually take for granted that $S$ inherits certain properties from its parent, $\mathbb{R}^2$.  So if $a\in\mathbb{R}$, $v\in S$, we calculate $av$ according to the multiplication rule on $\mathbb{R}^2$, namely $av=a(v_0,v_1)=(av_0,av_1)$.  And when we add two vectors, we add them as elements on $\mathbb{R}^2$.
So yes, $S$ is only given as a set, and a set is not a vector space without some additional structure, and it is only a subspace when that structure is compatible with the structure on $\mathbb{R}^2$.  It is by convention, however, that we always assume that $S$ is granted exactly the structure that makes it compatible.  Your conclusion at the very end is exactly the solution; we do assume these things, and this is a good point to recognize.
