Proving commutativity of addition for vector spaces I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:

$u+v+u+v = 2(u+v) = 2u + 2v = u+u+v+v = u+(u+v)+v$

Here I just wanna claim that $u+(v+u)+v = u+(u+v)+v$ 

$\Rightarrow -u+u+(v+u)+v+(-v) = -u+u+(u+v)+v+(-v)$ : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why? 

Quick help proof: $-v+v = (-1)v+(1)v = (-1+1)v = 0v = 0 = v-v$

And another: $ 0+v = v+(-v) + v = (1)v + (-1)v + v = (1-1)v + v = v = v+0$

We have $0 + (u+v) + 0 = 0+(v+u)+0 \Rightarrow u+v = v+u$

This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way? 

//not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :)
Thanks! 

EDIT: corrected notation a little.
Normally commutativity is taken as an axiom, but you can deduce it from associativity, distributivity and from the existence of inverses as follows:

$(u + v) - (v + u) = (u + v) - v - u$ (by distributivity) 
$ = u + (v - v) - u$ (by associativity) 
$ = u + 0 - u = (u + 0) - u = u - u = 0$

So $u + v = v + u$
I agree that commutativity is an axiom for Vector Spaces, so what you might need to show is that the ordinary vector spaces you know about satisfy the axioms of Vector Spaces.  Then you can forget about my capital letters.

You know numbers commute.  Use that to show vectors commute.

$\begin{align}u+v&amp;=&amp;(u_1,u_2,...,v_n)+(v_1,v_2,...,v_n)\\
&amp;=&amp;(u_1+v_1,u_2+v_2,...,u_n+v_n)\\
&amp;=&amp;(v_1+u_1,v_2+u_2,...,v_n+u_n)\\
&amp;=&amp;v+u
\end{align}$