Approach to finding basis of vector space I am given the following vector space:
$$U=\{(x,y,z)\in\mathbb{R^3}:x+y+z=0\}$$
So $\forall u\in U$, we can write  $u=(x,y,-x-y)$. 
I understand I can use trial and error to obtain the basis, but is there a better/more methodical approach?
 A: There is no "the" basis. All you need to do to find "a" basis is find a set maximal of linearly independent vectors that are in $U$.
You've already written a form for everything in $U$, and there are two free variables. So, this subspace has dimension $2$. You can get a basis as follows: first let $x=1,y=0$, second, let $x=0,y=1$. These give you basis $(1,0,-1),(0,1,-1)$.
This process will work because of what Bernard says, but I do not know if you have covered isomorphisms yet. Essentially, since this subspace has dimension $2$, it "looks like" any vector space with dimension $2$. For example, it "looks like" $\mathbb{R}^{2}$. The standard basis of $\mathbb{R}^{2}$ is $(1,0),(0,1)$, so above, when I let $x=1,y=0$ and then $x=0,y=1$, I "sent" the basis of $\mathbb{R}^{2}$ to that of $U$ through what is called an "isomorphism". You don't need to know what any of this means yet, in general, when you get that everything in a subspace over the field of reals can be represented with $n$ free variables, it will always work to pick the first variable $1$ and the rest $0$, the second variable $1$ and the rest $0$, and so on.
A: Hint: Note that $$u = (x,y,-x - y) = x(1,0,-1) + y(0,1,-1).$$
A: Hint:
The linear map
\begin{align}\mathbf R^2&\longrightarrow \mathbf R^3\\
(x,)&\longmapsto (x,y,-x-y)
\end{align}
induces an isomorphism from $\mathbf R^2$ onto $U$.
