Basis of a subspace in $R^4$ 
$U$ is a subspace $ U= \{(x_1,x_2,x_3,x_4) \in \mathbb{R} \mid x_1+x_2+x_3+x_4=0\} $
  Find a Basis of $U$.

I know a basis of the whole $\mathbb{R}^4$ needs at least 4 independent vectors, since it's only a subspace, I don't know how many I need. Are 2 vectors, fullfilling the equation, sufficient?
 A: There is only one condition, which as a rule of thumb removes only one from the dimension.
Formally:
$U$ is the kernel of the surjective linear map $\mathbb R^4\to \mathbb R$, $(x_1,x_2,x_3,x_4)\mapsto x_1+x_2+x_3+x_4$. Therefore $\dim U + \dim \mathbb R=\dim \mathbb R^4$. In other words: you need three independant vectors.
One vector $\in U$ of particularly simple kind is $(1,-1,0,0)$. Can you see a few others like that? Are there enough linearly independant among them? 
A: 2 vectors aren't enough. It is a 3D-plane in a 4D space. Its basis need to have 3, 4-dimensional vectors.
3 vectors: (1,0,0,-1), (0,1,0,-1), (0,0,1,-1).
A: $U=\{(x_1,x_2,x_3,x_4)\in \mathbb {R^4}|x_1+x_2+x_3+x_4=0\}$
$=\{(x_1,x_2,x_3,x_4)\in \mathbb {R^4}|x_4=-x_1-x_2-x_3\}$
$=\{(x_1,x_2,x_3,-x_1-x_2-x_3)\in \mathbb {R^4}|x_1,x_2,x_3\in\mathbb R\}$
$=\{x_1(1,0,0,0,-1)+x_2(0,1,0,-1)+x_3(0,0,1,-1)|x_1,x_2,x_3\in\mathbb R\}$
$=span\{(1,0,0,0,-1),(0,1,0,-1),(0,0,1,-1)\}$
We also see that $\{(1,0,0,0,-1),(0,1,0,-1),(0,0,1,-1)\}$ is linearly independent since
$c_1(1,0,0,0,-1)+c_2(0,1,0,-1)+c_3(0,0,1,-1)=(0,0,0,0)$ (for some scalars $c_1,c_2,c_3$)
$\Rightarrow (c_1,c_2,c_3,-c_1-c_2-c_3)=(0,0,0,0)$
$\Rightarrow c_1=0,c_2=0,c_3=0$
So $dimU=3$
