Basis for a subspace of $\mathbb R^4$ How do I find the basis for the subspace of $\mathbb R^4$: $(w,x,y,z)$ such that  $x + y + z = 0$.
 A: Since we have one equation and 4 variables, the solution space should be 3 dimensional. A three dimensional solution space can be written as the span of 3 basis vectors.
$z = - x - y$ and we have a fourth variable $w$ that is free. It appears that $x$ and $y$ are free but that $z$ is determined from $x$ and $y$. Best of luck.
$$w(1,0,0,0) + x(0,1,0,-1) + y(0,0,1,-1)$$
So the basis for the solution space is: 
$$(1,0,0,0), (0,1,0,-1),(0,0,1,-1)$$
A: First, we are going to calculate the dimension of the subspace generated by $x+y+z = 0$. 
$$dim(P) = \#variables - rank\ of\ the \ system = 4 - 1 = 3 \Rightarrow \text{The base has 3 elements.
}$$ 
Now you have 3 free variables so you only need to solve the equation with "trivial values"
For example, you can choose as free variables $$(y, z, t) \ y,z,t \in \mathbb{R}$$
Now we take $(1, 0, 0), (0,1,0), (0,0,1)$ as $(y, z, t)$  to solve the equation. (Note that this values are linearly independent).
Case $(1, 0, 0)$: $x = -1 \Rightarrow (-1,1,0,0) \in P$
Case $(0, 1, 0)$: $x = -1\Rightarrow (-1,0,1,0) \in P$
Case $(0, 0, 1)$: $x = 0\Rightarrow (0,0,0,1) \in P$
$\{(-1,1,0,0), (-1,0,1,0), (0,0,0,1) \}$ is a base of the subspace $P$ of  $\mathbb{R}^4$ 
