Basis for subspace in $\mathbb{R}^4$ How would I start to answer this:

Show that the vectors $(1,0,0,1)$, $(0,1,0,1)$, and $(0,0,1,1)$ form a basis for the subspace $V$ of $\mathbb{R}^4$ which is defined by the equation $x_1+x_2+x_3-x_4=0$.

I understand that to be a basis the vectors must be linearly independent and generate $V$, however, I do not know how to set this up. 
Also, all of the examples I have encountered have the same number of elements and dimensions and this one has $3$ elements and $4$ dimensions. Does this make a difference?
 A: Since
$$
\{ (x_{1},x_{2},x_{3},x_{4}) \in \mathbb{R}^{4} \mid x_{1}+x_{2}+x_{3} - x_{4} = 0\} = \{ (s,t,u,s+t+u) \mid s,t,u \in \mathbb{R} \},
$$
since
$
(s,t,u,s+t+u) = s(1,0,0,1) + t(0,1,0,1) + u(0,0,1,1)
$
for all $s,t,u \in \mathbb{R}$,
and since $(1,0,0,1), (0,1,0,1), (0,0,1,1)$ are linearly independent in $\mathbb{R}^{4}$,
we are done!
A: Hint: Consider the vector $v = (x_1,x_2,x_3,x_4)$ and the matrix $A$ whose columns are the vectors given. Then $Av = 0$ implies what about $x_1,x_2,x_3,$ and $x_4$?
A: Made an answer since I ran out of comment space:
Let $W$ be the subspace spanned by these vectors.
Well, first you should determine that these are linearly independent (which is simple). Then, verify that each of these vectors satisfy the equation, so $W$ will be a subspace of $V$. From here, you can verify that $V$ has $3$ dimensions (your previous proof showed that $\dim V\geq 3$, so it suffices to show $V\not=\mathbb{R}^4$). Then the result will follow since $W\subset V$ is a subspace with the same dimension as the whole space, so $W=V$.
