Determine whether the given set forms a basis for the indicated subspace Determine whether the given set forms a basis for the indicated subspace:
$\{(1,-1,0),(0,1,-1)\}$ for the subspace of $\mathbb{R}^3$ consisting of all $(x,y,z)$ such that $x+y+z=0$.
I know linear independence and span is involved but I'd like to see how you properly show it if possible.
I feel that because we are dealing with a subspace of $\mathbb{R}^3$, three vectors are needed to span it and because there are only two vectors then it can't span the subspace and thus isn't a basis for the subspace.

First, you would want to show that each vector is in the subspace. This is easy to verify.
Next, show the set is linearly independent. Since there are only two vectors, and they are not scalar multiples of each other, the set is independent. 
Finally, we need to show that the set spans our subspace. So suppose $x+y+z = 0$, so we may write a general vector in this subspace as $(x, y, -x-y)$. To (try to) express this general vector as a linear combination of our two given vectors $b_1$ and $b_2$, write $ab_1+cb_2 = (x,y,-x-y)$. This gives us the system
$$a = x$$
$$-a + c = y$$
$$-c = -x-y$$
So we have $a = x, c = x+y$, and so $$(x,y,-x-y) = x\cdot (1,-1,0)+ (x+y)\cdot (0, 1, -1)$$
Therefore, the two vectors do indeed span the subspace, as any such vector can be written as a linear combination of the two given vectors. (If they didn't, then this system of equations would have no solutions.)
If you have a subspace of $\Bbb{R^3}$, then its dimension is either $0, 1, 2$ or $3$, so you can't assume three vectors are necessary to span the subspace (indeed, this would only be true if the subspace is $\Bbb{R}^3$ itself). Here, you can see that you can choose the first two components freely, but in order for the vector to be in the subspace, this "forces" the value of the third component. This should give you a hint that the subspace has dimension 2. 
