How to determine the dimension of a subspace I am working on a problem for my math class that I'm just not quite grasping:

Let S denote the set of all vectors (x, y, z) in V$_3$ whose components satisfy the equation x + y = 0. Determine whether S is a subspace of V$_3$. If S is a subspace, compute dim S.

I have determined S is a subspace of V$_3$ because the set is closed under addition and multiplication. Thus I must compute the dimension of S.

How do I go about finding the dimension of S? The only way I know of is to start with dimension n = 1 and increase n by one recursively if a set of n linearly independent elements of S spans V$_n$, but what does it mean for a three dimensional vector to span a two dimensional space? Does this mean the x and y components span two dimensions, the y and z components span two dimensions, x and z, any of the above, etc?
Since you know $x$ and $y$ must satisfy the relation $x + y = 0$, then you can rewrite your restriction as: $$x = -y$$
And your vector space will be defined as: $(-y, y, z)$. Now, there are several ways to justify that this vector space is bidimensional. I believe that the most simple one is to notice that it is defined by only two independent variables, so you need only two coordinates to uniquely define a vector in this vector space, thus making it bidimensional.
You can also show this by finding the canonical base to this vector space, and showing that it has only two vectors - $(-1, 1, 0)$ and $(0, 0, 1)$.
The geometrical meaning of a subspace of a three dimensional space being a two dimensional space is that all the vectors from that subspace are contained on a plane in the three dimensional space - besides the meaning of needing only 2 coordinates do be uniquely defined even on a three dimensional space, because the third coordinate is defined as a function of those two.
The set of $(x,-x,z)$ for all $x,z$, is a two dimensional subspace.  A basis is $B=\{(1,-1,0), (0,0,1)\}$.  The order of a basis is the dimension of the subspace. ..

You can also tell it's $2$-dimensional because there are $2$ free parameters $x$ and $z$.  

Another way is that you start with a $3$-d space and impose $1$ restriction, $x+y=0$.  $3-1=2$...