Can a basis have dimension less than the space that spans? 
Let S be a subspace of $\mathbb{R^2}$, such that $S=\{(x,y):2x+3y=0 \}$.
  Find a basis,$B$, for $S$ and write $u=(-9,6)$ in the $B$ basis.

So, I started to solve $2x+3y=0$ for $x$ and I got $x=-\frac{3}{2}y$. Then I could write,
$$\left[ \begin{matrix} x \\  y \end{matrix}\right] = \left[ \begin{matrix} -\frac{3}{2} y \\  y \end{matrix} \right] = y\left[ \begin{matrix}  -\frac{3}{2} \\  1 \end{matrix} \right]$$
So I was force to concluded that S is spaned by the vector $(-\frac{3}{2},1)$.
My doubt is if it is correct to assume that $\{(-\frac{3}{2},1) \}$ is a basis of $S$. 
Because, in this circumstance, I don't know how to write the vector $u$ in the basis $B$.
Thanks for the help.
 A: You've obtained the correct basis. And note that we can now write $\vec u$ in terms of the basis: $$\vec u = (-9, 6) = 6\left(-\frac 32, 1\right)$$
We can figure out that $6$ is the needed scalar by solving for the unknown scalar $c$:
$$c\left(-\frac 32, 1\right) = \left(-\frac 32c, c\right) = (-9, 6) \iff -\frac 32 c = -9\;\text{and }\; c = 6 \iff c = 6$$

ADDED: Concerning your title, recall that $S$ is a subspace of $\mathbb R^2$. While $\mathbb R^2$ has dimension two, $S$ is a subspace of dimension one. So we should expect that, as we found, $S$ has a basis of cardinality one. (Indeed, a set of $n$ vectors  cannot (by definition) form a basis for a vector subspace of dimension greater than $n$.)
A: Your basis is correct. $(-9,6)=6(-\frac{3}{2},1)$
A: You can see all points above throughout Maple's environment visibly. 
> with(Student[LinearAlgebra]); with(plots):
> infolevel[Student[LinearAlgebra]] := 1
> a := ProjectionPlot(<-3/2, 1>, <-9, 6>, color = red):
> b:=arrow(<-9, 6>, width = .2, color = green):
> display(a,b);

               Vector:     <-3/2, 1>
               Projection: <-1.500, 1.000>
               Orthogonal complement: <.3000e-12, -.2001e-12>
               Norm of orthogonal complement: .3606e-12


