Finding the Dimension of a given space $V$ I am unsure how to solve this problem:

If $\vec{v}$ is any nonzero vector in $\mathbb{R}^2$, what is the
  dimension of the space $V$ of all $2 \times 2$ matrices for which
  $\vec{v}$ is an eigenvector?

What I have so far is: 
$$ \left[
  \begin{array}{ c c }
     a & b \\
     c & d
  \end{array} \right]
  \left[
  \begin{array}{ c }
     v_{1}  \\
     v_{2} 
  \end{array} \right] =  \left[
  \begin{array}{ c }
     \lambda v_{1} \\
     \lambda v_{2}
  \end{array} \right]  $$
And solving for this I get two equations with four unkowns. 
$$ av_{1} + bv_{2} = \lambda v_{1}$$ 
$$cv_{1} + dv_{2} = \lambda v_{2} $$ 
I am not sure where to go from here. At first I solved for $a$ and $c$ in terms of $b$, $v_1$ and $v_2$ and $d$, $v_1$, and $v_2$ respectively and got dim = 2, but the answer is dim = 3. Any hints why this is?

Another solution (using the fact that the vector is non-zero, and we are in dimension 2) is to express the fact that 
$ \left[\begin{array}{ c c } a & b \\ c & d \end{array} \right]
  \left[\begin{array}{ c } v_1 \\ v_2 \end{array} \right]$
and $\left[\begin{array}{ c } v_1 \\ v_2 \end{array} \right]$ 
are colinear by them having the same orthogonal complement, which is spanned by $\left[\begin{array}{ c } v_2 \\ -v_1 \end{array} \right]$. This means  that $a,b,c,d$ must satisfy the single equation (parametric in $v_1,v_2$)
$$ 0 = \left[\begin{array}{ c c } v_2 & -v_1 \end{array} \right]
       \left[\begin{array}{ c c } a & b \\ c & d \end{array} \right]
       \left[\begin{array}{ c } v_1 \\ v_2 \end{array} \right]
     = v_2^2b+v_1v_2(a-d)-v_1^2c,$$
which is non-trivial because $v_1,v_2$ are not both $0$. The dimension of the space of solutions is therefore $4-1=3$.
