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?
 A: Form a basis of $\mathbb R^2$ consisting of your given eigenvector, $v$, and another vector, $w \in \mathbb R^2$ (such that $w \notin \operatorname{span} v$). If $T$ is a linear map with eigenvector $v$ and corresponding eigenvalue $\lambda$, then the matrix of $T$ in the basis $(v, w)$ is 
$$
\operatorname{mat}(T) = \begin{matrix}
 & \begin{matrix} v & w \end{matrix} \\
\begin{matrix} v \\ w \end{matrix} & \left[ \begin{matrix} \lambda & * \\ 0 & * \end{matrix} \right]
\end{matrix}
$$
[why?]. There are three numbers that we have to pick to specify $T$: $\lambda$ and the two $*$s. Hence the set of all such $T$ is $3$-dimensional [why?].
A: Rewrite your system as
$$
\begin{pmatrix}
v_1 & v_2 & 0   & 0   & -v_1 \\
0   & 0   & v_1 & v_2 & -v_2
\end{pmatrix} \cdot 
\begin{pmatrix}
a\\b\\c\\d\\\lambda
\end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \end{pmatrix}
$$
Assuming that $v\ne 0$, the matrix has at least one nonzero maximal minor, so it has rank $2$. Hence, the space $W$ of pairs $(\lambda,a)\in\mathbb R\times\mathbb R^{2\times 2}\cong \mathbb R^5$ such that $av=\lambda v$ has dimension $2$. Now, let $p\colon \mathbb R\times\mathbb R^{2\times 2}\to\mathbb R^{2\times 2}$ be the projection, then $p(W)=V$. Note that $\ker(p)=\mathbb R\times\{ 0\}$ is not contained in $W$, so we get $\dim(V)=3$.
A: Let me give a constructive proof as follows. 
If $v_1 \neq 0, v_2 = 0$, we can choose the below basis:
$$E_1 = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, E_2 = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}, E_3 = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}.$$
If $v_1 = 0, v_2 \neq 0$, similarly, choose
$$E_1 = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, E_2 = \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, E_3 = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}.$$
If $v_1 \neq 0$ and $v_2 \neq 0$, choose
$$E_1 = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}, E_2 = \begin{bmatrix}0 & 0 \\ -v_2/v_1 & 1\end{bmatrix}, E_3 = \begin{bmatrix}-v_2/v_1 & 1 \\ 0 & 0\end{bmatrix}.$$
In each of above three cases, $E_1, E_2, E_3$ are linearly independent. Additionally, for each $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in V$,
$A$ can be expressed as linear combination of $E_1, E_2$ and $E_3$. 
For example, if $v_1 \neq 0, v_2 \neq 0$, and $Av = \lambda v$, then it can be verified that 
$$A = \lambda E_1 + bE_3 - \frac{v_1}{v_2}cE_2.$$
In conclusion, $\dim(V) = 3$.
A: 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 $4-1=3$.
