All real values $a$ for a $2$-dimensional vector? Find all real numbers $a$ for which there exists a $2D$, nonzero vector $v$ such that:
$\begin{pmatrix} 2 & 12 \\ 2 & -3 \end{pmatrix} {v} = a {v}$.
I substituted $v$ with $\begin{pmatrix} c \\ d \end{pmatrix}$ and multiplied to obtain the system of equations: 
$2x+12y = ax$
$2x-3y= ay$
Since the value $a$ is only of importance, I added the two equations to obtain $4x+9y = ax + ay$. Would that mean that $a = 4, 9$ is correct? 

When I wolframmed it, I got -6 and 5. It is usually harder to find the two dimensional vector, without finding the a's. What you can do, if your matrix is A and your looking for a such that there exists $v\not = \vec{0}$ such that $Av=av$ is to notice that ($(A-aI)(v)=\vec{0}$, since $\vec{v} \not = 0$ then the determininant of $A-aI$ must be zero! And you will get to solve a degree two equation in a, which is, in this case, $a^{2}+a-30=0$
