Let $A = \begin{bmatrix} 1 & 0\\ 0 & 0\\ \end{bmatrix}$. The task is to find a constant $c$ and a vector $v$ (not zero vector) such that $Av = cv$
My attempt is to let $v = \begin{bmatrix} v_1 \\ v_2 \\ \end{bmatrix}$.
$Av=cv$ then gives
$\begin{bmatrix} v_1 \\ 0 \\ \end{bmatrix} = c\begin{bmatrix} v_1 \\ v_2 \\ \end{bmatrix}$.
If $v_1=0$ we can have $c=0$ and $v_2 = t$ where $t \in \mathbb{R}$ yielding the zero vector.
If $v_1$ is not equal to $0$, $c$ is not zero, we must have $v_2=0$. $v_1=c*v_1$ and $c=1$, $v_1 \in \mathbb{R}$
and we get $\begin{bmatrix} t \\ 0 \\ \end{bmatrix}$.
Have I done this right? The question also states "for which choices of $c$ does such a zero-vector not exist". I'm not sure what they mean here, since it's very obvious any $c$ not equal to zero will yield a non zero $v$ if $v_1$ and $v_2$ are not both equal to zero.
I wonder if my solution is wrong, because I need to explain the geometric interpretation of what happens when I take the product $Av$ for the different $c$'s. I don't see anything special with my two answers, so I think I might not have solved it correctly.
It looks as if you're on the right track. The problem really wants you to look at this in terms of $c$, so restating what you've said,
For $c = 0$, the vector $\pmatrix{0 \\ 1}$ is a solution (as is any multiple of it).
For $c = 1$, the vector $\pmatrix{1 \\ 0}$ is a solution (as is any multiple of it).
For any other $c$, $Av = cv$ has no nonzero solutions (indeed, no solutions at all).
In short: you're on exactly the right track, and a little reorganizing of what you wrote gets things to be expressed the way the question-asked was probably hoping they'd be.