linear system straight line solution For a matrix

$
A = \begin{bmatrix}
-1 &amp; 3 \\
0 &amp; -2
\end{bmatrix}
$

I have obtained the Eigenvalues $\lambda = -2$ and $\lambda = -1$.

Now, after trying to find the straight line solution for $\lambda = -1$, I get:

$
\begin{bmatrix}
0 &amp; 3 \\
0 &amp; -1
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} =
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$

... 

Now what does it mean?
$ 3y = 0, -y = 0$, but what is the eigenvector ? ? ? 

Can it be a vector (0,1) ?

Thanks.
You showed that $A\begin{pmatrix}x\\ y\end{pmatrix}=-\begin{pmatrix}x\\ y\end{pmatrix}$ if and only if $y=0$. 

Thus the set of eigenvectors for the eigenvalue $-1$ is $\{(x,0)\mid x\ne0\}$ and the eigenspace for the eigenvalue $-1$ is $E_{-1}=\{(x,0)\mid x\in\mathbb R\}$. Note that $E_{-1}$ is the subspace generated by the vector $(1,0)$.
$3y=0$ and $-y=0$ implies $y=0$. For a linear transformation $T$ representing the matrix, $\IM T = \{(0,0)\}$ for every $x$.
So for every $x,y$ you get $(0,0)$.
Therefore you get as the $\KER T=\{(1,0)\}$.
$\DeclareMathOperator{\IM}{Im}\DeclareMathOperator{\KER}{Ker}$