Let's say I have a matrix \begin{bmatrix}7&3\\3&-1\end{bmatrix}
I get the corresponding eigenvalues $8$ and $-2$. Without showing all the work, if I plug in $-2$, I eventually get to one reduced equation: $$ \begin{cases} 3x + y = 0 \\ 3x = -y \end{cases} $$ In this situation, $x$ is a basic variable, but I can make so many different numbers $y$ and get a ton of different results. Is that ok? For example, I could get $(-3, 1)$, but I also could get $(1, -3)$.
By plugging $-2$, we get to $$3x+y=0$$ Writing it out as a vector: $$\vec{v} = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ -3x \end{bmatrix} = x \begin{bmatrix} 1 \\ -3 \end{bmatrix}$$
Hence, all multiples of the vector $$\begin{bmatrix} 1 \\ -3 \end{bmatrix}$$ Are solutions to the equation. We say, then, that this vector forms the eigenspace of this eigenvalue.
So answering your question: Yes, that is ok. Every multiple of the vector will also be an eigenvector with the same eigenvalue.
Also notice that, in your case, neither $(-3,1)$ nor $(1,-3)$ are solutions to the system and hence are not eigenvectors with eigenvalue $2$.