I am getting two different solutions for the eigenvector corresponding to an eigenvalue when using two different numeric methods. I believe I can say both are valid through the following logic:

$$
\lambda \mathbf{x}=A\mathbf{x}\\
\therefore -\lambda \mathbf{x}=-A\mathbf{x}\\
\therefore \lambda(-\mathbf{x})=A(-\mathbf{x})\\
$$

Hence, $-\mathbf{x}$ is also an eigenvector of $A$.

Am I thinking correctly? Can I make this conclusion? I couldn't find anything about this, probably because I don't know the search terms to use.