When row-reducing the matrix I have with the eigenvalue, this is the result:
$\begin{bmatrix} 1 &0 &1 &|0 \\ 0 &0 &0 &|0 \\ 0 &0 &0 &|0 \end{bmatrix}.$
The eigenvectors will be $\begin{bmatrix}-1\\0\\1\end{bmatrix}$ and $\begin{bmatrix}0\\1\\0\end{bmatrix}.$
Can someone explain to me why, because I thought the only eigenvector would be $\begin{bmatrix}-1\\0\\1\end{bmatrix}$ ?
You get $x+z=0$ but no constraint on $y$. Your eigenvector is $(x~y~-x)=x(1~0~-1)+y(0~1~0)$. Note that this generates two linearly independent eigenvectors: $(-1,0,1)$ for $x=-1,y=0$ and $(0,1,0)$ for $x=0,y=1$. Any other eigenvector for this eigenvalue is a linear combination of these two eigenvectors.