I have the following matrix given and am asked to find all its eigenvalues, eigenvectors and the geometric multiplicity of each eigenvalue:
$A_{n\times n} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$
I used the characteristic polynom to find the eigenvalues $λ_{1} = λ_{2} = 0$ and $λ_{3} = 1$. I then attempted to find the eigenvectors by solving the system $(A-λI)v=0$ for $λ_{1}=λ_{2}=0$ but once I get the system
$ \begin{array}{ccc|c} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} $
I am not sure how to get the eigenvectors. I know that there need to be 2, because $\dim (E_{λ_{1}}) = n- \operatorname{Rang}(A) = 2$ and MatLab tells me they are
$ v_{1} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\quad v_{2} = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} $
but I don't quite understand why. The way I learned it you are supposed to set one of the free variable to one and all the others to zero, then calculate a solution and repeat for every free variable there is. However, here there is only one free variable as far as I can see (the one in the first column). How can I get two eigenvectors out of it and why is the $-1$ trick valid here but not in other cases?