I have a specific example that I am hoping to use to learn something more general. The matrix $$A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 3 & -1 & 0 \\ -1 & 0 & 2 & 1 \\ 0 & 1 & -1 & 2 \end{bmatrix}$$ has a single eigenvalue $\lambda = 2$, with eigenspace spanned by $e_1 = (1, 0, 0, 1)$ and $e_2 = (0, 1, 1, 0)$. Thus, two generalized eigenvectors must be found to complete a Jordan basis. Yet, as far as I can tell, neither $e_1$ nor $e_2$ can be used to produce a generalized eigenvector. (I tried to solve $(A - 2I)x = e_1$ and derived a contradiction; similar for $(A - 2I)x = e_2$.) On the other hand, the eigenvector $e_3 = (1,1,1,1)$ does produce generalized eigenvectors.
So, how do I identify which eigenvectors correspond to generalized eigenvectors? Is there something "special" about $e_3 = (1,1,1,1)$ that I should be able to recognize?
Yes, there is. Note that$$A.(a,b,c,d)-2(a,b,c,d)=(-a+d,b-c,-a+d,b-c).$$This is a vector for which the first coordinate is equal to the third one and the second coordinate is equal to the fourth one. Neither $e_1$ nor $e_2$ have this form. But $e_3$ has.