Length-two Jordan chain solution ($e^{\lambda t}(v+te)$)

If $(M-\lambda I)e=0$ and $(M-\lambda I)v=e$, the vector $v$ is a generalized eigenvector and $e^{\lambda t}(v+te)$ solves $\dot{x}=Mx$. The polynomial factor distinguishes a nontrivial Jordan block from an ordinary repeated eigenvalue with a full eigenvector basis.