Consider a system of linear delay differential equations:

$$ \dot{z_1}(t) = z_1(t) + z_2(t-1) $$ $$ \dot{z_2}(t) = z_2(t) + z_3(t-1) $$ $$ \dot{z_3}(t) = z_3(t) - z_1(t-1) $$

The characteristic matrix is: $\Delta(\sigma) = \sigma \cdot Id - (Id + J e^{-\sigma})$, where $Id$ is the $3\times 3$ identity matrix, and $$ J = \left(\begin{array}{rrr} 0 & 1 & 0; \\ 0 & 0 & 1; \\ -1 & 0 & 0 \end{array}\right) $$ Clearly, the characteristic equation is: $p(\sigma) = \det(\Delta(\sigma)) = 0$, i.e. $p(\sigma) = (\sigma - 1)^3 + e^{-3\sigma} = 0$. It is easy to see that $\sigma=0$ is a characteristic root of algebraic multiplicity 2, as $$ p(0) = (-1)^3 + e^0 = -1 + 1 = 0, $$ $$ p'(\sigma) = 3(\sigma - 1)^2 - 3 e^{-3\sigma}, p'(0) = 3 - 3 = 0, $$ and $$ p"(\sigma) = 6(\sigma - 1) + 9 e^{-3\sigma}, p"(0) = -6 + 9 = 3 \not= 0 $$

However, when I tried to find the two generalized eigenvectors by solving $\Delta(0) \phi_2 = \phi_1$, where $\phi_1 = (1 -1 1)^T$, and $\phi_1$ is derived by solving $\Delta(0)\phi_1 = 0$, I found that the equation $\Delta(0) \phi_2 = \phi_1$ is inconsistent, i.e., there is no solution!

I did realize that $\Delta(0)$ is a matrix of rank 2, that is, the null space of $\Delta(0)$ is only one dimensional. But unfortunately, the null space of $(\Delta(0))^2$ is one dimensional too! This makes me unable to find $\phi_2$. I believe I must have missed something, or have misunderstood something. Any comment or suggestion would be highly appreciated!