Generalized Eigenvector in Dynamical System in Infinite Dimensions  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! 
 A: Ordinary generalized eigenvector has nothing to do with the problem.
Ordinary characteristic polynomial of the matrix $\Delta(0)$
is $\det(\lambda I-\Delta(0))=(\lambda+1)^3-1$ has simple root at $0$,
and no generalized eigenvector.
To solve your equation, you proceed as follows.


*

*Look for a solution of the form $e^{\sigma t}c$, where $c$ is a vector.
You obtain $((\sigma-1)I-e^{-\sigma}J)c=0$. So $\sigma$ is a root
of the "characteristic" equation, same as yours.
This root $\sigma=0$ gives you a constant solution $(1,-1,1)^T$.
But this root is of multiplicity $2$, so you want a second linearly independent solution.

*This second linearly independent solution must in general be of the form
$Y(t)=e^{\sigma t}(c_0+c_1t)$, where $c_0,c_1$ are vectors.
In your case, $\sigma=0$,
but I will keep it and make computation in the general case,
for the equation
$$Y'(t)=Y(t)+JY(t-1),$$
where $J$ can be arbitrary matrix. Substitute our form of the solution
to the equation, divide by $e^{\sigma t}$ and group the terms with $t$ and
without $t$. You obtain
$$(\sigma I-I-e^{-\sigma}J)c_1=0,$$
and
$$(\sigma I-I-e^{-\sigma}J)c_0=-(I+e^{-\sigma}J)c_1.$$
First equation means that $c_1$ is an "eigenvector" which we found on step 1. Using the first equation,
we can transform the right hand side of the second equation:
$$(\sigma I-I-e^{-\sigma}J)c_0=-\sigma c_1.$$
THIS is the equation of the "generalized eigenvector", adapted to our
differential-difference equation.
In our case $\sigma=0$, so
we obtain the second solution in the form
$$Y(t)=t(1,-1,1)^T.$$
You can verify that this is a solution.
