Consider the following PDE: $$ u_t=u_{xx}+f(u)-w,~~~~~w_t=\varepsilon (u-\gamma w),~~~~~~~~~(1) $$ where $f(u)=u(u-a)(1-u), 0<a<\frac{1}{2}, \varepsilon,\gamma >0, \varepsilon\ll 1,\gamma\ll 1$.
A travelling wave for (1) is a solution that is a function of the single variable $\xi=x-ct$, i.e., $(u(\xi),w(\xi))$ satisfies $$ -cu'=u''+f(u)-w,~~~~~~-cw'=\varepsilon (u-\gamma w)~~~~~~~('=d/d\xi).~~~~~~(2) $$
If (1) is recast in a moving coordinate frame, i.e., in terms of variables $\xi=x-ct$ and $t$, it becomes $$ u_t=u_{\xi\xi}+cu_{\xi}+f(u)-w,~~~~~~w_t=cw_{\xi}+\varepsilon (u-\gamma w).~~~~~~~~(3) $$
One can see by (2), that a travelling wave is an equilibrium solution of (3). So one nearby question is to ask whether a travelling wave is stable.
For a fixed $\varepsilon$, let $U_{\varepsilon}(\xi)=(u_{\varepsilon}(\xi),w_{\varepsilon}(\xi))$ be a travelling wave for (1), i.e., an equilibrium solution for (3).
A standard technique for determining stability of $U_{\varepsilon}(\xi)$ is to linearize: If the right-hand side of (3) is linearized along $U_{\varepsilon}(\xi)$, the resulting operator is $$ L\begin{pmatrix}p\\r\end{pmatrix}=\begin{pmatrix}p_{\xi\xi}+cp_{\xi}+f'(u_{\varepsilon})p-r\\cr_{\xi}+\varepsilon (p-\gamma r)\end{pmatrix}, $$ where $\begin{pmatrix}p\\r\end{pmatrix}(\xi)\in\text{BC}(\mathbb{R},\mathbb{R}^2):=\left\{u\colon\mathbb{R}\to\mathbb{R}^2 | u\text{ is bounded and uniformly continuous}\right\}$.
(The linearized criterion for stability of the travelling wave is that the spectrum of $L$ (except for $0$) lies in a left half-plane and $0$ is a simple eigenvalue.)
Thats the background.
Now, there is a statement that I could not understand yet.
Note, that $0$ must be in the spectrum because the translate of a travelling wave is another travelling wave.
I know that a translate of a travelling wave is again a travelling wave. But why does this imply that $0$ is in the spectrum of $L$?
Edit
After some looks at definitions, I have an idea:
Consider a solution $U(\xi,t)$ that is a translate of the travelling wave $U_{\varepsilon}(\xi)=(u_{\varepsilon}(\xi),w_{\varepsilon}(\xi))$, i.e. $$ U(\xi,t)=U_{\varepsilon}(\xi-k)=(u_{\varepsilon}(\xi-k),w_{\varepsilon}(\xi -k)) $$ for some $k$. Then $U$ is an equilibrium solution of (3), too.
If I am not totally wrong, then we have $$ p=u_{\varepsilon}(\xi-k)-u_{\varepsilon}(\xi),~~~~r=w_{\varepsilon}(\xi-k)-w_{\varepsilon}(\xi) $$ and $$ p_t=0,~~~w_t=0, $$ because travelling waves are equilibria for (3) and a translate of a travelling wave is a travelling wave.
Hence $$ L\begin{pmatrix}p\\r\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}, $$ meaning that $$ \text{ker}(L-0\text{ id })\neq\left\{\begin{pmatrix}0\\0\end{pmatrix}\right\}, $$ because for each translate of the travelling wave, $p,r\neq 0$ and $\begin{pmatrix}p\\r\end{pmatrix}\in\text{ker}(L-0\text{ id})$.
Hence $0$ is in the spectrum (point spectrum) of $L$.
Is this what was meant?