Why is $0$ an eigenvalue when we have translation invariance? Edit:
Hope to have improved my question:
Consider 
$$
u_t=u_{xx}+f(u)-w,\qquad w_t=\varepsilon (u-\gamma w)~~~~(1)
$$
with
$$
f(u)=u(u-a)(1-u),\qquad a<\frac{1}{2},~~ 0<\varepsilon\ll 1,~~0<\gamma\ll 1.
$$
We are now looking at travelling wave solutions $(u(\xi),w(\xi))$ with $\xi=x-ct$, i.e.
$$
-cu'=u''+f(u)-w,\qquad -cw'=\varepsilon (u-\gamma w)~~~\text{ with }'=d/d\xi
$$
If we recast (1) in terms of variables $\xi=x-ct$ and $t$, we get
$$
u_t=u_{\xi\xi}+cu_{\xi}+f(u)-w,\qquad w_t=cw_{\xi}+\varepsilon (u-\gamma w)~~~(2)
$$
and the travelling wave solution is an equilibrium (time independent) solution of (2).
If the right-hand side of system (2) is linearized in the travelling wave solution $(u(\xi),w(\xi))$, we get the operator
$$
L\begin{pmatrix}p\\r\end{pmatrix}=\begin{pmatrix}p_{\xi\xi}+cp_{\xi}+f'(u)p-r\\cr_{\xi}+\varepsilon (p-\gamma r)\end{pmatrix}.
$$


Note that $0$ must be in the spectrum because the translate of a travelling wave is another travelling wave. In other words, zero is of necessity an eigenvalue, due to translation of waves.


So, this is now my question: Of course a translate of a travelling wave is again a travelling wave; but why does this imply that $0$ is in the spectrum of $L$?
 A: There is some major confusion in your question. The notion of eigenvalues of $L$ only makes sense if $L$ is a linear operator. If $L$ is a linear operator then (assuming the notation means what it seems to mean) the formula $L(x+a)=L(x)$ doesn't make any sense - that would make sense if $L$ were a function on the line, which is not a linear operator.
If $L$ is a linear operator mapping functions on the line to functions on the line then saying $L$ is translation-invariant means that $L$ commutes with translations:


If $g(x)=f(x+a)$ then $(Lg)(x)=(Lf)(x+a)$.


And now no, a translation-invariant operator certainly need not have $0$ as an eigenvalue. For example the identity operator $L$, defined by $$Lf=f,$$is translation-invariant but has only one eigenvalue, namely $1$.
If you read somewhere that a translation-invariant operator must have $0$ as an eigenvalue that was under some assumptions on said operator; you need to say much more about the context for the question to become sensible. 
A: I think it is a bad habbit to write $u$ also for the equilibrium solution.
So let $U$ be the solution to the stationary problem and $u(x,t)=U(x-ct)$ the wave solution. 
To simplify matters let me omit the $w$ term and set $F(U)=U''+cU'+f(U)$ (function of $\xi$ only). Let $F(U+\delta U)=F(U)+L_U \delta U +...$ describe the linearization around $U$ and let $T_s U(\xi)=U(\xi+s)$ be the translation of $U$. It is a symmetry of the problem so $F(T_s U)=0$ for all $s$. Taking the derivative wrt $s$:
$$ 0=\frac{d}{ds}_{|s=0} F(T_s U) = L_U \frac{d}{ds}_{|s=0} T_s U = L_U U'$$
which shows that $U'$ (which is non-zero)  is in the kernel of $L_U$. With a bit more notation you may carry out the same analysis when including $w$.
