Find the spectrum of an operator I am trying to learn some basic stuff about spectral theory, and I am a little bit lost. Please, could you help me and tell me how to find  $\sigma(T)$ and $\sigma_p(T)$ of the operator $T:C([0,1]) 
\rightarrow C([0,1])$  $$Tf=f+f(0)-f(1).$$
I have some result where is written that the right answer is  $\sigma(T)=\sigma_p(T)= \{0\}$ but I do not know how to prove it. 
And I also do not understand why $\sigma_p(T)$ does not contain $1$. 
Because of $$Tf=f+f(0)-f(1)=\lambda f$$
if $\lambda =1$ we can take in account e.g. constant function, so $Tf=f $.
But from some reason, this is not in my teacher´s results.
Thank you for your help!!!
 A: I think you are right. If $Tf=\lambda f$, we have
$
\lambda f = f + f(1) - f(0),
$
or 
$$\tag{1}
(\lambda - 1) \, f= f(1)-f(0).
$$
If $\lambda=1$, then any $f$ with $f(1)=f(0)$ satisfies the equation, so $1\in\sigma_p(T)$. In particular, as you mention, constant functions are eigenfunctions for the eigenvalue $1$. 
When $\lambda\ne1$, the equation $(1)$ has no solution: when $t=1$ and $t=0$, we get respectively
$$
(\lambda-1)f(1)=f(1)-f(0),\ \ \ \ \ (\lambda-1)f(0)=f(1)-f(0),
$$
which imply that $f(1)=f(0)$; but then we go back to $(1)$ and get $f(t)=0$. So $\lambda\ne1$ cannot be an eigenvalue. 
Consider the operator $Sf=f+f(0)-f(1)$. Then 
\begin{align}
STf&=S(f+f(1)-f(0))=(f+f(1)-f(0))+(f(0)+f(1)-f(0))-(f(1)+f(1)-f(0))\\ \ \\
&=f+f(1)-f(0)+f(1)-2f(1)+f(0)\\ \ \\
&=f.
\end{align}
And
\begin{align}
TSf&=T(f+f(0)-f(1))=f+f(0)-f(1)+(f(1)+f(0)-f(1))-(f(0)+f(0)-f(1))\\ \ \\
&=f+f(0)-f(1)+f(0)-2f(0)+f(1)\\ \ \\
&=f.
\end{align}
So $T$ is invertible, and $0\not\in\sigma(T)$. 
More generally, for $\lambda\ne1$ one can define 
$$
Sf=\frac{f+\alpha f(1)+\beta f(0)}{1-\lambda}
$$
for appropriate $\alpha,\beta$ to get an inverse for $T-\lambda I$. 
In summary, $\sigma(T)=\sigma_p(T)=\{1\}$.
