Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$ I'm considering the bounded linear operator $T$ on $l^1$ (the space of all absolutely convergent complex sequences) given by (with $e_k=(\delta_{kj})_{j=1,2,...}$)
$$T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$$
for $(a_j) \in l^1$.
I'm trying to find the spectrum of $T$, $\sigma(T)$, and I read an idea would be to find its approximate point spectrum and the point spectrum of its adjoint.
I have that its adjoint is $T':l^{\infty} \to l^{\infty}$ given by
$$T'((b_j))=b_2 e_1+ \sum_{j=2}^{\infty} (b_1+b_{j+1})e_j$$
I could find the point spectrum of $T$ but not its point approximate spectrum, how would you go at that?
I also have a problem with dealing with the point spectrum of $T'$. If $k \in \mathbb{C}$ is one of its elements there must exist a non-zero sequence $(b_j) \in l^{\infty}$ such that $b_j=\frac{b_1}{k-1} +ck^j$ for some $c \in \mathbb{C}$ (for $j=2,3,...$). But I can't go further...
Could you give me a hand at this, I'm new with spectra and trying to get a feeling of how to deal with such problems. Thank you so much!
 A: I think you can find the spectrum directly. First,
$$T((a_j))_i=\sum_{j=2}^{\infty}a_j e_1+\sum_{i=2}^{\infty}a_{i-1} e_i=\sum_{j=2}^{\infty}a_j e_1+a_{i-1}$$
Then 
$$(T-\lambda I)((a_j))_1=\sum_{j=2}^{\infty}a_j -\lambda a_1,$$
$$(T-\lambda I)((a_j))_i=a_{i-1}-\lambda a_i$$
So if $\lambda$ is an eigenvalue, the second equation implies $a_i=\frac{1}{\lambda}a_{i-1}$. The first implies that 
\begin{eqnarray}
\lambda &=& \frac{1}{a_1}\sum_{j=2}^{\infty}\left(\frac{1}{\lambda}\right)^ja_1
\\
&=&\sum_{j=2}^{\infty}\left(\frac{1}{\lambda}\right)^j
\\
&=&\left(\frac{1}{\lambda}\right)^2\frac{\lambda}{\lambda-1}
\end{eqnarray}
Check the values $\lambda=0$ and $\lambda=1$. Then for all other values of $\lambda$, the equation defines a cubic polynomial equation which you can factor to find the roots. 
To get the rest of the spectrum, we first calculate the eigenvalues of the adjoint
$$T'((b_j))=b_2 e_1+ \sum_{j=2}^{\infty} (b_1+b_{j+1})e_j.$$
We have
$$(T^{\prime}-\lambda I)_1=b_2-\lambda b_1$$
$$(T^{\prime}-\lambda I)_i=b_1+b_{i+1}-\lambda b_i.$$
The last equation implies $b_{i+1}=\lambda b_i-b_1$. This and the first equation imply
$$b_i=\lambda^{i-2}b_2-\sum_{j=0}^{i-3}\lambda^{j}b_1=b_1\left(\lambda^{i-1}-\sum_{j=0}^{i-3}\lambda^{j}\right)=b_1\left(\lambda^{i-1}-(1-\lambda^{i-2})/(1-\lambda)\right)=b_1\left(\frac{\lambda^{i-2}(\lambda-\lambda^2+1)-1}{1-\lambda}\right)$$ 
The question is: when is this sequence bounded? It is bounded when $\vert \lambda\vert\leq 1$ and when $\lambda$ is a root of $\lambda-\lambda^2+1$. It remains to check the case $\lambda=1$. 
To finish, we use the fact that for an operator with closed range, $ker(T^{\prime})=Ran(T)^{\perp}$. Hence, $T^{\prime}-\bar{\lambda}I$ has nontrivial kernel iff $T-\lambda I$ is not surjective. But these are exactly the Eigenvalues of $T^{\prime}$. 
