Find the spectrum of the operator $T: \ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ defined by $(Tx)_n = \frac{x_n}{n}$ Consider the linear operator $T:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ defined as
$$ (Tx)_n = \frac{x_n}{n}, \quad x \in \ell^2(\mathbb{C}). $$
I can show that it is bounded with norm $\|T\|=1$, which tells me that
$$ \sigma(T) \subseteq \{ \lambda \in \mathbb{C} : \,\, |\lambda| \le 1 \}. $$
I am also able to show that the point spectrum is
$$ \sigma_p(T) = \left\{ \frac{1}{n} \right\}_{n \in \mathbb{N} }.$$
Finally I know that the residual spectrum is empty because $T$ is self-adjoint.


How can I go on and find the continuous spectrum of $T$?
Is there a "more direct" way to show that the residual spectrum of $T$ is empty, without using the self-adjointness property?
One can show that $T$ is a compact operator: define $S_n:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ by
$$(S_nx)_m = \left\{ \begin{matrix}
\frac{x_m}{m} &amp; m \leq n \\
0 &amp; m &gt; n
\end{matrix}
\right.
$$
Note that the $S_n$ are finite-rank and that
$$((T-S_n)x)_m = \left\{ \begin{matrix}
0 &amp; m \leq n \\
\frac{x_m}{m} &amp; m &gt; n
\end{matrix}
\right.
$$
so $\lVert T-S_n\rVert_2^2 = \frac{1}{n+1} \to 0$ as $n \to \infty$. Hence, $T$ is compact. By the Fredholm Alternative, the non-zero spectrum of $T$ consists purely of eigenvalues (i.e. the point spectrum). Since the spectrum must be closed, it must also contain $0$; as $T$ is injective and bounded, $0$ cannot belong to the point or residual spectra so it belongs to the continuous spectrum.
For $\lambda \notin \{0\} \cup \sigma_p(T)$, you can explicitly write down
$(\lambda I - T)^{-1}$.