Determine the spectrum of a shift-operator Let $\{a_n\}_{n=1}^{\infty} \subset \mathbb C$ be a sequence such that $|a_n|=r$ for all $n \geq 1$ and $r \geq 0$
Define $T: \ell^2 \to \ell^2$ by
$T(x_1,x_2,x_3,...) = (0,a_1x_1,a_2x_2,....) $ with $(x_1,x_2,....) \in l^2$
Determine the spectrum of T.
The spectrum of an operator is defined as $\{\lambda \in \mathbb{C}: T-\lambda I  \ \text{is not invertible}\}$
I know how to calculate the eigenvalues of a matrix, but here I have really no idea to start.
Can anyone help?
 A: Note first that $\|T\|=r$, and so $\sigma(T)\subset\{\lambda:\ |\lambda|\leq r\}$. 
It is very easy to check that $T$ has no eigenvalues. This makes it usually easier to deal with $T^*$. We have 
$$
T^*x=(\overline{a_1}x_2,\overline{a_2}x_3,\ldots).
$$
Assume $|\lambda|<r$. Define a sequence $\{c_n\}$ by $c_1=1$ and  $c_{n+1}=\dfrac{c_n}{\overline{a_n}}$ and put $x=(\lambda c_1,\lambda^2c_2,\lambda^3c_3,\ldots).$ Then $x\in\ell^2$, since $|c_n|=r^{n-1}$ and so 
$$
\sum_{n=1}^\infty |x_n|^2=r\sum_{n=1}^\infty\left|\frac\lambda r\right|^n<\infty.
$$
Now
$$
T^*x=(\overline{a_1}x_2,\overline{a_2}x_3,\ldots)
=(\overline{a_1}c_2\lambda^2,\overline{a_2}c_3\lambda^3,\ldots)
=_(c_1\lambda^2,c_2,\lambda^3,\ldots)=\lambda x.
$$
So all such $\lambda$ are eigenvalues of $T^*$. We obtained
$$
\{\lambda:\ |\lambda|<r\}\subset\sigma(T^*)\subset\{\lambda: |\lambda|\leq r\}.
$$
As the spectrum is always closed, we get $\sigma(T^*)=\{\lambda:\ |\lambda|\leq r\}$. Finally, since the spectrum of $T^*$ is obtained from the spectrum of $T$ by conjugating the elements, we have 
$$
\sigma(T)=\{\lambda:\ |\lambda|\leq r\},
$$
that is, $\sigma(T)$ is the ball of radius $r$ centered at zero. 
