Spectrum of T in $B(\ell^2)$

Question

Let $T:\ell^2 \to \ell^2$ be an operator on $\ell^2$ is defined as follows: $$T\{a_1,a_2,\dots\}=\{0,a_1,a_2,\dots\}$$ What is spectrum of T in $B(\ell^2)$?

Answer 1

First note that as $\|T\|= 1$, we have $\sigma(T) \subseteq \{\lambda \in \mathbb C: |\lambda| \le 1\}$.

Now for $|\lambda| < 1$, $(T-\lambda)\colon (a_1, \ldots) \mapsto (-\lambda a_1, a_1 - \lambda a_2, \ldots)$. Suppose there were an $a \in \ell^2$ with $(T -\lambda)(a) = (1, 0, \ldots)$, then $-\lambda a_1 = 1$, so $a_1 = -\lambda^{-1}$, $a_2 = \lambda^{-1} a_1 = -\lambda^{-2}$, hence we must have $a = (-\lambda^{-n})_n$. But that's not possible as $a \in \ell^2$. That gives $\{\lambda \in \mathbb C : |\lambda| < 1\} \subseteq \sigma(T)$.

Now recall that $\sigma(T)$ is closed, and you are done.