Be $\boldsymbol{X}=\ell^{2}$ and operators $A, B: \ell^{2} \rightarrow \ell^{2}$ by $$ A x:=\left(\xi_{2}, \xi_{3}, \ldots\right), \quad B x:=\left(0, \xi_{1}, \xi_{2}, \xi_{3}, \ldots\right) $$ to $x=\left(\xi_{i}\right)_{i \in \mathbb{N}} \in \ell^{2}$. Show that $$ \sigma(A)=\sigma(B)=\mathbb{D}, $$ where $\mathbb{D}=\{z \in \mathbb C : |z| \leq 1\} .$
I'm learning Basic spectral theory and i don't understand how deal with the problems i saw in the book.
Here i was thinking in finding the resolvent set $\rho(A) = \{\lambda \in K : A-\lambda I \text{ is invertible} \}$ and $K$ is scalar field, so i just have to find that $\rho(A) = K - \mathbb{D}$ because the spectrum is the complement. If my thought is correct, how can i solve this? or another hint to solve?. Thanks
Edit: i found that $B$ is right shift operator and $\|B\| = 1$ also i know that the spectrum $\sigma(A)$ is contained in the circle with radius $\|T\|$ and center $0$. Here i have to questions: ¿Why the norm is $1$? ¿And the inverse, left shift operator has the same norma?
Then i have that $\sigma (A) \subset D$, now i have to proof the inverse $D \subset \sigma(A)$: So taking $|\lambda| < 1 $ and the vector $x = (1, \lambda, \lambda^2,...)$ clearly $Ax = \lambda x$ so we have that the open disc is in the spectrum and we know that the spectrum is closed so $\sigma(A) = D$