Finding spectrum of the operator A $A:\mathcal{l}_2\rightarrow \mathcal{l}_2:(x_n)_{n=1}^\infty \rightarrow (x_{n+1})_{n=1}^\infty$ (left shift)
Find the spectrum and all its parts for the operator A.
What should I do?
 A: Since $\|A\| = 1$, using the Neumann series we can indeed invert $A - \lambda I$ when $|\lambda|>1$. For $|\lambda|<1$, the sequence $x_n = \lambda^n$ lies in the kernel of $A-\lambda I$ and it is not invertible.
Then the conclusion follows by recalling that the spectrum of a bounded operator is always closed.
A: Notice that $\|Ax\| \le \|x\|$. So $\sigma(A)$ is contained in the closed unit disk $D^{c}$. To find the resolvent $(A-\lambda I)^{-1}$, try to solve $(A-\lambda I)x = y$ for $x$, and see if it can be done. First, see if there are any eigenfunctions $Ax=\lambda x$. If there were such an $x$, then $|\lambda| \le 1$ and $x=(x_1,x_2,x_3,\cdots)$ would satisfy
$$
        (x_2-\lambda x_1,x_3-\lambda x_2,x_4-\lambda x_3,\cdots)=0,\\
          x_2=\lambda x_1,\; x_3=\lambda x_2,\;x_4=\lambda x_3.
$$
Assuming $x_1=1$, there is a solution $(1,\lambda,\lambda^{2},\lambda^{3},\cdots)$. This solution is valid only for $|\lambda| < 1$ because, otherwise, $x \notin l^{2}$. In other words, $\sigma_{P}(A)=\{ \lambda \in \mathbb{C} : |\lambda| < 1\}$ is the open unit disk. And $\sigma(A)=\{ \lambda \in\mathbb{C} : |\lambda|\le 1\}$ because $\sigma(A)$ is closed, contains the open unit disk, and is contained in the closed unit disk.
The adjoint of $A$ is the shift $B(x_1,x_2,x_3,\cdots)=(0,x_1,x_2,x_3,\cdots)$. Notice that $\|Bx\|=\|x\|$ for all $x \in l^{2}$. $B$ has no point spectrum because $Bx=\lambda x$ implies $(x_1,x_2-\lambda x_1,x_3-\lambda x_2,\dots)=0$, which forces $x_1=0, x_2=\lambda x_1=0,etc.$. So the range of $(A-\lambda I)$ is dense for all $\lambda$, which puts the unit circle $\{ \lambda \in\mathbb{C} : |\lambda|=1\}$ in the continuous spectrum of $A$. $A$ has no residual spectrum.
A: First note that $A$ is an isometry, so $\|A\| = 1$. Hence $\sigma(A)$ lies in the closed unit ball of the complex plane $C$.
Define the adjoint of $A : X \rightarrow Y$ to be $A^* : Y^* \rightarrow X^*$ such that $A^*(\alpha)(x) = \alpha(Ax)$ where $\alpha \in Y^*$ and $x \in X$ (here $X,Y$ are Banach spaces). Then $\sigma(A) = \sigma(A^*)$. Then easy computation should give you the adjoint of left shift is the right shift (D'oh!). It is very easy to see that the spectrum of the right shift is all the closed unit disc by considering $x = (1,\lambda, \lambda^2, \lambda^3,...)$.
