Spectrum of left shift operator $L\in B(H)$

Question

Let $H$ be a Hilbert space with an orthonormal base $e_i$ and $L$ the left shift operator $L\in B(H)$: $(x_1, x_2, \dots) \mapsto (x_2, x_3, \dots)$.

I computed the spectrum could someone please tell me if this is right?

My work:

$\lambda \in \sigma (L)$ if and only if there exists $x \in H$ ($x\neq 0$) such that $$ (L-\lambda)(x) = (x_2 - \lambda x_1, x_3 - \lambda x_2, \dots) = 0$$

If $\lambda = 0$ then it quickly follows that $x=0$. For $\lambda \neq 0$ this is true if and only if $x$ is of the form $(x_1,\lambda x_1, \lambda^2 x_1, \dots)$.

Hence $\sigma (L) = \mathbb C \setminus \{0\}$.

Answer 1

First, in general case the spectrum consists of much more than eigenvalues: the resolvent set is $$R(L)=\{z\in\Bbb C: L-zI \text{ is continuously invertible}\}$$ and $$\sigma(L)=\overline{\Bbb C\setminus R(L)}.$$

Second, in your case, you can explicitly write $$(L-zI)x = (x_2-zx_1,x_3-zx_2,\ldots).$$

Fix $z\in \Bbb C$. If for an arbitrary $\epsilon>0$ you can find an $x \in H$ such that $\|x ,H\|=1$ and $\|Lx -zx ,H\|\le \epsilon$, then this value of $z$ is not in the resolvent set.

The trivial case $z=0$ is done by taking $x=(1,0,0,\ldots)$, which gives $\|Lx\|=0$, therefore $0\in\sigma(L)$.

Can you take it from here?

Answer 2

Note that $L$ is nothing but the adjoint of the unweighted shift $S$ defined by $$S(x_1,x_2,...):=(0,x_1,x_2,...),~(x=(x_1,x_2,...)\in H).$$ It is known that the spectrum of $S$ is the closed unit disc $\overline{\Bbb{D}}$, and so $$\sigma(L)=\overline{\sigma(L^*)}=\overline{\sigma(S)}=\overline{\Bbb{D}}.$$ If you are not familiar with shit operators, you may argue as follows. Note that $$\|Lx\|^2=\sum\limits_{k>1}{|x_k|^2}\leq \|x\|^2=\sum\limits_{k>0}{|x_k|^2}$$ for all $x=(x_1,x_2,...)\in H$. Hence, $\|L\|\leq1$ but since $$\|L(0,1,0,...)\|=\|(1,0,0,...)\|=1,$$ we see that $\|L\|=1$ and $$(*)~~~~~~~~~~~~~~~~\sigma(L)\subset\{\lambda\in\Bbb{C}:|\lambda|\leq\|L\|=1\}=\overline{\Bbb{D}}.$$ Now, let us solve $$(L-\lambda I)(x_1,x_2,...)=0.$$ This implies that $x_2-\lambda x_1=0$, $x_3-\lambda x_2=0$, $x_4-\lambda x_3=0$, ... Therefore, $x_2=\lambda x_1$, $x_3=\lambda x_2=\lambda^2x_1$, $x_4=\lambda x_3=\lambda^3 x_1=$, ... So, for any $n$, we have $$x_n=\lambda^{n-1}x_1.$$ Take $x_1=1$, to see that $$(L-\lambda I)(1,\lambda,\lambda^2,\lambda^3,...)=0.$$ Then the only thing that we need to check is when $(1,\lambda,\lambda^2,\lambda^3,...)$ is in $H$. This holds true only if $$\sum_{k\geq 0}|\lambda|^{2k}=\|(1,\lambda,\lambda^2,\lambda^3,...)\|^2<\infty.$$ But the radius of convergence of this series is 1, and \begin{equation} \Bbb{D}=\{\lambda:|\lambda|<1\}\subset\sigma(L). \end{equation} Since $\sigma(L)$ is compact (in particular closed), we get that \begin{equation} (**)~~~~~~~~~~~~~~~~\overline{\Bbb{D}}=\{\lambda:|\lambda|\leq1\}\subset\sigma(L). \end{equation} Clearly, by (*) and (**), we have \begin{equation} \sigma(L)=\overline{\Bbb{D}}=\{\lambda:|\lambda|\leq1\}. \end{equation}