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


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?

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  • $\begingroup$ I don't understand your first sentence. If $A$ is a unital banach algebra then the spectrum is defined to be $$ \sigma (T) = \{\lambda \in \mathbb C \mid T - \lambda \notin \mathrm{Inv}(A)\}$$ Could you elaborate on your first sentence? I think an operator is invertible if and only if its kernel is $\{0\}$. $\endgroup$ – user167889 Dec 15 '14 at 0:46
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    $\begingroup$ @student There is more to invertibility in infinite dimensions (i.e. in an infinite dimensional Banach space) than kernel=0. You also have to be surjective, which is a separate condition. $\endgroup$ – Olivier Bégassat Dec 15 '14 at 0:49
  • $\begingroup$ @OlivierBégassat Right, thank you for pointing it out! $\endgroup$ – user167889 Dec 15 '14 at 0:51
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    $\begingroup$ What is the notation $\|x,H\|=1$? I have never seen this before. $\endgroup$ – user167889 Dec 15 '14 at 1:02
  • $\begingroup$ Also, I see you are using a different definition of the spectrum. Is there a way to arrive at the same conclusions using my definition (i.e. without knowing $\sigma (L) = \overline{\mathbb C \setminus R(L)}$)? $\endgroup$ – user167889 Dec 15 '14 at 1:44

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