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Let $S:H\to H$ be a unilateral shift operator. I preferred in Example2.3.2 of Murphy's C*-algebras and operator theory that S has no eigenvalues. While $\{\lambda \in \Bbb C ; |\lambda|<1\} \subset \sigma(S)$. To find spectrum and eigenvalue, we should find a $\lambda$ that satisfies in the equality $S\xi = \lambda\xi$ for some $\xi \in H$. To find spectrum of $S$, the author says

$$S(\alpha_1,\alpha_2,...)=\lambda(\alpha_1,\alpha_2,...) \to $$ $$(\alpha_2,\alpha_3,...)=\lambda(\alpha_1,\alpha_2,...) \to$$ $$\alpha_{n+1} = \lambda\alpha_n~~~~for~every~n \to$$ $$(\alpha_1,\alpha_2,...)=\alpha_1(1,\lambda,\lambda^2,...)$$ So $\lambda \in \sigma(S)$ if $|\lambda|<1$.

I agree with above argument, but I think we can use this argument to show that $S$ has eigenvalue. Where is my mistake? I can not understand the difference between eigenvalue and spectrum. I really confused about it. Please help me. Thanks in advance.

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Your definition is not the unilateral shift. If your book gave you $Se_{n}=e_{n+1}$, then that is correct. The coefficient of $e_{1}$ is mapped to become the coefficient of $e_{2}$. So, $$ S(\alpha_{1},\alpha_{2},\alpha_{3},\cdots)=(0,\alpha_1,\alpha_2,\alpha_3,\cdots). $$ The operator $S$ never sends anything to $0$ except $0$ because $\|Sx\|=\|x\|$. The operator you defined in the problem, however, sends $(1,0,0,0,\cdots)$ to $0$. So your interpretation of the operator is not correct, and that's why you're getting a result that contradicts the one you were asked to show.

If the unilateral shift had an eigenvalue $\lambda$, then $\|Sx\|=\|x\|$ would for $|\lambda|=1$. But it has none, which you can show by assuming $$ (0,\alpha_1,\alpha_2,\alpha_3,\cdots)=(\lambda \alpha_1,\lambda \alpha_2,\lambda \alpha_3,\cdots),\\ (0,\overline{\lambda}\alpha_1,\overline{\lambda}\alpha_2,\overline{\lambda}\alpha_3,\cdots) = (\alpha_1,\alpha_2,\alpha_3,\cdots) $$ and concluding that $\alpha_1=0$, and then $\alpha_2=\overline{\lambda}\alpha_1=0$, etc.

You know $\sigma(S)$ is contained in the closed unit disk because $\|S\|=1$. What is interesting is that $\sigma(S)$ is the closed unit disk in $\mathbb{C}$. You have already shown this because your operator is $S^{\star}$, and $\sigma(S)=\sigma(S^{\star})$; you have done this by showing that every $|\lambda| < 1$ is an eigenvalue of $S^{\star}$. It then follows that $\sigma(S)=\sigma(S^{\star})$ is the closed unit disk because the spectrum is closed and contains the open unit disk, and the spectrum must be contained in the closed unit disk.

Finally, to see that your version of the shift operator--which is the backward shift--is the adjoint of the unilateral shift $S$, write $$ \begin{align} (Sx,y) & =(S\sum_{n}(x,e_n)e_n,y) \\ & =(\sum_{n}(x,e_n)e_{n+1},y) \\ & =\sum_{n}(x,e_n)(e_{n+1},y) \\ & = (x,\sum_{n}(y,e_{n+1})e_{n}). \end{align} $$ Hence, $S^{\star}y = \sum_{n}(y,e_{n+1})e_{n}$; this operator maps $e_{1}$ to $0$, $e_{2}$ to $e_{1}$, $e_3$ to $e_2$, etc..

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I think you are mixing up notation from two different sources. Your work shows that there are indeed eigenvalues (the open unit disc) for this definition of the unilateral shift. However there is another definition for the unilateral shift: $S(\alpha_1,\alpha_2,\ldots) = (0,\alpha_1,\alpha_2,\ldots)$. This incarnation has no eigenvalues. This incarnation is actually the adjoint of your choice for the unilateral shift.

For concreteness, you can consider the element $x=\left(1,\frac{1}{2},\frac{1}{4},\ldots\right)$, then

$$Sx = \left(\frac{1}{2},\frac{1}{4},\ldots\right) = \frac{1}{2}\left(1,\frac{1}{2},\ldots\right).$$

So your choice for the unilateral shift definitely has eigenvalues.

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  • $\begingroup$ Exactly your example is which I thought. But I checked the book and the definition of unilateral shift is $Se_n = e_{n+1}$ which I defined. $\endgroup$ – niki Nov 29 '14 at 18:57
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    $\begingroup$ Eigenvalues are part of the spectrum. In finite dimensional spaces, the spectrum and eigenvalues are synonymous. In infinite dimensional spaces, the eigenvalues may comprise all of the spectrum or only be part of it. However not every operator on infinite dimensions has eigenvalues. $\endgroup$ – Cameron Williams Nov 29 '14 at 19:01
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    $\begingroup$ I think either the text is wrong or maybe you're misinterpreting. Can you please tell me which text you are using? $\endgroup$ – Cameron Williams Nov 29 '14 at 19:01
  • $\begingroup$ Perhaps it would be better to redefine or label which is which by calling them different things. For example, $S_l e_n = e_{n-1}$ for $n\geq 2$ (Shift left) and $S_r e_n = e_{n+1}$ (shift right). The shift you used in the problem statement was $S_l$. As for the difference between eigenvalues and spectrum, the eigenvalues are the Point spectrum (commonly labeled as $P_\sigma$ or $\sigma_P$). The point spectrum is a subset of the spectrum itself($\sigma$), which contains also residual spectrum($\sigma_R$) and continuous spectrum($\sigma_C$). $\endgroup$ – JMoravitz Nov 29 '14 at 19:02
  • $\begingroup$ It's Example 2.3.2(page 49) of Murphy's C*-algebras and operator theory. $\endgroup$ – niki Nov 29 '14 at 19:06

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