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My friend and I are a little bit confused about this, the question we are working on asks us to find the eigenvalues of $R$, the right shift operator. The main point of contention is that $\sigma(R)$ is non-empty, and so should surely(?) contain an eigenvalue of $R$. Does the non-emptiness of $\sigma(R)$ only apply in finite dimensional Hilbert spaces? A detailed reply to clear up the confusion would be appreciated.

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    $\begingroup$ $\sigma(R)$ is the set of points $\lambda\in\Bbb C$ so that $R-\lambda$ is not invertible. Unlike in finite dimensional spaces, $\lambda\in\sigma(R)$ does not imply that $\lambda$ is an eigenvalue of $R$. Indeed $R$ has no eigenvalues, but $\sigma(R)$ is the entire (closed) unit disk in $\Bbb C$. $\endgroup$ – s.harp Feb 5 '17 at 19:10
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For a bounded linear operator on a complex Hilbert space, the spectrum is always non-empty. But it may lack eigenvalues.

If $H=\ell^2(\mathbb N)$ and $R$ is the shift $$ R(a_1,a_2,\ldots)=(0,a_1,s_2,\ldots), $$ then $R$ has no eigenvalues. It's spectrum, though, is $\{\lambda:\ |\lambda|\leq1\}$. The easiest way to see it is to consider the left shift $$ L(a_1,a_2,\ldots)=(a_2,a_3,\ldots). $$ For any $\lambda$ with $|\lambda|<1$, one can check that $\lambda$ is an eigenvalue of $L$ with eigenvector $(\lambda,\lambda^2,\lambda^3,\ldots)$. As the spectrum is closed and $\|L\|\leq1$, one deduces that $$\sigma(L)=\{\lambda:\ |\lambda|\leq1\}.$$ Then $$ \sigma(R)=\{\bar\lambda:\ \lambda\in\sigma(L)\}=\{\lambda:\ |\lambda|\leq1\}. $$

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