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Is there a normal operator which has no eigenvalues?

If your answer is yes, give an example.

Thanks.

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I think "shift operator or translation operator" is one of them. –  Ali Qurbani Feb 6 '13 at 20:50
    
this is a bit abstract to me and I'm not sure I know what you mean. But the translation operator has the so called momentum states as eigenvalues. –  Guest 86 Feb 6 '13 at 20:54
    
The "momentum states" (i.e., plane waves) are not square-integrable, so they are not in the Hilbert space $L^2$. –  mjqxxxx Feb 6 '13 at 20:56
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2 Answers

up vote 1 down vote accepted

The answer is yes, if you allow the normal operator to be non-compact. The spectral theorem for compact normal and self-adjoint operators guarantees eigenvalues, but in general, for general bounded (or even unbounded) operators, the "big" spectral theorem is much more complicated and does not guarantee eigenvalues.

To find a counterexample, pick your Banach space to be an appropriate function space, and make your operator a multiplication operator. e.g. make your operator $T(f(x)) = g(x) f(x)$, for some fixed function $g(x)$.

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The operator $L(\mathbb R^{2})$ defined by $T(x,y)=(-y,x)$ is normal, but it has no eigenvalue

$$T: \mathbb R^{2} \rightarrow \mathbb R^{2}~ ,~(x,y)‎ \mapsto (-y,x)~;~x,y \in \mathbb R $$

T is normal : $$TT^{*}=T^{*}T$$

T has not eigenvalue in $\mathbb R^{2}$.

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This operator has complex eigenvalues. –  Christopher A. Wong Feb 6 '13 at 21:25
    
This operator in $R^{2}$ has not eigenvalue –  Ali Qurbani Feb 6 '13 at 21:29
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