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Is there a normal operator which has no eigenvalues?

If your answer is yes, give an example.


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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
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

Example 1
'I think "shift operator or translation operator" is one of them.' – Ali Qurbani

Indeed, the bilateral shift operator on $\ell^2$, the Hilbert space of square-summable two-sided sequences, is normal but has no eigenvalues. Let $L:\ell^2 \to \ell^2$ be the left shift operator, $R:\ell^2 \to \ell^2$ the right shift operator and $\langle\cdot,\cdot\rangle$ denote the inner product. Take $x=(x_n)_{n\in \mathbb{Z}}$ and $y=(y_n)_{n\in \mathbb{Z}}$ two sequences in $\ell^2$: $$\langle Lx, y\rangle = \sum_{n \in \mathbb{Z}} x_{n+1}\cdot y_n = \sum_{n \in \mathbb{Z}} x_n\cdot y_{n-1} = \langle x, Ry\rangle,$$

hence $L^*=R=L^{-1}$, i.e. $L$ is unitary.
Now let $\lambda$ be a scalar and $x\in\ell^2$ such that $Lx = \lambda x$ then $x_n = \lambda^n x_0$ holds for $n \in \mathbb{Z}$ and we have $$\|x\|^2=\sum_{n \in \mathbb{Z}} x_n^2 = x_0\left( \sum_{n=1}^\infty \lambda^{2n} + \sum_{n=0}^{-\infty} \lambda^{2n} \right).$$ The first sum diverges for $|\lambda|\geq 1$ and the second sum diverges for $|\lambda|\leq 1$ so the only $x\in\ell^2$ solving the equation must be the zero sequence which cannot be an eigenvector, hence $L$ has no eigenvalues.

Example 2
As Christopher A. Wong pointed out you can construct another example with a multiplication operator. Let $L^2$ be the Hilbert space of Lebesgue-square-integrable functions on $\mathbb{R}$ and $M:L^2\to L^2,\ f \mapsto f \cdot h$ where $$h(x) = \begin{cases} 1, &x \in [0,1] \\ 0, &\text{else} \end{cases}.$$ For $f,g\in L^2$ we have $$\langle Mf,g\rangle = \int_\mathbb{R} f\cdot h\cdot g \ dx = \langle f,Mg\rangle,$$ i.e. $M$ is self adjoint.
Now let $\lambda$ be a scalar and $f\in L^2$ such that $(M-\lambda)f = 0$ which is only true if $f=0$ or if $h=\lambda$ hence there are no eigenvalues.

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