Invertible linear operators without eigenvalues

Asked May 16 '22 at 14:37

Active May 16 '22 at 14:37

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Suppose $V$ is a vector space over the complex numbers, and let $\alpha: V \mapsto V$ be an invertible linear operator.

If the dimension of $V$ is finite, we know that $\alpha$ has (nonzero) eigenvalues, due to the fact that $\mathbb{C}$ is algebraically closed.

Are there interesting examples in the infinite-dimensional case (and still working over the complex numbers) without eigenvalues ?

Just to be clear: $\alpha$ is supposed to be invertible.

asked May 16 '22 at 14:37

Boccherini

Sorry, this post is a better duplicate. It explains how to obtain an invertible counterexample from a counterexample. – Dietrich Burde May 16 '22 at 14:53
A direct example: Multiplication with the function $\exp(i2\pi x)$ is an example of an invertible operator without eigenvalues on the Hilbert space $L^2([0,1])$. – s.harp May 17 '22 at 11:35
@s.harp: but this operator is not linear ? – Boccherini May 17 '22 at 15:29
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The operator is the following: $$L^2([0,1])\to L^2([0,1]),\qquad f\mapsto [x\mapsto e^{i2\pi x}f(x)]$$ this is linear. – s.harp May 17 '22 at 16:18
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Does this answer your question? Does every invertible $\mathbb{C}$-linear operator have an eigenvalue? (This is @DietrichBurde's second suggestion) – Theo Bendit May 17 '22 at 21:07

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