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Let $T:X\rightarrow X$ be an invertible linear operator over a complex vector space $X$ (possibly infinite-dimensional), then does $T$ always have an eigenvalue? We may assume that $X$ is a separable Hilbert space if necessary.

I know this is true for finite dimensions by the fundamental theorem of algebra, but how about for infinite dimensions?

Andrew Yuan
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1 Answers1

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Take any bounded operator $S$ with no eigenvalue and choose $N$ such that $\|S\| <N$. Let $T=I+\frac S N$. Then $T$ is invertible but it has no eigenvalue.

metamorphy
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Kavi Rama Murthy
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  • I don't understand your answer. Why would $S$ be bounded. Why not give an example $T \sum_{n=-\infty}^\infty c_n e_n = \sum_{n=-\infty}^\infty c_{n+1} e_n $ whose spectrum is $\Bbb{C}^*$ with eigenvectors the distributions $\sum_n e^{zn} e_n$ – reuns Jul 19 '19 at 01:20
  • I am giving a counterexample. So I can start with a bounded operator $S$ with no eigen value. There are plenty of such operators on a separable Hilbert space. – Kavi Rama Murthy Jul 19 '19 at 05:40
  • @KaviRamaMurthy: why is this operator invertible ? I see that it is injective, but why is it surjective ? – Boccherini May 16 '22 at 16:47
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    @Boccherini $I+U$ is invertible with inverse $I-U+U^{2}-U^{3}+\cdots$ whenever $|U| <1$ and the space is any complete normed linear space. This is a standard result found in any book on FA. – Kavi Rama Murthy May 16 '22 at 23:14