2
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $\sum_n e^{zn} e_n$ $\endgroup$ – reuns Jul 19 '19 at 1:20
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Jul 19 '19 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.