# Does every invertible $\mathbb{C}$-linear operator have an eigenvalue?

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?

Take any bounded operator $$S$$ with no eigen value and choose $$N$$ such that $$\|S\| . Let $$T=I+\frac S N$$. Then $$T$$ is invertible but it has no eigen value.
• 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 1:20
• 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 5:40