Finite number of eigenvalues outside circle $\|\lambda\|>\delta>0$

Question

I know that, for any $\delta>0$, a compact operator $A$ defined on a linear variety of a Banach space has only a finite number of linearly independent eigenvectors corresponding to the eigenvalues of absolute value $>\delta$.

I read, on the Italian language translation of Kolmogorov-Fomin's, that such a theorem implies that the number of eigenvalues $\lambda_n$ of such an operator outside circle $\|\lambda\|>\delta>0$ is always finite, but I don't understand the implication, since, from theorem 5, I only see that the quantity of the eigenvalues $\lambda_n$ such that $\|\lambda_n\|>\delta$ corresponding to linearly independent eigenvectors is finite, but there can be eigenvalues corresponding to a linear combination of such linearly independent eigenvectors...

Thank you very much for any help!

Answer 1

If $A x_k = \lambda _k x_k$, $x_k \ne 0$, $k \in \{1,...,n \}$, and $\lambda _i \ne \lambda _j$, $i \ne j$, then a vector of the form

$$ x = \sum _{k=1}^n a_k x_k, \ \ \ a_k \in \mathbb{R} $$ can be an eigenvector of $A$ only if $a_k=0$ for all $k \in \{1,...,n \}$ but one. Indeed, if $A x = \lambda x$, then the linear independence of $x_1,...,x_n$ implies $a_k \lambda _k = a_k \lambda$, $k \in \{1,...,n \}$. I hope I have understood correctly the question.

Answer 2

I understand your confusion. There's an error in the critical equation of the proof that, once corrected, makes everything obvious. I'll rewrite it for you $$ \left\|A\left(\frac{y_{q}}{\lambda_{q}}\right)-A\left(\frac{y_{p}}{\lambda_{p}}\right)\right\| = \left\|\left[A\left(\frac{y_{q}}{\lambda_{q}}\right)+y_{p}-A\left(\frac{y_{p}}{\lambda_{p}}\right)\right]-y_{p}\right\| \ge \frac{1}{2}. $$ The above holds because $y_{p}-A(\frac{1}{\lambda_{p}}y_{p})\in E_{p-1}$ and $A(\frac{1}{\lambda_{q}}y_{q})\in E_{q}\subseteq E_{p-1}$ for $q < p$.