Why are nonzero eigenvalues of a compact operator poles of its resolvent?

Question

Let $X$ be a complex Banach space and $T$ a compact operator on $X$. I read on Wikipedia that nonzero elements of the spectrum of $T$ are poles of its resolvent. It says "by functional calculus", but I don't see how.

I'm looking for a reference or a sketch of the proof. At the moment I have no idea how it goes.

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By reading Yosida's "Functional Analysis" I know that an eigenvalue is a pole when its residue has finite rank (VIII.8, Theorem 4). But I don't see why / know if residues have finite rank when $T$ is compact.

I can show that those isolated singularities are poles when $T$ is normal and trace class, using the spectral theorem and by investigating its Laurent-coefficients. But I don't want to assume that $T$ is normal.

Answer 1

A proof that the residue of the resolvent at an eigenvalue $\lambda \neq 0$ has finite rank, is sketched at that same wikipedia page, in the section "Invariant subspaces". The residue is called the Riesz projection $E(\lambda)$.

Combinated with that result from Yosida's book, the conclusion follows.

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Since the Wikipedia page does not give much details, here are some more: The invertibility of the restriction of $C:=T$ to $E(\lambda)$ follows from the properties of holomorphic functional calculus: $$\int \frac{C}{\xi - C} d\xi = \int \frac{\xi}{\xi - C} d\xi = \left(\int \frac{\xi^{-1}}{\xi - C} d\xi \right)^{-1}$$ Thus $C$ is compact and invertible on the range of $E(\lambda)$, hence this space has finite dimension. We can stop here and use Yosida's Theorem 4, but there is a shortcut:

That the spectrum of the restriction is $\{\lambda\}$ follows from the spectral mapping theorem for holomorphic functional calculus.

Finally, by finite-dimensional linear algebra, we have $(C-\lambda)^nE(\lambda) = 0$ for $n$ sufficiently large, and from the following formula for the Laurent coefficients of the resolvent: $$A_{-(n+1)} = (C-\lambda I)^n A_{-1}$$ (Cf. Yosida VIII.8 Theorem 2, but this is elementary) we conclude that $\lambda$ is a pole.

Answer 2

If a bounded operator $T$ has an isolated point $\lambda_0$ of its spectrum $\sigma(T)$, then the resolvent is holomorphic in a punctured neighborhood of $\lambda_0$, which ensures a Laurent series expansion with operator coefficients $C_n$: $$ (T-\lambda I)^{-1} = \sum_{n=-\infty}^{\infty}(\lambda-\lambda_0)^n C_n,\;\;r < |\lambda-\lambda_0| < R $$ If you multiply by $T-\lambda_0 I$, you find $$ (T-\lambda_0 I)(T-\lambda I)^{-1}=I+(\lambda-\lambda_0)(T-\lambda I)^{-1}\\ \sum_{n=-\infty}^{\infty}(\lambda-\lambda_0)^n(T-\lambda_0 I)C_n = I+\sum_{n=-\infty}^{\infty}(\lambda-\lambda_0)^{n+1}C_n. $$ Equating coefficients gives $$ (T-\lambda_0 I)C_n=C_{n-1},\;\; n \ne 0. $$ Therefore $(T-\lambda_0I)^kC_{-1}=C_{-k-1}$.

The following is a projection: $$ C_{-1} = \frac{1}{2\pi i}\oint_{|\lambda-\lambda_0|=r}(\lambda I-T)^{-1}d\lambda. $$ That is, $C_{-1}^2=C_{-1}$. Furthermore, $(T-\lambda_0I)$ is nilpotent on $C_{-1}X$, provided $r$ is chosen so that the punctured closed disk $\{ \lambda : 0 < |\lambda-\lambda_0| \le r\}$ is contained in the resolvent set of $T$. Because $C_{-N-1}=(T-\lambda_0 I)^{N}C_{-1}=0$ for some $N$, it follows that the resolvent has a pole at $\lambda_0$.