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Algebraic multiplicity of a nonzero compact-operator eigenvalue

Definition

Assume the Axiom of Choice (The Axiom of Choice). Let $T$ be a compact operator on a complex Hilbert space $H$ , and let $λ \neq 0$ belong to its spectrum. Riesz–Schauder stabilization supplies an integer $m_{0}$ for which the kernels of $(I - T / λ)^{m}$ are constant for $m \geq m_{0}$ . Define the generalized eigenspace and algebraic multiplicity by

$G_{λ} (T) : = \ker (T - λ I)^{m_{0}}, m_{a l g} (λ; T) : = \dim G_{λ} (T) .$

The value is independent of the choice of stabilized exponent. If $P_{λ}$ is the Riesz spectral projection of $T$ at $λ$ , then $G_{λ} (T) = ran P_{λ}$ ; in particular the algebraic multiplicity is the finite rank of that projection.

Facts & Assumptions

Given: AC; a complex Hilbert space $H$ ; a compact $T : H \to H$ ; and $λ \neq 0$ in $σ (T)$ .

[A1]

For compact $T$ , each nonzero spectral value is an eigenvalue with finite-dimensional generalized eigenspace (Riesz schauder spectrum of a compact operator).

[A2]

For every $ε > 0$ , only finitely many spectral values of a compact operator have modulus at least $ε$ (Riesz schauder spectrum of a compact operator).

[A3]

If $A = I - K$ with $K$ compact on a Banach space and DC holds, the kernels of $A^{m}$ stabilize at a finite index (Riesz Schauder ascent and descent stabilize).

[A4]

For the isolated spectral set $E = {λ}$ , the Riesz spectral projection is defined as $P_{λ} = χ_{E} (T)$ (Riesz spectral projection).

[A5]

The Riesz projection is idempotent, commutes with $T$ , and splits $H$ into its closed invariant range and kernel (Riesz spectral projection properties).

[A6]

If nonzero, the restriction spectra on $ran P_{λ}$ and $\ker P_{λ}$ are respectively ${λ}$ and $σ (T) ∖ {λ}$ (Riesz spectral projection properties).

[A7]

A subspace $W$ is $T$ -invariant when $T (W) \subseteq W$ (Invariant subspaces, restrictions, and induced quotient operators).

[A8]

A compact operator on an infinite-dimensional Banach space has $0$ in its spectrum (Riesz schauder spectrum of a compact operator).

[A9]

Every nonconstant complex polynomial has a root (Fundamental theorem of algebra by Liouville's theorem); an endomorphism of a finite-dimensional space whose characteristic polynomial splits has a Jordan form (Jordan form over the base field exists exactly when the characteristic polynomial splits).

[A10]

In ZF, AC implies DC (AC implies DC implies countable choice).

Proof

technique · direct

1.1A1A3A10

With $A = I - T / λ$ , we have $T - λ I = - λ A$ . The declared AC assumption supplies DC by [A10], so [A3] gives an $m_{0}$ for which $\ker (T - λ I)^{m} = \ker A^{m}$ is constant for every $m \geq m_{0}$ ; [A1] makes this stabilized space finite-dimensional.

1.2A2A4construct

Put $ε = ∣ λ ∣ / 2$ . By [A2] the set $S_{ε} = {μ \in σ (T) : ∣ μ ∣ \geq ε}$ is finite. Every spectral point within distance $∣ λ ∣ / 2$ of $λ$ lies in $S_{ε}$ ; because $λ \in S_{ε}$ , choose a smaller positive radius excluding the finitely many other points of $S_{ε}$ . Thus $λ$ is isolated and [A4] defines $P_{λ}$ .

2.1A4A5A6A7A8step 1.2

Let $P = P_{λ}$ from [A4]. By [A5] and the meaning of invariant subspace in [A7], $H = ran P \oplus \ker P$ with both summands closed and $T$ -invariant. The restriction of $T$ to $ran P$ is compact. If this range were infinite-dimensional, [A8] would put $0$ in its restriction spectrum, contrary to [A6], which gives that spectrum as ${λ}$ and $λ \neq 0$ . Hence $ran P$ is finite-dimensional.

3.1A6A9step 2.1

If $ran P = {0}$ , then $P = 0$ and [A6] would give $σ (T) = σ (T ∣_{\ker P}) = σ (T) ∖ {λ}$ , impossible since $λ \in σ (T)$ . Thus the range is nonzero. Its finite-dimensional restriction has spectrum ${λ}$ by [A6]. By [A9], its characteristic polynomial splits over $C$ and it has a Jordan form; all Jordan blocks have eigenvalue $λ$ . Therefore $(T - λ I)^{d}$ vanishes on $ran P$ , where $d = \dim ran P$ , so $ran P \subseteq G_{λ} (T)$ .

4.1A5A6step 1.1step 2.1∎

Conversely, take $x \in G_{λ} (T)$ and write $x = P x + (I - P) x$ using [A5]. Since $P$ commutes with $T$ , the second term lies in $\ker P$ and is killed by a power of $T - λ I$ . If $\ker P \neq {0}$ , [A6] says $λ$ is outside the spectrum of $T ∣_{\ker P}$ , so $T - λ I$ is invertible there and the second term is zero. If $\ker P = {0}$ it is zero directly. Hence $x \in ran P$ , and $G_{λ} (T) = ran P$ . The rank and dimension are finite by step 2.1, and step 1.1 proves independence of the stabilized exponent.

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Sources

Kostenko, Trace Ideals with Applications, §3.4 (standard reference, not scraped)
van Neerven, Functional Analysis, §14.5.a (standard reference, not scraped)
Dyatlov–Zworski, Mathematical Theory of Scattering Resonances, Appendix B §§B.5–B.6 (standard reference, not scraped)