CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Lambda identity minus compact has index zero

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a Banach space over $R$ or $C$ , let $K : X \to X$ be a compact operator (Compact linear operator) and let $λ \neq 0$ be a scalar. Then $λ I - K$ is a Fredholm operator and $ind (λ I - K) = 0$ (Fredholm operator cokernel and index).

Facts & Assumptions

[A1]

If $A$ is Fredholm and $C$ is compact then $A + C$ is Fredholm with $ind (A + C) = ind A$ (Fredholm index is stable under compact perturbations).

[A2]

A scalar multiple of a compact operator is compact (Linear combinations of compact operators are compact); the identity is bounded linear, $λ I$ is invertible with inverse $λ^{- 1} I$ for $λ \neq 0$ , and an invertible bounded operator is Fredholm of index $0$ , because its kernel and cokernel are the zero spaces (Linear map between vector spaces over the same field, A bounded linear operator between normed spaces, Fredholm operator cokernel and index, The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), Banach space).

Proof

technique · direct

Given: $A C$ , a Banach space $X$ over $R$ or $C$ , a compact $K : X \to X$ and a scalar $λ \neq 0$ .

1.1

The operator $λ I$ is bounded and invertible with inverse $λ^{- 1} I$ , hence Fredholm with $ind (λ I) = \dim {0} - \dim {0} = 0$ .

1.2

The operator $- K$ is compact by [A2], and $λ I - K = λ I + (- K)$ is a compact perturbation of the Fredholm operator $λ I$ .

A2algebra

2.1

By [A1] the operator $λ I - K$ is Fredholm and $ind (λ I - K) = ind (λ I) = 0$ , which is the claim.

step 1.1step 1.2A1∎

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Sources

Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §6.5 pp.187–188, Theorem 6.27 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis — §4.4, index of lambda I minus compact (standard reference, not scraped)