Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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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→X be a compact operator (Compact linear operator) and let λ≠0 be a scalar. Then λI−K is a Fredholm operator and ind⁡(λI−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, λI is invertible with inverse λ−1I for λ≠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→X/M, Banach space).

Proof

technique · direct

Given: AC, a Banach space X over R or C, a compact K:X→X and a scalar λ≠0.

1.1A2

The operator λI is bounded and invertible with inverse λ−1I, hence Fredholm with ind⁡(λI)=dim⁡{0}−dim⁡{0}=0.

1.2A2algebra

The operator −K is compact by [A2], and λI−K=λI+(−K) is a compact perturbation of the Fredholm operator λI.

2.1step 1.1step 1.2A1∎

By [A1] the operator λI−K is Fredholm and ind⁡(λI−K)=ind⁡(λI)=0, which is the claim.

Depends on

Used by

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Sources