CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A normal operator need not have any eigenvectors

Statement refuted

Assume AC. Every bounded normal operator on a nonzero complex Hilbert space has a nonzero eigenvector.

Facts & Assumptions

[A1]

$T = M_{x}$ on $L^{2} ([0, 1], λ)$ is bounded self-adjoint with $σ (T) = [0, 1]$ , spectral projections $E (B) = M_{1_{B}}$ , and the eigenvector identity is available in the form $E ({μ}) H = \ker (T - μ I)$ (Pvm of a multiplication operator, Spectral projections and resolution of the identity, Spectral theorem for bounded normal operators pvm form).

[A2]

In $L^{2}$ , a class $h$ satisfies $x h = μ h$ if and only if $(x - μ) h = 0$ almost everywhere; a product of a bounded measurable function with $h$ vanishes almost everywhere exactly when $h = 0$ almost everywhere off the zero set of the factor, and every representative of a nonzero $L^{2}$ class is nonzero on a set of positive measure (The space $L^{p} (μ)$ as the quotient by null functions, Borel functional calculus for a bounded normal operator).

[A3]

AC is the declared choice hypothesis of this page from the construction item onward (The Axiom of Choice).

Counterexample

technique · direct

Given: Lebesgue measure on $[0, 1]$ , the operator $T = M_{x}$ , and the spectral projection $E ({μ})$ of a point $μ \in C$ .

1.1

$T$ is normal, indeed bounded self-adjoint, with $σ (T) = [0, 1]$ .

1.2

Let $h \neq 0$ in $L^{2} ([0, 1])$ and suppose $T h = μ h$ . Then $(x - μ) h = 0$ almost everywhere, so $h = 0$ almost everywhere on the set ${x \neq μ}$ ; since that set is $[0, 1]$ minus at most one point, hence of full measure, $h = 0$ almost everywhere on $[0, 1]$ .

2.1

A class vanishing almost everywhere is the zero class, contradicting $h \neq 0$ ; hence no nonzero eigenvector exists at any $μ$ .

step 1.2A2

2.2

Consistently with the eigenvector identity $E ({μ}) H = \ker (T - μ I)$ , the spectral projection of each singleton is zero, since the measure $x \mapsto λ$ is nonatomic: $E ({μ}) = M_{1_{{μ}}} = 0$ in $L^{2}$ .

step 1.2A1

3.1

The bounded normal operator $M_{x}$ on $L^{2} ([0, 1])$ therefore has spectrum $[0, 1]$ and no nonzero eigenvectors, refuting the statement that every bounded normal operator has one.

step 1.1step 2.1step 2.2A3∎

Depends on

Used by

Direct integrals and general multiplicity theory Remark

Dependency tree · two levels

48 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.7, printed pp.293–296 (standard reference, not scraped)
Gerald Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., §4.1, printed pp.113–115 (standard reference, not scraped)