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 quasinilpotent operator need not be zero

Statement refuted

Assume AC. Every nonzero bounded operator has nonzero spectrum; equivalently, vanishing of the spectrum forces an operator to be zero.

Facts & Assumptions

[A1]

A nonzero normal operator with spectrum ${0}$ is zero (Normal operator with zero spectrum is zero).

[A2]

$T$ is normal when $T^{*} T = T T^{*}$ , and $z \in ρ (T)$ exactly when $z I - T$ is bijective with bounded inverse (Self-adjoint, positive, unitary and normal operators, Spectrum and resolvent of a bounded operator).

[A3]

The adjoint is characterised by $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ and depends conjugate-linearly on the entries of a matrix in an orthonormal basis (Hilbert-adjoint identities).

[A4]

AC is the hypothesis of the zero-spectrum corollary (The Axiom of Choice).

Counterexample

technique · direct

Given: The two-dimensional complex inner-product space with orthonormal basis $(e_{1}, e_{2})$ and the Jordan block $J = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ , so $J e_{1} = 0$ , $J e_{2} = e_{1}$ .

1.1

$J^{2} = 0$ and $J \neq 0$ , and $z I - J$ is invertible for every $z \neq 0$ with inverse $z^{- 1} (I + z^{- 1} J)$ , because $(z I - J) z^{- 1} (I + z^{- 1} J) = z^{- 1} (z I + J - J - z^{- 1} J^{2}) = z^{- 1} (z I - z^{- 1} J^{2}) = I$ using $J^{2} = 0$ .

A2algebra

1.2

$J$ is not normal: $J^{*} = (\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array})$ , so $J J^{*} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ while $J^{*} J = (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ , and these are different operators.

A2A3

2.1

$0 \in σ (J)$ because $J$ is not injective: $J e_{1} = 0$ while $e_{1} \neq 0$ , so $z I - J$ is not bijective at $z = 0$ ; hence $σ (J) = {0}$ .

step 1.1A2

3.1

The operator $J$ is therefore a nonzero bounded operator whose spectrum is the singleton ${0}$ — the property the title calls quasinilpotence — and it is not normal; by the zero-spectrum corollary this is only possible because normality fails.

step 2.1step 1.2A1

4.1

The statement that nonzero operators have nonzero spectrum, equivalently that zero spectrum forces vanishing, is refuted by the witness $J$ ; dropping the normality hypothesis from the zero-spectrum corollary is therefore not legitimate.

step 2.1step 3.1A4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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, §5.3, printed pp.235–245 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem, §4, pp.10–13 (standard reference, not scraped)