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

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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 $J e_{1} = 0$ , $J e_{2} = e_{1}$ .

1.1A2algebra

$J^{2} = 0$ 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 $J^{2} = 0$ .

1.2A2A3

$J$ 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.

2.1step 1.1A2

$0 \in σ (J)$ 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 $σ (J) = {0}$ .

3.1step 2.1step 1.2A1

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.

4.1step 2.1step 3.1A4∎

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.

Depends on

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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)