Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Normal operator with zero spectrum is zero

Statement

Assume AC. A bounded normal operator whose spectrum is {0} is the zero operator.

Facts & Assumptions

[A1]

For a bounded normal operator T on a nonzero complex Hilbert space, T=r(T)=max{λ:λσ(T)} (Normal operator norm equals spectral radius).

[A2]

The operator norm is the least bound of T, so T=0 forces Tx=0 for every x (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A3]

The spectrum is a subset of C; a normal operator is one with TT=TT, and the zero operator is normal (Spectrum and resolvent of a bounded operator, Self-adjoint, positive, unitary and normal operators).

[A4]

AC is the hypothesis of the norm-and-spectral-radius supplier (The Axiom of Choice).

Proof

technique · direct

Given: A nonzero complex Hilbert space H and a bounded normal TB(H) with σ(T)={0}.

1.1

The spectral radius is r(T)=max{λ:λσ(T)}=0=0.

A3
2.1

Hence T=r(T)=0 by the spectral-radius identity for normal operators.

step 1.1A1A4
3.1

Since T=0 is a bound for T, Tx0x=0 for every x, so Tx=0 for every x and T=0.

step 2.1A2

Depends on

Used by

Dependency tree · two levels

24 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