Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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.

Trace preservation is Kraus normalization

Statement

Let H,K be finite-dimensional complex inner-product spaces and let Φ(X)=∑a∈AKaXKa† be a finite Kraus representation, where each Ka:H→K. The map Φ is trace-preserving on all of L(H), meaning tr⁡K(Φ(X))=tr⁡H(X) for every X∈L(H), if and only if ∑a∈AKa†Ka=IH.

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces H,K, a finite family Ka:H→K, and the indicated Kraus formula for every X∈L(H).

[F1]

Finite Kraus representation: a finite Kraus representation satisfies Φ(X)=∑aKaXKa† on every input operator.

[F2]

For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA): for A:H→K and B:K→H, tr⁡K(AB)=tr⁡H(BA), including zero dimensions.

[F3]

Trace is a linear functional on Mn(F): trace is linear over finite sums.

[F4]

The trace pairing on a finite endomorphism space is nondegenerate: if tr⁡H(AX)=0 for every X∈L(H), then A=0, including H=0.

Proof

1.1givenF1F2F3algebra

Put S=∑aKa†Ka∈L(H). For every X∈L(H), [F1], trace linearity [F3], and rectangular cyclicity [F2] give tr⁡K(Φ(X))=∑atr⁡K(KaXKa†)=∑atr⁡H(Ka†KaX)=tr⁡H(SX). If S=IH, this equals tr⁡H(X) for every X, so Φ is trace-preserving by the Statement. This includes an empty family when H=0, where IH=0.

2.1givenstep 1.1F4algebra∎

Conversely, suppose Φ is trace-preserving on all of L(H). The identity in step 1.1 gives tr⁡H((S−IH)X)=0 for every X∈L(H). Apply [F4] with A=S−IH to obtain S−IH=0. Thus the normalization holds. In particular, if H=0 both sides are the unique zero endomorphism and the argument still applies.

Depends on

Used by

Dependency tree · two levels

25 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