LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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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.

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Isometric invariance of finite quantum relative entropy

Statement

Let $V : H \to K$ be a linear map between finite-dimensional complex inner-product spaces satisfying $V^{*} V = I_{H}$ , and let $A, B$ be non-negative on $H$ . Then

$D (V A V^{*} ∥ V B V^{*}) = D (A ∥ B) .$

Facts & Assumptions

Given: $V, A, B$ as in the Statement.

[F1]

Finite-dimensional quantum relative entropy: support inclusion determines the finite branch of $D$ ; its value is computed from spectral logarithms on the supports.

[F2]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ : $V^{*} V = I_{H}$ implies $⟨ V x, V y ⟩ = ⟨ x, y ⟩$ for every $x, y \in H$ .

[F3]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: non-negative operators have orthonormal eigenbases with nonnegative eigenvalues.

[F4]

For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ : trace is cyclic for compatible finite-dimensional operators.

[F5]

Every linear map between finite-dimensional inner product spaces has a unique adjoint: every linear map between finite-dimensional inner-product spaces has a unique linear adjoint, including zero-dimensional spaces.

Proof

1.1givenF5F2F3algebra

All adjoints of the finite-dimensional linear maps used below exist uniquely by [F5]. Let $M = ran V$ . Since $V$ is an isometry, it maps an orthonormal eigenbasis of $A$ to an orthonormal eigenbasis of $V A V^{*}$ on $M$ with the same nonzero eigenvalues; $V A V^{*}$ is zero on $M^{⊥}$ . The same holds for $B$ . Consequently $supp (V A V^{*}) = V (supp A)$ and $supp (V B V^{*}) = V (supp B)$ , so support inclusion holds on one side exactly when it holds on the other.

2.1givenF1F3F4step 1.1casesalgebra∎

On $M$ , spectral calculus gives $\ln (V A V^{*}) = V (\ln A) V^{*}$ on the positive support of $A$ , with the analogous identity for $B$ ; on $M^{⊥}$ both conjugated operators contribute zero to the support-restricted traces. Thus in the finite branch, cyclicity [F4] gives $tr (V A V^{*} \ln (V A V^{*})) = tr (A \ln A)$ and $tr (V A V^{*} \ln (V B V^{*})) = tr (A \ln B)$ . By [F1] their difference is $D (A ∥ B)$ . If support inclusion fails, step 1.1 shows both sides are $+ \infty$ by [F1].

Depends on

Used by

Dependency tree · two levels

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Sources

John Watrous, The Theory of Quantum Information, Proposition 5.19 (standard reference, not scraped)
Mark M. Wilde, From Classical to Quantum Shannon Theory, arXiv:1106.1445v8, Exercise 11.8.6 (standard reference, not scraped)