LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-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.

Inverse compression for positive resolvents

Statement

Let $H, K$ be finite-dimensional complex inner-product spaces, let $T : K \to K$ be positive definite, and let $V : H \to K$ satisfy $V^{*} V = I_{H}$ . For every $s \geq 0$ ,

$V^{*} (T + s I_{K})^{- 1} V ⪰ (V^{*} T V + s I_{H})^{- 1},$

where $A ⪰ B$ means that $A - B$ is non-negative. If $H = {0}$ , both sides are the unique endomorphism of $H$ .

Facts & Assumptions

Given: The spaces $H, K$ , the positive definite operator $T$ , the isometry $V$ , and $s \geq 0$ in the Statement.

[F1]

Real and complex inner-product spaces and their induced length: the inner product is linear in its first argument, conjugate-linear in its second, and positive definite.

[F2]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ : $⟨ V x, z ⟩ = ⟨ x, V^{*} z ⟩$ .

[F3]

Non-negative and positive operators: positive definiteness means self-adjointness and strictly positive quadratic form on nonzero vectors.

[F4]

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 are self-adjoint with nonnegative quadratic form.

[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

1.1givenF5F1F2F3algebra

All adjoints of the finite-dimensional linear maps used below exist uniquely by [F5]. Let $C$ be positive definite on a finite-dimensional space and $y$ any vector. The operator $C$ is injective by [F3], hence invertible in finite dimension; put $z_{0} = C^{- 1} y$ . Expanding with [F1]–[F2] gives $2 Re ⟨ y, z ⟩ - ⟨ C z, z ⟩ = ⟨ C^{- 1} y, y ⟩ - ⟨ C (z - z_{0}), z - z_{0} ⟩$ . The last term is nonnegative by [F3] and vanishes at $z = z_{0}$ , so $⟨ C^{- 1} y, y ⟩ = \max_{z} {2 Re ⟨ y, z ⟩ - ⟨ C z, z ⟩}$ .

2.1givenF2F3F4step 1.1algebra∎

If $H = {0}$ the assertion is immediate. Otherwise fix $x \in H$ , put $C = T + s I_{K}$ and $y = V x$ . By [F2] and $V^{*} V = I_{H}$ , $V^{*} C V = V^{*} T V + s I_{H}$ , which is positive definite. Restrict the maximum in step 1.1 to $z = V w$ ; the isometry identity gives $⟨ V x, V w ⟩ = ⟨ x, w ⟩$ and $⟨ C V w, V w ⟩ = ⟨ (V^{*} T V + s I_{H}) w, w ⟩$ . Hence $⟨ (T + s I_{K})^{- 1} V x, V x ⟩ \geq ⟨ (V^{*} T V + s I_{H})^{- 1} x, x ⟩$ . This holds for every $x$ , so the self-adjoint operator difference is non-negative by [F4], proving the order inequality.

Depends on

Used by

The compression Jensen inequality for the negative logarithm Lemma

Dependency tree · two levels

16 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

John Watrous, The Theory of Quantum Information, §5.2.3 (standard reference, not scraped)
Mark M. Wilde, From Classical to Quantum Shannon Theory, arXiv:1106.1445v8, §11.8 (standard reference, not scraped)