LemmaStatement: AI-adaptedProof: 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.

A compact Hilbert-space Gram operator is positive and has the forward kernel

Statement

Assume Countable Choice. Let $X, Y$ be real or complex Hilbert spaces and let $A : X \to Y$ be compact and bounded, with adjoint $A^{*} : Y \to X$ . Then both $A^{*} A : X \to X$ and $A A^{*} : Y \to Y$ are compact, self-adjoint and positive, and $\ker (A^{*} A) = \ker A, \ker (A A^{*}) = \ker A^{*} .$

Facts & Assumptions

Given: Countable Choice, real or complex Hilbert spaces $X, Y$ , a compact bounded linear operator $A : X \to Y$ , and its Hilbert adjoint $A^{*}$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is sufficient for the Hilbert adjoint used here; the proof does not assume full Choice.

[F1]

The Hilbert-space adjoint of a bounded operator: The adjoint satisfies $⟨ A x, y ⟩_{Y} = ⟨ x, A^{*} y ⟩_{X}$ .

[F2]

Compact linear operator: A compact operator maps bounded sets to sets with compact closure.

[F3]

Compositions with a compact operator are compact: A composition of a compact operator with a bounded operator is compact.

[F4]

Hilbert-adjoint identities: The adjoint of a composition reverses order, and $(A^{*})^{*} = A$ .

[F5]

Self-adjoint, positive, unitary and normal operators: Self-adjoint means $T^{*} = T$ ; positive means $⟨ T x, x ⟩ \geq 0$ for every $x$ .

[F6]

Real and complex inner product spaces, with the inner product linear in the first argument: The inner product is positive definite.

[F7]

The induced length is a norm: $∥ x ∥^{2} = ⟨ x, x ⟩$ .

Proof

1.1A1F2F3F4

Since $A$ is compact and $A^{*}$ is bounded, [F3] makes $A^{*} A$ compact. The same lemma makes $A A^{*}$ compact because it is the composition of the compact map $A$ after the bounded map $A^{*}$ . By [F4], $(A^{*} A)^{*} = A^{*} A$ and $(A A^{*})^{*} = A A^{*}$ , so both are self-adjoint.

1.2F1F5F7algebra

For every $x \in X$ , [F1] and [F7] give $⟨ A^{*} A x, x ⟩_{X} = ⟨ A x, A x ⟩_{Y} = ∥ A x ∥^{2} \geq 0$ . For every $y \in Y$ , the same identity gives $⟨ A A^{*} y, y ⟩_{Y} = ⟨ A^{*} y, A^{*} y ⟩_{X} = ∥ A^{*} y ∥^{2} \geq 0$ . Thus both operators are positive.

2.1F1F6step 1.2algebra∎

If $A x = 0$ , then $A^{*} A x = 0$ . Conversely, if $A^{*} A x = 0$ , step 1.2 implies $∥ A x ∥^{2} = ⟨ A^{*} A x, x ⟩ = 0$ , so positive definiteness [F6] gives $A x = 0$ . Hence $\ker (A^{*} A) = \ker A$ . The same argument with $A^{*}$ in place of $A$ proves $\ker (A A^{*}) = \ker A^{*}$ .

Depends on

Used by

Dependency tree · two levels

40 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

Guido Clason, Regularization of Inverse Problems, full lecture notes (standard reference, not scraped)
Per Christian Hansen, Introduction to Inverse Problems, Chapter 4: Regularization Methods (standard reference, not scraped)