Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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.

Galerkin orthogonality turns residual orthogonality into A-orthogonality of the error, and the Galerkin iterate is unique

Statement

Let A be Hermitian positive definite, let Ax=b, let x0 be an initial guess, let r0=b−Ax0, let x∗ be the exact solution of Ax=b, and let xm∈x0+Km(A,r0). Then the following are equivalent:

  1. b−Axm is orthogonal to Km(A,r0) in the standard inner product.
  2. x∗−xm is orthogonal to Km(A,r0) in the energy inner product.

When these conditions hold, the vector xm is the unique element of x0+Km(A,r0) with that property.

Facts & Assumptions

Given: A Hermitian positive-definite system Ax=b, an initial guess x0, r0=b−Ax0, the exact solution x∗, and a vector xm∈x0+Km(A,r0).

[F1]

A Krylov Galerkin iterate is an element of x0+Km(A,r0) whose residual is orthogonal to Km(A,r0) (The mth Krylov Galerkin iterate in the affine space x0+Km(A,r0) for a Hermitian positive-definite system).

Proof

technique · direct
1.1F1L1algebra

For every v∈Km(A,r0), the identity Ax∗=b gives ⟨b−Axm,v⟩=⟨A(x∗−xm),v⟩=⟨x∗−xm,v⟩A by [L1]. Hence the residual is orthogonal to Km(A,r0) if and only if the error is A-orthogonal to Km(A,r0).

2.1L1step 1.1algebra∎

Suppose xm and ym both lie in x0+Km(A,r0) and both satisfy the equivalent conditions from step 1.1. Then d:=xm−ym lies in Km(A,r0), and both errors are A-orthogonal to d. Subtracting the two orthogonality relations gives 0=⟨x∗−xm,d⟩A−⟨x∗−ym,d⟩A=⟨ym−xm,d⟩A=−⟨d,d⟩A. Positive definiteness of the energy inner product therefore forces d=0, so xm=ym.

Depends on

Used by

Dependency tree · two levels

6 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