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=bAx0, let x be the exact solution of Ax=b, and let xmx0+Km(A,r0). Then the following are equivalent:

  1. bAxm is orthogonal to Km(A,r0) in the standard inner product.
  2. xxm 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=bAx0, the exact solution x, and a vector xmx0+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
1.1

For every vKm(A,r0), the identity Ax=b gives bAxm,v=A(xxm),v=xxm,vA by [L1]. Hence the residual is orthogonal to Km(A,r0) if and only if the error is A-orthogonal to Km(A,r0).

F1L1algebra
2.1

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

L1step 1.1algebra

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.

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