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PropositionStatement: Literature-sourcedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)
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Before convergence, every CG denominator pk∗Apk is positive

Statement

Assume the conjugate-gradient recurrence of The conjugate-gradient recurrence is defined through step k for a Hermitian positive-definite matrix A. If rk≠0, then pk≠0 and

pk∗Apk=∥pk∥A2>0.

Facts & Assumptions

Given: A Hermitian positive-definite system Ax=b, a CG run through step k, and a nonzero residual rk.

[F1]

CG uses the recurrence αj=⟨rj,rj⟩⟨Apj,pj⟩,rj+1=rj−αjApj,βj=⟨rj+1,rj+1⟩⟨rj,rj⟩,pj+1=rj+1+βjpj with p0=r0 (The conjugate-gradient recurrence).

[L1]

For Hermitian positive-definite A, the energy norm satisfies ∥u∥A2=⟨Au,u⟩, and it is positive on nonzero vectors (The energy inner product ⟨u,v⟩A=⟨Au,v⟩ and energy norm ∥u∥A for a Hermitian positive-definite matrix).

Proof

technique · direct
1.1F1inductionalgebra

We first show by induction on j≤k that ⟨pj,rj⟩=⟨rj,rj⟩. For j=0 this is immediate from p0=r0 in [F1]. If it holds at j<k, conjugate symmetry gives ⟨rj,pj⟩=⟨rj,rj⟩, since the latter is real. The inner product is linear in its first argument, so [F1] gives ⟨rj+1,pj⟩=⟨rj,pj⟩−αj⟨Apj,pj⟩=0. Conjugate symmetry then gives ⟨pj,rj+1⟩=0, whence ⟨pj+1,rj+1⟩=⟨rj+1,rj+1⟩+βj⟨pj,rj+1⟩=⟨rj+1,rj+1⟩. Thus the identity holds for every j≤k.

2.1L1step 1.1algebra∎

At j=k, step 1.1 gives ⟨pk,rk⟩=⟨rk,rk⟩>0 because rk≠0. Hence pk≠0. By [L1], pk∗Apk=⟨Apk,pk⟩=∥pk∥A2>0, which is the desired positivity of the denominator.

Depends on

Used by

Cited to discharge well-definedness by The conjugate-gradient recurrence.

Dependency tree · two levels

5 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