Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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An Arnoldi Ritz pair has residual norm controlled by the last Hessenberg subdiagonal entry

Statement

Assume Arnoldi runs through step m without breakdown, let Hm=QmAQm, and let Hmy=θy with y2=1. If u=Qmy is the associated Ritz vector, then

Auθu=hm+1,m(emy)vm+1.

In particular,

Auθu2=hm+1,memy.

Facts & Assumptions

Given: An Arnoldi run through step m without breakdown, a unit eigenvector y of Hm, and the Ritz vector u=Qmy.

[F1]

In Arnoldi, Ritz values are eigenvalues of Hm=QmAQm and Ritz vectors have the form u=Qmy (Ritz values and Ritz vectors extracted from the Arnoldi Hessenberg reduction).

[L1]

Arnoldi gives AQm=Qm+1Hˉm, where Hˉm=[Hmhm+1,mem] (Before breakdown, Arnoldi produces an orthonormal Krylov basis and a rectangular upper-Hessenberg factorization).

Proof

technique · direct
1.1

Using [F1] and [L1], Au=AQmy=Qm+1Hˉmy=QmHmy+hm+1,m(emy)vm+1. Since Hmy=θy, this becomes Au=θQmy+hm+1,m(emy)vm+1.

F1L1algebra
2.1

Because Qmy=u, step 1.1 is exactly Auθu=hm+1,m(emy)vm+1. The Arnoldi vectors are orthonormal, so vm+12=1, and the norm formula follows.

step 1.1L1algebra

Depends on

Used by

Dependency tree · two levels

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