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=Qm∗AQm, and let Hmy=θy with ∥y∥2=1. If u=Qmy is the associated Ritz vector, then

Au−θu=hm+1,m(em∗y)vm+1.

In particular,

∥Au−θu∥2=∣hm+1,m∣ ∣em∗y∣.

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=Qm∗AQm 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
1.1F1L1algebra

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

2.1step 1.1L1algebra∎

Because Qmy=u, step 1.1 is exactly Au−θu=hm+1,m(em∗y)vm+1. The Arnoldi vectors are orthonormal, so ∥vm+1∥2=1, and the norm formula follows.

Depends on

Used by

Dependency tree · two levels

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Sources