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TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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Arnoldi reduces GMRES to a least-squares problem for the small Hessenberg matrix

Statement

Let r0=b−Ax0≠0 and set β=∥r0∥2. Run Arnoldi on (A,r0) through step m without breakdown, so that v1=r0/β. Then every vector x∈x0+Km(A,r0) has the form x=x0+Qmy for some y∈Fm, and

b−Ax=Qm+1(βe1−Hˉmy).

Consequently,

min⁡x∈x0+Km(A,r0)∥b−Ax∥2=min⁡y∈Fm∥βe1−Hˉmy∥2.

Facts & Assumptions

Given: A linear system Ax=b, an initial guess x0, the nonzero initial residual r0=b−Ax0, and an Arnoldi run on (A,r0) through step m without breakdown.

[F1]

A GMRES iterate minimizes the residual norm over x0+Km(A,r0) (The GMRES iterate as the residual minimizer over an affine Krylov space).

[L1]

Arnoldi gives an orthonormal basis of Km(A,r0) and the factorization AQm=Qm+1Hˉm with Qm+1∗Qm+1=I (Before breakdown, Arnoldi produces an orthonormal Krylov basis and a rectangular upper-Hessenberg factorization).

Proof

technique · direct
1.1L1constructalgebra

By [L1], the columns of Qm form a basis of Km(A,r0), so every x∈x0+Km(A,r0) can be written uniquely as x=x0+Qmy. Since r0=βv1=βQm+1e1 and b−Ax0=r0, one has b−Ax=r0−AQmy=Qm+1βe1−Qm+1Hˉmy=Qm+1(βe1−Hˉmy).

2.1F1L1step 1.1∎

The columns of Qm+1 are orthonormal by [L1], so multiplying by Qm+1 preserves the Euclidean norm. Therefore ∥b−Ax∥2=∥βe1−Hˉmy∥2. Taking minima over the corresponding parametrized sets gives the stated small least-squares problem. This is the residual norm whose minimizers [F1] calls GMRES iterates.

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