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PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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The GMRES residual has the form rm=pm(A)r0 with pm(0)=1 and degpmm

Statement

Let xm be an mth GMRES iterate for Ax=b with initial residual r0=bAx0, and let rm=bAxm. Then there is a polynomial pm such that

rm=pm(A)r0,pm(0)=1,degpmm.

Moreover, among all polynomials p with degpm and p(0)=1, GMRES chooses one minimizing p(A)r02.

Facts & Assumptions

Given: A GMRES iterate xm with residual rm=bAxm.

[L1]

Every vector in Km(A,r0) has the form q(A)r0 with degq<m (The Krylov subspace consists exactly of the vectors p(A)b for zero polynomials or polynomials of degree less than m).

[L2]

GMRES minimizes the residual norm over x0+Km(A,r0) (GMRES minimizes the residual norm over the affine Krylov space x0+Km(A,r0)).

Proof

technique · direct
1.1

Since xmx0Km(A,r0), [L1] gives a polynomial q with degq<m and xm=x0+q(A)r0. Therefore rm=bAxm=r0Aq(A)r0=pm(A)r0, where pm(z):=1zq(z). Then pm(0)=1 and degpmm.

L1constructalgebra
2.1

Conversely, every polynomial p with degpm and p(0)=1 can be written as p(z)=1zq(z) with degq<m. The corresponding vector x=x0+q(A)r0 lies in x0+Km(A,r0) by [L1], and its residual is p(A)r0. Since [L2] makes xm minimize that residual norm over all such x, the polynomial attached in step 1.1 minimizes p(A)r02 among all admissible p.

L1L2step 1.1

Depends on

Used by

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