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Arnoldi action error is the semigroup-weighted residual integral

Statement

Let A∈Mn(C), n≥1, v∈Cn, β≥0, V∈Mn×m(C), H∈Mm(C), and 1≤m≤n. For exact Arnoldi, β=∥v∥2; in a certified finite-precision run, β may be the backend's returned nonnegative scalar, with its effect included in the initial mismatch. Put c(s):=esHβe1 and u(s):=Vc(s). For every real t, the exact defect identity is

etAv−u(t)=etA(v−βVe1)+∫0te(t−s)A(AV−VH)c(s) ds,

where the integral is oriented. Let It=[0,t] for t≥0 and It=[t,0] for t<0; let σt=1 for t≥0 and σt=−1 for t<0. If t=0 the integral is zero. Suppose the Arnoldi relation is

AV=VH+hvm+1em∗+D+Ω,

where h≥0, ∥vm+1∥2=1 when h>0, D=[d1,…,dm], ηj≥0, ∥dj∥2≤ηj, ηorth≥0, and ∥Ω∥2≤ηorth. In exact happy breakdown, interpret the rank-one term as zero and set D=Ω=0. If a reduced computation returns c^(t) with ∥c^(t)−c(t)∥2≤ϵred(t) for ϵred(t)≥0 and its final projection returns x^ with ∥x^−Vc^(t)∥2≤ηproj for ηproj≥0, then

∥etAv−x^∥2≤e∣t∣μ2(σtA)∥v−βVe1∥2+∫Ite∣t−s∣μ2(σtA)(h∣em∗c(s)∣+(∑j=1mηj2+ηorth)∥c(s)∥2)ds+∥V∥2ϵred(t)+ηproj.

For exact Arnoldi from a nonzero start with no breakdown through m, D=Ω=0, V has orthonormal columns, and v=βVe1, recovering the standard residual integral. Whenever AV=VH and v=βVe1, the invariant-subspace relation gives φk(tA)v=βVφk(tH)e1 for every k≥0 and every real t; exact happy breakdown of Arnoldi from a nonzero start is a sufficient case.

Facts & Assumptions

Given: The matrices and vectors in the Statement; for the numerical relation clause, the stated decomposition and columnwise error bounds; for the reduced term, a certified bound ϵred(t) and a final projection certificate ∥x^−Vc^(t)∥2≤ηproj.

[F1]

Variation of constants for a finite-matrix action defect: for a continuously differentiable approximate trajectory u, the exact action error is its initial error plus the oriented integral of e(t−s)A(Au−u′).

[F2]

The logarithmic norm bounds forward and reverse matrix exponentials: for τ≥0, ∥eτB∥2≤eτμ2(B) for every finite complex matrix B.

[F3]

Before breakdown, Arnoldi produces an orthonormal Krylov basis and a rectangular upper-Hessenberg factorization: before breakdown, Arnoldi columns are orthonormal and AVm=VmHm+hm+1,mvm+1em∗.

[F4]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: for a finite-dimensional linear map T, ∥T∥2=max⁡∥x∥2=1∥Tx∥2; rescaling gives ∥Tx∥2≤∥T∥2∥x∥2 for all x.

[F5]

Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors: ∣⟨x,y⟩∣≤∥x∥2∥y∥2 in a complex Euclidean space.

[F6]

The Arnoldi process for building an orthonormal basis of a Krylov subspace: Arnoldi starts with v1=v/∥v∥2, defines wj=Avj−∑i=1jhijvi, and exact breakdown at step j means hj+1,j=∥wj∥2=0.

[F7]

The exponential series of a bounded operator: for a bounded operator on a Banach space, its exponential is differentiable with derivative AetA.

[F8]

The phi-function family by everywhere-convergent series and The scalar and matrix phi series converge absolutely: every finite-matrix phi function is defined by an absolutely convergent power series.

[F9]

Every finite-dimensional normed space is Banach: each finite-dimensional normed space over C is complete.

[A1]

For a finite-precision relation, AV=VH+hvm+1em∗+D+Ω and all displayed norm bounds are certified data; the theorem does not infer them from a small computed subdiagonal.

[A2]

The final computed projection satisfies ∥x^−Vc^(t)∥2≤ηproj for a nonnegative bound ηproj, with ηproj=0 when the projection is exact.

Proof

technique · direct
1.1givenF4F7F9algebra

For any nonzero x, rescaling x/∥x∥2 in [F4] shows that the finite matrix H is bounded; the inequality is immediate at x=0. By [F9], Cm is complete, so [F7] applies and gives c′(s)=Hc(s). Hence u(s)=Vc(s) is continuously differentiable and its defect is Au(s)−u′(s)=(AV−VH)c(s).

1.2givenF2algebra

For s∈It, ∣t−s∣≥0 and e(t−s)A=e∣t−s∣σtA; the same identity gives etA=e∣t∣σtA. Thus [F2] bounds the initial factor by e∣t∣μ2(σtA) and the integral kernel by e∣t−s∣μ2(σtA).

2.1step 1.1F1

Apply [F1] to this u. It gives the displayed exact integral identity, including the initial mismatch v−u(0)=v−βVe1.

3.1givenA1F4F5step 2.1step 1.2algebra

Substitute the certified relation [A1] into the defect in step 2.1. The rank-one term has norm h∣em∗c(s)∣ because ∥vm+1∥2=1. For any unit vector x∈Cm, Cauchy–Schwarz [F5] gives ∥Dx∥2≤∑j∣xj∣ηj≤(∑jηj2)1/2; hence ∥D∥2≤(∑jηj2)1/2. The remaining term is at most ηorth∥c(s)∥2. Applying the kernel bound of step 1.2 inside the integral proves the first two terms of the claimed estimate.

3.2givenF6F8step 2.1algebrainduction

Under the exact invariant-subspace hypotheses AV=VH and v=βVe1 in the Statement, induction gives AjV=VHj for every j≥0: it holds at j=0, and left multiplication by A gives Aj+1V=AVHj=VHHj=VHj+1. Applying the convergent series [F8] term by term yields φk(tA)v=βVφk(tH)e1 for every k≥0. To verify the stated exact Arnoldi case, each earlier column satisfies Avℓ=∑i=1ℓ+1hiℓvi by the recurrence, while the last satisfies Avm=∑i=1mhimvi because [F6] gives wm=0; hence AV=VH, and the initialization in [F6] gives v=βVe1. This proves exact invariant-space projection for every phi action; the separate ϵred(t) hypothesis in this theorem concerns the exponential trajectory c(s).

4.1givenA2F4step 3.1algebra

The reduced-action error satisfies ∥V(c^(t)−c(t))∥2≤∥V∥2ϵred(t) by [F4]. By [A2] and the triangle inequality, replacing Vc^(t) by x^ adds at most ηproj. Adding both errors to the exact projected-action error from step 3.1 proves the claimed estimate.

5.1givenF3F6algebra∎

In exact Arnoldi without breakdown, [F3] supplies the orthonormal basis and the relation AV=VH+hvm+1em∗, so D=Ω=0; [F6] gives v=βVe1. Orthonormality gives ∥V∥2=1, and an exact reduced action has ϵred=0. Substitution in the proved identity and bound leaves exactly the usual initial-zero residual integral and its semigroup-weighted estimate.

Remarks

The residual integral is exact; its upper bound is conditional on the displayed semigroup, matvec, orthogonalization, initial-vector, reduced-action, and final-projection data. A near-zero computed Arnoldi residual without an exact-zero or relation certificate never establishes happy breakdown. No mesh-uniform estimate follows automatically from this finite-dimensional result.

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