ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Unrestarted GMRES solves a two-dimensional diagonal system in two steps while GMRES(1) restarts after one

Example

Let $A = diag (1, 2)$ , $b = (1, 1)^{T}$ , and $x_{0} = 0$ . Verify that unrestarted GMRES reaches the exact solution by step $2$ , whereas the first GMRES $(1)$ cycle leaves residual $(2 / 5, - 1 / 5)^{T} \neq 0$ .

Facts & Assumptions

Given: The real diagonal system and starting vector in the Example.

[F1]

The GMRES iterate as the residual minimizer over an affine Krylov space: GMRES minimizes $∥ b - A x ∥_{2}$ over $x_{0} + K_{m} (A, b)$ .

[F2]

The Krylov subspace $K_{m} (A, b) = span {b, A b, \dots, A^{m - 1} b}$ : $K_{m} (A, b) = span {b, A b, \dots, A^{m - 1} b}$ .

[F3]

Restarted GMRES $(m)$ as repeated GMRES cycles on fixed-size Krylov spaces: GMRES $(1)$ completes one GMRES step at the current guess and then starts a new cycle from the resulting residual.

Verification

1.1givenF1F2algebra

We have $A b = (1, 2)^{T}$ and the determinant of the matrix with columns $b, A b$ is $1 \cdot 2 - 1 \cdot 1 = 1 \neq 0$ . Thus $K_{2} (A, b) = R^{2}$ . The exact solution is $x_{*} = A^{- 1} b = (1, 1 / 2)^{T} \in K_{2} (A, b)$ , so the minimum residual in [F1] is zero at step $2$ .

2.1givenF1F3algebra∎

In the first GMRES $(1)$ cycle, [F1] restricts $x_{1}$ to $x_{0} + y b = y b$ for a scalar $y$ . Its residual is $b - y A b = (1 - y, 1 - 2 y)^{T}$ , whose squared norm is $(1 - y)^{2} + (1 - 2 y)^{2} = 5 (y - 3 / 5)^{2} + 1 / 5$ . The unique minimizer is $y = 3 / 5$ , and the residual is $(2 / 5, - 1 / 5)^{T}$ , which is nonzero. By [F3] this is the first-cycle residual before any restart.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

Yousef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., §§6.4–6.5, printed pp. 179–193 (standard reference, not scraped)