PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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Exact Arnoldi breakdown is equivalent to invariance of the current Krylov space

Statement

Let $m \geq 1$ . Suppose Arnoldi has run through step $m$ without earlier breakdown. Then the next Arnoldi residual vanishes, equivalently $h_{m + 1, m} = 0$ , if and only if the current Krylov space $K_{m} (A, b)$ is $A$ -invariant.

Facts & Assumptions

Given: An integer $m \geq 1$ and an Arnoldi run through step $m$ with basis vectors $v_{1}, \dots, v_{m}$ .

[L1]

Before breakdown, Arnoldi produces an orthonormal basis of $K_{m} (A, b)$ and the relations $A v_{j} \in span {v_{1}, \dots, v_{j + 1}}$ for $1 \leq j \leq m$ (Before breakdown, Arnoldi produces an orthonormal Krylov basis and a rectangular upper-Hessenberg factorization).

Proof

technique · direct

1.1

Assume $h_{m + 1, m} = 0$ . Then the Arnoldi step gives $A v_{m} \in span {v_{1}, \dots, v_{m}} = K_{m} (A, b)$ . For every $j < m$ , [L1] already gives $A v_{j} \in K_{j + 1} (A, b) \subseteq K_{m} (A, b)$ . Since the $v_{j}$ form a basis of $K_{m} (A, b)$ , linearity shows $A (K_{m} (A, b)) \subseteq K_{m} (A, b)$ .

L1algebra

2.1

Conversely, assume $K_{m} (A, b)$ is $A$ -invariant. Because $v_{m} \in K_{m} (A, b)$ , one has $A v_{m} \in K_{m} (A, b) = span {v_{1}, \dots, v_{m}}$ . Therefore the part of $A v_{m}$ orthogonal to that span is zero, so the Arnoldi residual $w_{m}$ vanishes and hence $h_{m + 1, m} = ∥ w_{m} ∥ = 0$ .

L1algebra∎

Depends on

Used by

Lucky breakdown occurs when the current Krylov space is already invariant Example

Dependency tree · two levels

7 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

Lloyd N. Trefethen and David Bau III, Numerical Linear Algebra (standard reference, not scraped)