ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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A Ritz pair from Arnoldi has a residual given by the last Hessenberg column entry

Example

Continue the Arnoldi run from the previous example, where

$H_{2} = (\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}), h_{3, 2} = 1.$

Take the unit eigenvector $y = \frac{1}{\sqrt{2}} (1, 1)^{T}$ of $H_{2}$ with eigenvalue $θ = 3$ . Then the Ritz vector is

$u = Q_{2} y = \frac{1}{\sqrt{2}} (e_{1} + e_{2}),$

and its residual is

$A u - θ u = \frac{1}{\sqrt{2}} e_{3},$

whose norm is $1 / \sqrt{2}$ .

Facts & Assumptions

Given: The Arnoldi data $H_{2} = (\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array})$ , $h_{3, 2} = 1$ , $v_{3} = e_{3}$ , and $y = \frac{1}{\sqrt{2}} (1, 1)^{T}$ .

[L1]

For an Arnoldi Ritz pair, $A u - θ u = h_{m + 1, m} (e_{m}^{*} y) v_{m + 1}$ (An Arnoldi Ritz pair has residual norm controlled by the last Hessenberg subdiagonal entry).

Verification

technique · direct

1.1

One has $H_{2} y = 3 y$ , so $θ = 3$ is a Ritz value and $u = Q_{2} y = \frac{1}{\sqrt{2}} (e_{1} + e_{2})$ is the associated Ritz vector. Also $e_{2}^{*} y = 1 / \sqrt{2}$ .

algebra

2.1

Applying [L1] with $m = 2$ gives $A u - θ u = h_{3, 2} (e_{2}^{*} y) v_{3} = 1 \cdot \frac{1}{\sqrt{2}} e_{3}$ . Therefore $∥ A u - θ u ∥_{2} = 1 / \sqrt{2}$ .

L1step 1.1algebra∎

Depends on

An Arnoldi Ritz pair has residual norm controlled by the last Hessenberg subdiagonal entry

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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)