LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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.

Local unitary-similarity perturbations accumulate as a backward perturbation

Statement

Let $A \in C^{n \times n}$ , let $U_{0}$ be unitary, and set $H_{0} = U_{0}^{*} A U_{0}$ . For $j = 1, \dots, m$ , suppose

$H_{j} = P_{j}^{*} (H_{j - 1} + E_{j} + D_{j}) P_{j},$

where each $P_{j}$ is unitary, $E_{j}$ is a local arithmetic perturbation, and $D_{j}$ is a local deflation perturbation. Then for $U_{m} = U_{0} P_{1} \dots P_{m}$ there is a matrix $Δ$ such that

$H_{m} = U_{m}^{*} (A + Δ) U_{m}, ∥ Δ ∥_{F} \leq \sum_{j = 1}^{m} (∥ E_{j} ∥_{F} + ∥ D_{j} ∥_{F}) .$

The assertion also holds over $R$ with orthogonal factors.

Facts & Assumptions

Given: The matrices and update relations in the statement.

[F1]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces: each $P_{j}$ and $U_{0}$ is unitary.

[F2]

Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison: unitary multiplication on either side preserves the Frobenius norm.

Proof

1.1givenF1algebra

Set $Δ_{0} = 0$ and $U_{j} = U_{j - 1} P_{j}$ . If $H_{j - 1} = U_{j - 1}^{*} (A + Δ_{j - 1}) U_{j - 1}$ , then the stated update gives $H_{j} = U_{j}^{*} (A + Δ_{j - 1} + U_{j - 1} (E_{j} + D_{j}) U_{j - 1}^{*}) U_{j}$ . Thus define $Δ_{j} = Δ_{j - 1} + U_{j - 1} (E_{j} + D_{j}) U_{j - 1}^{*}$ . The base case holds by the definition of $H_{0}$ , so induction gives the displayed representation for $j = m$ .

2.1step 1.1F2algebra∎

By [F2], $∥ U_{j - 1} (E_{j} + D_{j}) U_{j - 1}^{*} ∥_{F} = ∥ E_{j} + D_{j} ∥_{F} \leq ∥ E_{j} ∥_{F} + ∥ D_{j} ∥_{F}$ . Applying the triangle inequality to the sum defining $Δ_{m}$ proves the stated bound.

Depends on

Used by

Dependency tree · two levels

15 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

James W. Demmel, Applied Numerical Linear Algebra, §4.4 (standard reference, not scraped)
David Bindel, CS 6210 Matrix Computations, Hessenberg QR iteration (standard reference, not scraped)