LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

A symmetric matrix operator norm is its largest absolute Rayleigh quotient

Statement

Let $A \in R^{p \times p}$ be symmetric, where $p \geq 1$ . Then $∥ A ∥_{o p} = \max_{∥ v ∥_{2} = 1} ∣ v^{T} A v ∣ = \max_{j} ∣ λ_{j} ∣,$ where $λ_{1}, \dots, λ_{p}$ are the eigenvalues of $A$ .

Facts & Assumptions

Given: A symmetric real matrix $A \in R^{p \times p}$ with $p \geq 1$ .

[F1]

A symmetric real matrix has an orthonormal eigenbasis (Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis).

[F2]

The operator norm is the maximum of $∥ A v ∥_{2}$ over unit vectors (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

Proof

1.1F1F2algebrachoose

Choose an orthonormal eigenbasis $e_{1}, \dots, e_{p}$ with eigenvalues $λ_{1}, \dots, λ_{p}$ by [F1], and put $M = \max_{j} ∣ λ_{j} ∣$ . For a unit vector $v = \sum_{j} c_{j} e_{j}$ , orthonormality gives $\sum_{j} c_{j}^{2} = 1$ and $∥ A v ∥_{2}^{2} = \sum_{j} λ_{j}^{2} c_{j}^{2} \leq M^{2}$ . Equality is attained at an eigenvector whose eigenvalue has absolute value $M$ . Thus [F2] gives $∥ A ∥_{o p} = M$ .

2.1F1step 1.1algebra∎

For the same unit vector, $v^{T} A v = \sum_{j} λ_{j} c_{j}^{2}$ , so $∣ v^{T} A v ∣ \leq \sum_{j} ∣ λ_{j} ∣ c_{j}^{2} \leq M$ . At an eigenvector with absolute eigenvalue $M$ , equality holds. Therefore the largest absolute Rayleigh quotient equals $M = ∥ A ∥_{o p}$ .

Depends on

Used by

Dependency tree · two levels

6 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

Philippe Rigollet and Jan-Christian Hütter, High-Dimensional Statistics: Lecture Notes, Chapter 1, Lemma 1.18 (standard reference, not scraped)
Sheldon Axler, Linear Algebra Done Right, fourth edition, Chapter 7 (standard reference, not scraped)