PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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.

Matrix quadratic forms have the expected first derivative and Hessian

Statement

Fix a matrix $C$ and consider the real-valued quadratic form

$q (x) : = \frac{1}{2} ∥ C x ∥_{2}^{2} = \frac{1}{2} x^{*} C^{*} C x .$

Then

$D q (x) [h] = Re ⟨ C^{*} C x, h ⟩,$

so the gradient is $C^{*} C x$ , and the Hessian is the constant Hermitian map $h \mapsto C^{*} C h$ .

Facts & Assumptions

Given: A fixed matrix $C$ , a vector $x$ , and a perturbation direction $h$ .

[L1]

Matrix differentials satisfy the product and adjoint rules (Matrix differentials obey the sum rule, product rule, and adjoint rule).

[L2]

Frobenius-linear functionals differentiate by inspection (Trace and Frobenius-linear matrix functionals differentiate by inspection).

Proof

technique · direct

1.1

Expanding at $x + h$ gives $q (x + h) = \frac{1}{2} ∥ C x + C h ∥_{2}^{2} = \frac{1}{2} ∥ C x ∥_{2}^{2} + Re ⟨ C x, C h ⟩ + \frac{1}{2} ∥ C h ∥_{2}^{2} .$ Hence $D q (x) [h] = Re ⟨ C^{*} C x, h ⟩,$ so the gradient is $C^{*} C x$ .

L1L2algebra

2.1

The derivative of the gradient map $x \mapsto C^{*} C x$ is the constant linear map $h \mapsto C^{*} C h$ . Therefore the Hessian is $C^{*} C$ .

step 1.1algebra∎

Depends on

Used by

The Frobenius least-squares objective has gradient A^*(Ax-b) and Hessian A^*A in the vector variable Theorem

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

Alan Edelman and Steven G. Johnson, Matrix Calculus for Machine Learning and Beyond (standard reference, not scraped)