Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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The Frobenius least-squares objective has gradient A∗(Ax−b) and Hessian A∗A in the vector variable

Statement

For fixed A and b, let

f(x):=12∥Ax−b∥22.

Then

Df(x)[h]=Re⁡⟨A∗(Ax−b),h⟩,

so the gradient is ∇f(x)=A∗(Ax−b) and the Hessian is the constant map h↦A∗Ah.

Facts & Assumptions

Given: A fixed matrix A, a fixed vector b, a vector x, and a direction h.

[L1]

The quadratic form x↦12∥Cx∥22 has gradient C∗Cx and Hessian C∗C (Matrix quadratic forms have the expected first derivative and Hessian).

Proof

technique · direct
1.1L1algebra

Write f(x)=12∥A(x−A+b)∥22+constant only heuristically; directly, f(x+h)−f(x)=12∥Ax−b+Ah∥22−12∥Ax−b∥22. Expanding yields Df(x)[h]=Re⁡⟨Ax−b,Ah⟩=Re⁡⟨A∗(Ax−b),h⟩.

2.1step 1.1algebra∎

The gradient map from step 1.1 is x↦A∗(Ax−b), whose derivative is the constant linear map h↦A∗Ah. Thus the Hessian is A∗A.

Depends on

Used by

Dependency tree · two levels

13 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