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 Incoming links: Hessian of a sum of functions of linear forms

Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 6 / c / ii / Solution 2026-10-25

Differentiating the gradient term by term gives the Hessian matrix

\nabla^{2} l (x) = \sum_{i = 1}^{n} y_{i} f^{''} (a_{i}^{T} x) a_{i} a_{i}^{T} .

Take row

i

of

A

to be

a_{i}^{T}

. Then

\nabla^{2} l = A^{T} Δ A, Δ_{ii} = y_{i} f^{''} (a_{i}^{T} x), Δ_{ij} = 0 (i \neq = j) .

Indeed, entry

(j, k)

of this product is

\sum_{i} Δ_{ii} a_{ij} a_{ik}

, exactly the entry of the displayed sum of outer products. Explicitly,

Δ_{ii} = y_{i} ((a_{i}^{T} x)^{2} - 1) e^{- (a_{i}^{T} x)^{2} /2} .

These coefficients may have either sign, so a positive-definite Hessian matrix cannot be assumed. This is an instance of the Hessian of a sum of functions of linear forms.

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