 Solution

Differentiating the gradient term by term gives the Hessian matrixTake row $i$ of $A$ to be ${\mathbf{a}}_i^{\mathsf{T}}$. ThenIndeed, entry $(j,k)$ of this product is ${\sum }_i{\Delta }_{ii}{a}_{ij}{a}_{ik}$, exactly the entry of the displayed sum of outer products. Explicitly,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.