What is the gradient of a scalar function of a symmetric matrix ?

TL;DR

This paper clarifies the correct form of the gradient of a scalar function of a symmetric matrix, resolving conflicting approaches in different scientific communities by proving the symmetric gradient equals the symmetric part of the general gradient.

Contribution

It demonstrates that the symmetric gradient is equal to the symmetric part of the general gradient, correcting a common misconception in matrix calculus.

Findings

The relation $G_s = G + G^T - G \

The symmetric gradient $G_s$ equals the symmetric part of the general gradient $G$.

Reconciles different approaches to matrix calculus in various scientific fields.

Abstract

Perusal of research articles that deal with the topic of matrix calculus reveal two different approaches to calculation of the gradient of a real-valued function of a symmetric matrix leading to two different results. In the mechanics and physics communities, the gradient is calculated using the definition of a \frechet derivative, irrespective of whether the argument is symmetric or not. However, members of the statistics, economics, and electrical engineering communities use another notion of the gradient that explicitly takes into account the symmetry of the matrix, and this "symmetric gradient" $G_s$ is reported to be related to the gradient $G$ computed from the \frechet derivative with respect to a general matrix as $G_s = G + G^T - G \circ I$, where $\circ$ denotes the elementwise Hadamard product of the two matrices. We demonstrate that this relation is incorrect, and reconcile both these viewpoints by proving that $G_s = \mathrm{sym}(G)$.