Prove that $\nabla_X tr(X^TB)= B $ where $B \in \mathbb{C^{m*n}}$ and $X \in \mathbb{R^{m*n}} $

Question

Prove that $\nabla_X tr(X^TB)= B $ where $B \in \mathbb{C^{m*n}}$ and $X \in \mathbb{R^{m*n}} $ and $\nabla_X$ is the derivative with respect to X.

How can I prove the above?

Answer 1

In your function, replace the trace with the Frobenius product, then take its differential to get $$ \eqalign { f &= {\rm tr}(X^TB) \cr &= B:\!X \cr \cr df &= B:dX \cr &= \bigg(\frac {\partial f} {\partial X}\bigg):dX \cr } $$ Which means that the derivative must be $$ \eqalign { \frac {\partial f} {\partial X} &= B \cr } $$