Given a matrix $X\in \Bbb{R}^{m\times n}$, and its singular value decomposition is \begin{align} X = U \Sigma V^T \end{align} where $U\in \Bbb{R}^{m\times r}$, $\Sigma\in \Bbb{R}^{r\times r}$ and $V\in \Bbb{R}^{n\times r}$.
My question is how to calculate the following sub-differential: $$\frac{\partial \operatorname{tr}(\Sigma^{2/3})}{\partial X}.$$
I want to use the gradient-based method to solve an optimization problem with $\operatorname{tr}(\Sigma^{2/3})$. Any idea is appreciated!
$ \def\p{\partial} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\vec#1{\operatorname{vec}\LR{#1}} \def\diag#1{\operatorname{diag}\LR{#1}} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\grad#1#2{\frac{\p #1}{\p #2}} $For ease of typing, rename $\,\Sigma\to S\,$ so that $$\eqalign{ X &= USV^T \quad\iff\quad S = U^TXV \\ }$$ Start by finding the differential of the function with respect to $S,\,$ then change the independent variable from $S\to X,\,$ and finally recover the gradient with respect to $X$. $$\eqalign{ \phi &= \trace{S^{2/3}} \\ d\phi &= \tfrac 23\,S^{-1/3}:dS \\ &= \tfrac 23\,S^{-1/3}:U^TdX\,V \\ &= \tfrac 23\,US^{-1/3}V^T:dX \\ \grad{\phi}{X} &= \tfrac 23\,US^{-1/3}V^T \\ }$$ where $(:)$ denotes the Frobenius product, which is a convenient notation for the trace $$\eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{AB^T} \\ A:A &= \big\|A\big\|^2_F \\ }$$ The properties of the underlying trace function allow the terms in a Frobenius to be rearranged in many different ways, e.g. $$\eqalign{ &A:B = B:A = B^T:A^T \\ &A:BC = B^TA:C = AC^T:B \\ }$$ Since you plan on using this in a gradient-based method, you'll need to update $X$ and recompute its SVD at every step of the iteration.