Does there exist a function $f(X)$ with the following gradient $\mathrm{Tr}[(I-X)^{-1}]\cdot g(X)$?

Question

Let $f: \mathbb{R}^{n\times n} \to \mathbb{R}$ be a function that receives a square matrix and spits out a scalar. Does there exist a function $f$ such that the gradient $\nabla_Xf(X) = \mathrm{Tr}[(I-X)^{-1}] \cdot g(X)$? Where $I$ is the identity matrix, $\mathrm{Tr}$ denotes the trace, and $g(X)$ is a function that returns a matrix.

Answer 1

_{This is an answer to the question as originally formulated, whether there exist a function $f(X)$ with $\nabla_X f(X)=\mathrm{Tr}[(I-X)^{-1}g(X)]$.}

The gradient $\nabla_X$ will give you an $n\times n$ matrix, while the trace is a scalar, so $\nabla_Xf(X)$ cannot be equal to a trace; to get a scalar derivative you can let $X$ depend on a parameter $t$, and take $$f(X(t))=-\log\,{\rm det}(I-X(t))\Rightarrow \frac{d}{dt}f(X(t))={\rm tr}\,[(I-X(t))^{-1}X'(t)].$$