Proximal Operator of $ f \left( U \right) = -\log \det \left( U \right) $

Question

This is an assignment problem which I failed to solve in a couple of days.

Denote the set of all $n \times n$ symmetric matrices and the set of all $n \times n$ symmetric positive definite matrices by $\mathbb{S}^n$ and $\mathbb{S}^n_{++}$ respectively. Let $f: \mathbb{S}^n_{++} \rightarrow \mathbb{R}$ be defined by $$f(U) = -\log \det (U) \text{ for } U \in \mathbb{S}^n_{++}$$ We are asked to find the proximal mapping $\text{prox}_f$ of $f$.

My question is:

1. What is the domain of $\text{prox}_f$? It is not mentioned in the problem. I suppose it is $\mathbb{S}^n$ and I will explain it later.

2. The function $g(U;X) = -\log \det(U) + \frac{1}{2}\|U - X\|_F^2$ is differentiable in each entry $U_{ij}$ of $U$. If I did not make mistakes in the calculation, the minimization problem $$ \text{prox}_f(X) = \text{argmin}_{U \in \mathbb{S}^n_{++}} g(U;X) $$ can be formulated as $$ 0 = \frac{\partial g(U;X)}{\partial U_{ij}} = -U^{-1}_{ji} + U_{ij} - X_{ij} = -U^{-1}_{ij} + U_{ij} - X_{ij}$$ and therefore $U = \text{prox}_f(X) \in \mathbb{S}^n_{++}$ should satisfy $$U - U^{-1} = X$$ This also sees why $X$ has to be symmetric. But I cannot proceed from here: How do we calculate $U$ from $X$ in practice?

Thanks in advance. Any comments or hints are welcome.

Answer 1

Thanks to the hints from @Michael Grant, we let $X = QDQ^T$ is the eigenvalue decomposition of $X$ with $Q$ being the orthogonal matrix consisting of eigenvectors of $X$. Let $w^*_i$ be the positive root of $$w_i^2 - D_{ii}w_i - 1 = 0$$ The existence and uniqueness of the positive root is an analogue of the case $n = 1$. Let $W^* = \text{diag}(w^*_1, w^*_2, \ldots, w^*_n)$ and it satisfies $$W^* - (W^*)^{-1} = D$$ Letting $U^* = QW^*Q^T$ yields $$U^* - (U^*)^{-1} = X$$ and it is straightforward to check $U^* \in \mathbb{S}^n_{++}$.