I am looking if it is possible to simplify the expression \begin{align} e^{-u^T \log(A) v} \end{align} whre $A \in \mathbb{R}^{n\times n}$ is a positive definite matrix and $u,v \in \mathbb{R}^m $. Here $\log(\cdot)$ is the matrix logarithm.
For example, can this be written in the form of a rational form or some kind of determinant of a matrix?
Edit: Example for the diagonal case. If $A$ is a diagonal matrix, then \begin{align} e^{-u^T \log(A) v}&= e^{- u^T {\rm diag}([\log(a_{11}), ...,\log( a_{nn})] ) v}\\ &=e^{- \sum_{i=1}^n u_i v_i \log(a_{ii})}\\ &=e^{- \log( \prod_{i=1}^na_{ii}^{u_i v_i})}\\ &= \frac{1}{ \prod_{i=1}^na_{ii}^{u_i v_i}} \end{align}
Edit2: Here is another approach using matrix-matrix power \begin{align} e^{-u^T \log(A) v}&= e^{- Tr( \log(A) v u^T)}\\ &=det( e^{- \log(A) v u^T} ) \end{align} Now I found on wiki that $ e^{\log(A) B}= A^B$, so \begin{align} e^{-u^T \log(A) v}= det( A^{-v u^T} )=\frac{1}{det( A^{v u^T} )} \end{align} Now can a term $det( A^{v u^T} )$ be factor somehow? For a scalar exponent, we know that $det(A^k)= (det(A)^k)$, but here we are rather interested in the case of \begin{align} det(A^B) \end{align} where $B$ is some matrix. I guess this can be treated as the main question of the post.
At the very least, we can get an expression in terms of the eigendecomposition. In particular, since $A$ is (presumably symmetric and) positive definite, we have $A = PDP^T$ for some orthogonal matrix $P$ and the diagonal matrix $D = \operatorname{diag}(\lambda_1,\dots,\lambda_n)$. It follows that $$ \exp[-u^T \log(A) v] = \exp[-u^T \log(PDP^T) v] = \exp[-u^T P\log(D)P^T v] = \\ \exp[-(P^Tu)^T \log(D) (P^Tv)]. $$ Now, using your work for the diagonal case: let $\hat u = P^Tu$ and let $\hat v = P^Tv$. We have $$ \exp[-u^T \log(A) v] = \exp[-\hat u^T \log(D) \hat v] = \prod_{i=1}^n \lambda_i^{-\hat u_i \hat v_i}. $$