Show that $$\operatorname{trace}(A\ln A)=\operatorname{trace}(UAU^\dagger\ln[UAU^\dagger])$$ where $U$ is a unitary matrix, i.e. $U^\dagger U=I\iff U^\dagger=U^{-1}$, and $A$ is a hermitian matrix, i.e. $A^\dagger=A$.
Attempt:
I read this question but do not fully understand the approach taken by the accepted answer - namely that I don't know what series is being used for the logarithm function. If there were such a series, then $$\ln(X)=\sum_n a_nX^n$$ which gives $\ln[UAU^\dagger]=\sum_na_nUA^nU^\dagger=U\ln [A]U^\dagger$, and so $$\operatorname{trace}(UAU^\dagger\ln[UAU^\dagger])=\operatorname{trace}(UAU^\dagger U\ln [A]U^\dagger)=\operatorname{trace}(UA\ln[A]U^\dagger)=\operatorname{trace}(A\ln [A])$$ where the last step is due to the cyclic property of trace. So this approach seems to work, except for the step where the series was introduced.
Question:
What is the series that can be used here? The only series I am aware of is for $\ln(1+x)$, and I am not sure if that can be useful here. Also would there be any convergence issues with this?
I also tried $\ln(x)=(x-1)-\frac12(x-1)^2+\frac13(x-1)^3-\cdots$, but to no avail, since $(UAU^\dagger-I)^n$ doesn't simplify nicely to something like $(UA^nU^\dagger-I)$.
This is presumably using the functional calculus with the function $f(x) = x \ln x$ (defined to be $0$ when $x=0$; any branch of the logarithm is used for $x < 0$). For any function $f$ defined on the spectrum of the hermitian matrix $A$, $f(A) = p(A)$ if $p$ is a polynomial such that $f(\lambda) = p(\lambda)$ for all eigenvalues $\lambda$ of $A$. Thus there is no need for series: it's simply that $$\text{trace }p(A) = \text{trace }(U p(A) U^\dagger) = \text{trace }(p(U A U^\dagger))$$ for all polynomials $p$.