The von Neumann entropy is defined as $S(\rho)=-Tr({\rho \ln \rho})$, where $\rho$ is density matrix.
http://en.wikipedia.org/wiki/Von_Neumann_entropy
In the above article it says:
S(ρ) is invariant under changes in the basis of ρ, that is, S(ρ) = S(UρU†), with U a unitary transformation.
How can we prove this statement?
We have that the trace is independent of the choice of basis in which the matrix $\rho$ is expressed: $$Tr(\rho)=Tr(U \rho U^{\dagger})$$ But in the case of the von Neumann entropy we have the $\ln \rho$, so a change of basis for $\rho$ gives: $$Tr[U \rho U^{\dagger}\ln (U \rho U^{\dagger})]$$ How is this equal to $Tr(\rho \ln\rho)$?
I am not fully sure of the notations used here but the argument is standard for matrix computations.
Given that the von Neumann entropy can also be written as $\rho = -\sum_j \eta_j \log \eta_j $
where, $\eta_j$ are the eigenvalues of $\rho$, the only thing that remains to be proved is that eigenvalues are invariant under a change of basis. In other words,
$ U\rho U^\dagger = \sum_j \eta_j U|j><j|U^\dagger$
is an eigendecomposition with the same eigenvalues $\eta_j$. Therefore,
$ S(\rho) = S(U\rho U^\dagger) = - \sum_j \eta_j \log \eta_j$