von Neumann entropy and change of basis 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 $Tr(\rho \ln\rho)$?
 A: The matrix log is defined by 
$$\exp(A)=B \implies A = \log(B).$$
If $B=B^\dagger$, then $B=UDU^\dagger$ is diagonalizable by unitary matrices. We find that
$$\exp(A) = UDU^\dagger \implies A = \log(UDU\dagger).$$
Using $\exp(A)=\sum_nA^n/n!$ in the first equality below, we can rearrange the above to give
$$U^\dagger\exp(A)U = \exp(U^\dagger AU) = D \implies U^\dagger AU = \log(D).$$
Therefore, the unitary matrices "pop" out of the logarithm as desired:
$$A = U\log(D)U^\dagger.$$

Now the result for the entropy should be clear. The state matrix $\rho=\rho^\dagger$ is hermitian, and so it is diagonalizable $\rho = UDU^\dagger$. Then the entropy is given by
$$S = -Tr\left( \rho \log\rho\right) = -Tr\left(UDU^\dagger\log\left(UDU^\dagger\right)\right) = -Tr\left(UDU^\dagger U\log(D)U^\dagger\right)$$
$$=-Tr(UD\log(D)U^\dagger) = -Tr(D\log(D)U^\dagger U) = -Tr(D\log(D)).$$
The cyclic property of the trace is used in the last line. Finally, one can say that
$$S = -\sum_k \rho_{kk}\log(\rho_{kk})$$
where $\rho_{kk}$ is the $k$-th diagonal matrix element of $\rho$ in the diagonal basis.
A: 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$
A: To prove that the von Neumann entropy (defined as $S(\rho)=-\mathrm{Tr}({\rho \ln \rho})$ with $\rho$ being the density matrix) is invariant under unitary change of basis, one should first realize what $\ln(\rho)$ stands for.
The logarithm of a Hermitian matrix $\rho$ is defined as
$$ \ln(\rho) = V \ln( V^\dagger \rho V) V^\dagger,$$
where $V$ is the unitary transformation that diagonalises $\rho$, namely, $\rho_D=V^\dagger \rho V$ is diagonal. See also the Wikipedia page on the logarithm of a diagonalizable matrix.
Therefore, the von Neumann entropy reads
$$S(\rho)=-\mathrm{Tr}({\rho V   \ln (\rho_D ) V^\dagger}) = -\mathrm{Tr}({V^\dagger \rho V   \ln (\rho_D ) }) =  -\mathrm{Tr}({\rho_D   \ln (\rho_D ) }),$$
where the last step relies on the fact the trace is invariant under cyclic permutations, see the related Wikipedia page.
Note also that both $\rho$ and $U\rho U^\dagger$ are associated with the same diagonal matrix $\rho_D$ (unitary transformations do not change the eigenvalues).
Finally, because the expression of $S(\rho)$ depends only on the diagonal expression of the density matrix $\rho_D$, we conclude that $S(\rho)$ is invariant under unitary changes of basis.
From a physical point of view, this means that the von Neumann entropy of an isolated system remains constant in time because of its invariance under unitary time evolution.
