Derivative of Tr$(\rho\log\rho)$

The von Neumann entropy is the analogue of the Shannon entropy and is defined for positive semidefinite matrices as

$$S = \text{Tr}(\rho\log\rho)$$

It is computed in terms of the eigenvalues $\lambda_i$ as $S = \sum_i\lambda_i\log\lambda_i$ where we set $0\log(0) = 0$.

Is $S$ differentiable with respect to $\rho$? The answer to this seems to be yes (see for example the question here and links therein) and is given as $S'(\rho) = I + \log\rho$.
If the result above is true, what happens when $\rho$ has at least one eigenvalue of zero? The von Neumann entropy is defined for positive semidefinite matrices while the log is only defined for positive definite matrices.

asked May 28 at 20:45

user1936752

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$\begingroup$ The set of all positive semidefinite matrices is not an open set so it is not clear what is meant by the statement "$S$ is differentiable". Consider the case $n = 1$. The set of positive semidefinite matrices is the interval $[0,\infty)$ and $S(\rho) = \rho \log \rho$ is a function of one variable so you can interpret your question as "does $S$ has a one-sided derivative at $\rho = 0$" and the answer is no. $\endgroup$ – levap May 28 at 21:12
$\begingroup$ @levap, thank you for the comment and sorry for a belated response. Can you comment on the following - en.wikipedia.org/wiki/Binary_entropy_function#Derivative. It seems like the function is derivative at zero is allowed provided that one admits that it goes to $-\infty$. I'm not sure if this is mathematically sound. $\endgroup$ – user1936752 Jun 2 at 17:02

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