On matrices with zero von Neumann entropy I was indecisive about whether to post this problem in the Physics forum or in the Mathematics one. However, since I am mostly interested in the mathematical understanding of it, I am posting it here.

Suppose I have a matrix $A$ subject to the conditions (for its trace): 
$$\mbox{Tr}(A) = 1$$
$$A^2 = A$$
$$A \succeq 0 $$
$$A^H = A$$
Now from this conditions I need to prove the following:
$$-\mbox{Tr} (A \ln{A}) = 0$$ 
In physics, $A$ can be seen as a pure density matrix and $-\mbox{Tr}(A\ln{A})$ is the von Neumann entropy. How does that follow from the first four conditions?

Following Darij Grinberg in the comments we may assume WLOG that $A$ is diagonal. From $A^2 = A$ we then see that these diagonal entries are all either $1$ or $0$. You can fill in the rest from there
