Why does the density matrix $\rho$ obey a wrong-signed Heisenberg equation of motion? The density matrix is defined as
$$ \rho_\psi ~:=~ \frac{\lvert\psi(t)\rangle  \langle \psi(t)\vert}{ \langle \psi(t) |\psi(t)\rangle }$$
in the Schrödinger picture. $\rho_\psi$ is obviously a time dependent projector, and the equation of motion on these projectors become:
$$
  i\hbar\frac{d}{dt} \rho_\psi ~=~ [H,\rho_\psi]   \tag{S}
$$
but my book also reports that the Heisenberg equation of motion on the operators/observables is:
 $$
  \mathrm{i}\hbar\frac{d}{dt} A ~=~ [A,H] .  \tag{H}
$$
Why are the signs in eqs. (S) and (H) opposite? 
Isn't $A$ an operator like  $ \rho_\psi$, although time independent? They belong to the same operator space, so I don't think I can apply duality, but I know that $A$ operate on the states to give us the expectation value through the relation 
$$  \mathrm{Tr}( \rho_\psi A)$$
so it should be in the dual space of the observables.
 A: $\rho_\psi$, the density matrix, is not an observable/operator evolving in the sense of the Heisenberg equation of motion
$$ \mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t} A = - [H,A]$$
since it is defined, as you correctly write, as a projector on states, hence it is time-dependent in the Schrödinger picture (since there the states it projects on are time-dependent), obeying the von Neumann equation
$$ \mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t}\rho_\psi = [H,\rho_\psi]$$
as a direct consequence of the Schrödinger equation. The von Neumann equation indeed differs from the Heisenberg equation of motion by a sign because it is not an equation in the Heisenberg picture, but in the Schrödinger picture.
In the Heisenberg picture, the states are time-independent, and the density matrix does consequently not evolve. In particular, it does not obey the Heisenberg equation of motion.
You can also look at this by considering that $\mathrm{Tr}(\rho_\psi A)$ is the expectation value of a operator for a particular state. It takes to inputs - the "state input" $\rho_\psi$ and the "operator input" $A$. In the Schrödinger and Heisenberg pictures, only one of these should be time-dependent, even if they both "look" like operators.
A: Actually, you can use duality:
the normal states of quantum mechanics are objects of the (unique) predual of the von Neumann algebra of quantum observables.
Using a concrete example: if the algebra of observables are the bounded operators on a Hilbert space, the predual are the trace class operators. Of them, the normal states are the ones positive, self-adjoint and of trace norm one.
It is then clear that by mutual duality the evolution on observables/states induces the evolution of states/observables; and that takes into account of the "minus sign" in the generator that is different between the two.
A: Your question apparently stems from a lack of understanding of the different pictures in quantum mechanics, that are Schrödinger picture, Heisenberg picture and Interaction picture.
In the Schrödinger picture, states are time-evolving, while observables are time-independent. The density matrix is another (more general) way of writing the state vector; its time evolution follows from the von-Neumann equation, which can be derived from the Schrödinger equation and its Hermitian conjugate, given by
$$ \mathrm i \hbar \frac{\partial}{\partial t} | \psi(t) \rangle_\mathrm{S} = H | \psi \rangle_\mathrm{S},$$
and
$$ -\mathrm i \hbar \frac{\partial}{\partial t} {}_{\mathrm S}\langle \psi(t)| = {}_{\mathrm S}\langle \psi(t)|H .$$
Take the time derivative of the density matrix here for a pure state (beware, there are partial derivatives, see http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28Hamiltonian%29#Quantum_Liouville_equation), and use the product rule, you get
$$\mathrm i \hbar \frac{\partial}{\partial t}\rho_{\mathrm S} (t) = \mathrm i \hbar \frac{\partial}{\partial t} | \psi(t) \rangle_\mathrm{S}  \langle \psi(t)| = \mathrm i \hbar \Bigl( \frac{\partial}{\partial t} | \psi(t) \rangle_\mathrm{S} \Bigr) {}_\mathrm{S}  \langle \psi(t)| + \mathrm i \hbar  | \psi(t) \rangle_\mathrm{S} \Bigl( \frac{\partial}{\partial t} {}_\mathrm{S}  \langle \psi(t)|\Bigr) .$$
Now you can pull the $\mathrm i \hbar $ inside the bracket and substitute each  of the brackets by the correspondings left hand sides of the Schrödinger equation and its Hermitian conjugate, to obtain
$$ \mathrm i \hbar \frac{\partial}{\partial t} | \psi(t) \rangle_\mathrm{S}  \langle \psi(t)| = H | \psi \rangle_\mathrm{S} \langle \psi(t)| -  | \psi(t) \rangle_\mathrm{S} \langle \psi(t)|H  = [H,\rho_{\mathrm S}(t)]. $$
In the Heisenberg picture, the observables are evolving in time, while the states are constant.
The density matrix can be stated in any of these pictures, where you take the expectation value of an observable $A$ always via $\mathrm{Tr}[\rho_{\mathrm S}(t) A_{\mathrm S}] = \mathrm{Tr}[\rho_{\mathrm H} A_{\mathrm H}(t)] $.
Here $\mathrm S$
and $\mathrm H$ denote the Schrödinger and the Heisenberg picture.
Please note that
$$\rho_{\mathrm S}(0) = \rho_{\mathrm H} \text{ and } A_{\mathrm H} (0) = A_{\mathrm S} .$$
By using the unitary time evolution operator, we can show the equivalence of the pictures quite easy for the density matrix.
The unitary evolution operator is given (for time-independent Hamiltonian $H$)
$$ U(t) = \mathrm e^{-\mathrm i H t/\hbar} . $$
The density matrix at time $t$ is then given in the Schrödinger picture by
$$ \rho_\mathrm{S} (t) = U(t) \rho_{\mathrm S} (0) U^\dagger (t) ,$$
while the operators evolve in the Heisenberg picture as
$$ A_\mathrm{H} (t) = U^\dagger(t) A_{\mathrm H} (0) U (t) .$$
So we find for the expetation value for the observable $A$ the following:
$$ \langle A \rangle (t) = \mathrm{Tr}[\rho_{\mathrm S}(t) A_{\mathrm S}] = \mathrm{Tr} [ U(t) \rho_{\mathrm S}(0) U^\dagger (t)  A_{\mathrm S} ]$$
in the Schrödinger picture. We can now very easy switch to the Heisenberg picture by using the cyclic property of the trace, i.e.
$$\mathrm{Tr} [ABC] = \mathrm{Tr} [BCA] = \mathrm{Tr} [CAB] ,$$
by cycling the first unitary operator to the end, so we obtain
$$\langle A \rangle (t) = \mathrm{Tr} [ \rho_{\mathrm S}(0) U^\dagger (t)  A_{\mathrm S}  U(t)] .$$
Using the equivalence of the two pictures at $t=0$, we can reqrite this as
$$\langle A \rangle (t) = \mathrm{Tr} [ \rho_{\mathrm H} U^\dagger (t)  A_{\mathrm H}(0)  U(t)] = \mathrm{Tr} [ \rho_{\mathrm H} A_{\mathrm H}(t)].$$
