How do I prove that reduced density matrix is Hermitian? It is known that density matrix $\rho$ is hermitian. How do I prove that for a bipartite system $AB$, reduced density matrix of $A$, $\rho_A = Tr_B\{\rho_{AB}\}$, is hermitian, given that $\rho_{AB}$ is hermitian as well? 
 A: This can be worked out from the definition itself.
The partial trace $tr_B$ is defined as the linear extension of the mapping
$$tr_B : S \otimes T \rightarrow tr(T)S$$
for any matrix $S$ on $H_A$ and $T$ on $H_B$.
Let ${|a_i\rangle}$ be a basis of $H_A$, and ${|b_i\rangle}$ be a basis of $H_B$. Any density matrix $\rho_{AB}$ on $H_A \otimes H_B$ can then be decomposed as $\rho_{AB} = \sum_{ijkl} m_{ijkl}|a_i\rangle \langle a_j| \otimes |b_k\rangle \langle b_l|$.
We know that $\rho_{AB}^\dagger = \rho_{AB}$.
This implies $m_{ijkl} = m^*_{jilk}$.
The partial trace then reads $\rho_A = tr_B \rho_{AB} = \sum_{ijkl} m_{ijkl}|a_i\rangle \langle a_j| \langle b_l|b_k\rangle$.
Since $\langle b_l| b_k \rangle = \langle b_k| b_l\rangle$, ensuring $m_{ijkl} = m^*_{jilk}$ should imply that $\rho_A = \rho^\dagger_A$.
A: The other answer is overly complicated in my opinion. This answer is more simple:
The solution has only 2 pieces of prerequisite knowledge:
$$(pAB)^{\dagger} = p(A B)^{\dagger}=p(B^{\dagger} A^{\dagger})$$
...and...
$$(A+B)^{\dagger}=A^{\dagger}+B^{\dagger}$$
for any 2 matrices $A$ and $B$ and some real number $p$. These 2 rules aren't too difficult to prove, with some good effort. Just write out generic matrices of an arbitrary size, and you will see.
Consider the density matrix $\rho$ for a pure state $|\psi \rangle$. By definition $\rho = |\psi \rangle \langle \psi | $.
Taking the adjoint of this gives: $\rho^{\dagger} = (|\psi \rangle \langle \psi |)^{\dagger} = \langle \psi |^{\dagger} |\psi \rangle^{\dagger} = |\psi \rangle \langle \psi | = \rho$
Taking the more general definition of a density matrix:
$$\rho=\sum_{j} p_{j}\left|\psi_{j}\right\rangle\left\langle\psi_{j}\right|$$
we can apply the first rule I listed to each of the terms $p_j\left|\psi_{j}\right\rangle\left\langle\psi_{j}\right|$, as per the second rule I listed. The addition of all the resulting hermitian matrices is also a hermitian matrix (because this holds true for the addition of any 2 hermitian matrices).
