Partial traces of a finite bipartite state are states

By the second-factor formula [F2] and the positivity/trace theorem [F4], ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ is non-negative and has trace $\mathrm{tr}({\rho }_{AB})=1$. It is therefore a density operator by [F1].

Choose ordered orthonormal bases $(a_i)$ and $(b_j)$ of $H_A,H_B$ by [F7]. Formula [F3] and the product inner product in [F6] give $⟨{\rho }_{B}y,y⟩={\sum }_i⟨{\rho }_{AB}(a_i\otimes y),a_i\otimes y⟩$ for every $y\in H_B$; every summand is nonnegative by [F1] and [F5]. The matrix formula [F3] also shows ${\rho }_B$ is self-adjoint because ${\rho }_{AB}$ is self-adjoint. Hence ${\rho }_B$ is non-negative.

Summing the diagonal entries in [F3] gives $\mathrm{tr}({\rho }_B)={\sum }_{i,j}({\rho }_{AB}{)}_{(i,j),(i,j)}=\mathrm{tr}({\rho }_{AB})=1$ by [F8] and [F1]. Therefore ${\rho }_B$ is a density operator by [F1].