Prove that the partial trace preserves density operators Let $H_A$ and $H_B$ be two finite dimensional Hilbert spaces and $\rho_{AB}$ a density operator acting on $H_A\otimes H_B$. I am to show that $\rho_{A} = \operatorname{tr}_B\rho_{AB}$ is also a density operator.
I've been able to show $\operatorname{tr}\rho_A = 1$, but am struggling to show $\rho_A$ is a positive semi-definite operator. I feel like it should be simple, but just don't know where to take it. Here's an attempt:
The positive semi-definite-ness of $\rho_{AB}$ shows us that
$$\sum_{ijk\ell}\overline{\psi}_{ij}p_{ijk\ell}\psi_{k\ell} \geq 0$$
for all $|\psi\rangle = \sum_{ij}\psi_{ij}|a_i\rangle\otimes|b_j\rangle$ and $p_{ijk\ell}$ are the matrix elements in the tensor product basis. That said I have no idea how to extrapolate any information from this in order to apply it to the elements of $\rho_A$.
 A: Let $\mathscr{H}\equiv \mathscr{H}_{\mathrm{A}} \otimes  \mathscr{H}_\mathrm{B}$. Then consider a density operator $\rho$  on $\mathscr{H}$ in its spectral decomposition:
$$\rho = \sum\limits_k \lambda_k \, |\lambda_k\rangle \langle \lambda_k| \quad , $$
with $ \langle \lambda_k|\lambda_q\rangle = \delta_{kq}$, $\lambda_k \geq 0$ and $\sum\limits_k \lambda_k = 1$. We calculate the reduced density matrix of subsystem $\mathrm{A}$ as
$$ \rho_{\mathrm{A}} \equiv \mathrm{Tr}_{\mathrm{B}}(\rho) = \sum\limits_k \lambda_k \,  \mathrm{Tr}_{\mathrm{B}}(|\lambda_k\rangle \langle \lambda_k|) \quad , $$
where the second equality follows from the linearity of the partial trace. We then note that we can express each rank-one projection $ |\lambda_k\rangle \langle \lambda_k|$ in terms of the Schmidt decomposition of $|\lambda_k\rangle \in \mathscr{H}$. By doing so, we find that
$$\mathrm{Tr}_{\mathrm{B}}(|\lambda_k\rangle \langle \lambda_k|) = \sum\limits_j |\alpha^k_j|^2\, |a^k_j\rangle \langle a^k_j| \quad . $$
Here, $\{ |a^k_j\rangle \}_j$ denotes a complete and orthonormal basis of $\mathscr{H}_{\mathrm{A}}$. It is easy to see that $\mathrm{Tr}_{\mathrm{B}}(|\lambda_k\rangle \langle \lambda_k|)$ is positive semi-definite for all $k$ and since $\lambda_k \geq 0$  we find that $\rho_{\mathrm{A}} \geq 0$. Additionally, the normalization of $\rho_{\mathrm{A}}$ follows from the normalization of the Schmidt coefficients.
