Reduced density matrix and partial trace
# Reduced density matrix
Suppose we have two quantum systems $a, b$, with dimension $N_a, N_b$ respectively. Then the Hilbert space of $a+b$ is of dimension $N=N_aN_b$. Suppose we have a density matrix $$\hat\rho=\sum_{i,j}\rho_{ij}\lvert i\rangle \langle j\rvert=\sum_{i,j,k,l}\rho_{ijkl}\lvert i\rangle_a\lvert j\rangle_b \langle k\rvert_a\langle l\rvert_b$$
Then the reduced density matrix of $a$ is defined as $$\hat\rho_a=\mathrm{tr}_b\hat\rho=\sum_i \langle i\rvert_b\hat\rho\lvert i\rangle_b$$
i.e. reduced density matrix problem is equivalent to partial trace problem.
# Tensor
In fact, if we take $\hat\rho$ as a 4-tensor $\rho_{ijkl}$, then the reduced density matrix is $$\rho^{(a)}_{ij}=\delta^{\mu\nu}\rho_{i\mu k\nu}$$ For simple density matrix $\rho=\lvert \psi\rangle \langle \psi\rvert$, the reduced matrix is $$\rho^{(a)}_{ik}=\delta^{jl}\rho_{ijkl}=\delta^{jl}\psi_{ij}\psi^+_{lk}=\sum_i |\langle i_b\lvert \psi\rangle|^2=[\psi\psi^+]_{ik}$$ Here we are taking $\psi$ as an $N_a\times N_b$ matrix.
For general case, if we find decomposition $$\rho=\sum_c \lambda_c\lvert \psi_c\rangle \langle \psi_c\rvert,\quad \sum_c \lambda_c=1$$ then we have $$\rho^{(a)}_{ik}=\left[\sum_c\lambda_c\psi_c\psi^+_c\right]_{ik}$$