How does the trace acts over one of the two entangled qubits? If $|\psi\rangle$ is entangled, i.e, is in the subspace spanned by the states $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ how does the partial trace operates on the density operator $|\psi\rangle\langle\psi|$, if I only want it to operate on the first qubit. 
$$
Tr_{1}(|\psi\rangle\langle\psi|)
$$
 A: For general basis vectors $\{ | n_{1} \rangle, | n_{2} \rangle \} \subset \mathcal{H}$ and a state $|\psi\rangle \in \mathcal{H} \otimes \mathcal{H}$ then the partial trace over the first sector is schematically (forgive the bad notation)
$$
\mathrm{Tr}_{1}\left( |\psi\rangle\langle\psi| \right) = \sum_{j=1}^2  \big( \langle n_{j} |\otimes \mathbb{I}_{2} \big) |\psi\rangle\langle\psi| \big( | n_{j} \rangle \otimes \mathbb{I}_{2} \big) 
$$ 
ie. trace over the states in the first sector and leave the states in the second sector untouched. If $|\psi\rangle$ is a linear combination of $\{ | 0 \rangle \otimes | 0 \rangle, | 0 \rangle \otimes | 1 \rangle, | 1 \rangle \otimes | 0 \rangle, | 1 \rangle \otimes | 1 \rangle \} \subset\mathcal{H} \otimes \mathcal{H}$, meaning:
$$
|\psi\rangle = a_{00} | 0 \rangle \otimes | 0 \rangle + a_{01} | 0 \rangle \otimes | 1 \rangle + a_{10} | 1 \rangle \otimes | 0 \rangle + a_{11}  | 1 \rangle \otimes | 1 \rangle = \sum_{j,k\in\{0,1\}} a_{jk} |j\rangle\otimes |k\rangle
$$
Then the corresponding density matrix is
$$
|\psi\rangle\langle\psi| \ = \ \sum_{j,k,\ell,m \in\{0,1\}} a_{jk} a^{\ast}_{\ell m} |j\rangle\otimes |k\rangle \langle \ell|\otimes \langle m| =  \sum_{j,k,\ell,m \in\{0,1\}} a_{jk} a^{\ast}_{\ell m} |j\rangle\langle\ell | \otimes |k\rangle \langle m|
$$
Then a partial trace over the first sector is:
$$
\mathrm{Tr}_{1}\left( |\psi\rangle\langle\psi| \right) \ = \ \sum_{n\in\{0,1\}} \sum_{j,k,\ell,m \in\{0,1\}} a_{jk} a^{\ast}_{\ell m} \langle n|j\rangle\langle\ell | n \rangle |k\rangle \langle m| = \sum_{k,m,n \in \{0,1\}} a_{nk}a_{nm}^{\ast} |k\rangle\langle m|
$$
Picking a coordinate basis $| 0 \rangle \to \left[ \begin{matrix} 1 \\ 0  \end{matrix} \right]$ and $| 1 \rangle \to \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right]$ we have:
$$
|\psi\rangle\langle\psi| \to \left[ \begin{matrix} |a_{00}|^2 & a_{00} a_{01}^{\ast} & a_{00} a_{10}^{\ast} & a_{00} a_{11}^{\ast}  \\
a_{01}a_{00}^{\ast} & |a_{01}|^2 & a_{01} a_{10}^{\ast} & a_{01}a_{11}^{\ast} \\
a_{10}a_{00}^{\ast} & a_{10} a_{01}^{\ast} & |a_{10}|^2 & a_{10}a_{11}^{\ast} \\
a_{11}a_{00}^{\ast} & a_{11} a_{01}^{\ast} & a_{11} a_{10}^{\ast} & |a_{11}|^2 \end{matrix} \right]
$$
Which leaves us with:
$$
\mathrm{Tr}_{1}\left( |\psi\rangle\langle\psi| \right)  \to \left[ \begin{matrix} \sum_{n \in \{0,1\}}a_{n0} a_{n0}^{\ast} & \sum_{n \in \{0,1\}}a_{n0} a_{n1}^{\ast} \\
\sum_{n \in \{0,1\}}a_{n1} a_{n0}^{\ast} & \sum_{n \in \{0,1\}}a_{n1} a_{n1}^{\ast} \end{matrix} \right] \ = \ \left[ \begin{matrix} |a_{00}|^2 + |a_{10}|^2 & a_{00} a_{01}^{\ast}+a_{10} a_{11}^{\ast} \\
a_{01} a_{00}^{\ast}+a_{11} a_{10}^{\ast} & |a_{01}|^2 + |a_{11}|^2 \end{matrix} \right]
$$
