How to apply a 1-qubit gate to a single qubit from an entangled pair?
I came to a conclusion that we do not actually take away a single qubit from an entangled pair and operate it, but operate on both qubits with Identity operator on the one we do not want to change.

So basically to apply Pauli X gate to the first qubit from an entangled pair, we apply X tensored with $I$ to the entangled pair to get the result.
That is, to operate a not on the first qubit, we evolve the system by
$$ \sigma_x \otimes I = 
\begin{pmatrix}
0 &amp; 1 \\
1 &amp; 0 
\end{pmatrix}
\otimes
\begin{pmatrix}
1 &amp; 0 \\
0 &amp; 1 
\end{pmatrix} 
=
\begin{pmatrix}
0 &amp; 0 &amp; 1 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; 1\\
1 &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 1 &amp; 0&amp;  0
\end{pmatrix} 
$$

Intuitively it is like applying a single operation to the first qubit and identity to the second one (effectively not changing its state).

Thus if we start with the entangled state $|\psi\rangle \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ we end up with
$$ (\sigma_x \otimes I)|\psi\rangle = 
\begin{pmatrix}
0 &amp; 0 &amp; 1 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; 1\\
1 &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 1 &amp; 0&amp;  0
\end{pmatrix} 
\cdot \frac1{\sqrt{2}}
\begin{pmatrix}1 \\0 \\0 \\1 \end{pmatrix} = 
\frac1{\sqrt{2}}
\begin{pmatrix}0 \\1 \\1 \\0 \end{pmatrix} = 
\frac1{\sqrt{2}}(|01\rangle+|10\rangle)
$$