How Do I Use a Single Qubit Gate, On a State With Multiple Qubits? Say I have a vector, which is a combination of 2 qubits.
$V_1\lvert00\rangle + V_2\lvert01\rangle + V_3\lvert10\rangle + V_4\lvert11\rangle =\left( \begin{matrix} V_1\\V_2\\V_3\\ V_4\end{matrix} \right)$
How would I multiply this, by a 2x2 matrix, as to apply it to only one of the qubits?
i.e, if I wanted to apply a NOT gate, with the matrix:
$X = \left[ \begin{matrix} 0 & 1\\1 & 0\end{matrix} \right]$
To only one of the qubits, how would I do it?
 A: Let's put things in context
The value $V_1\lvert00\rangle + V_2\lvert01\rangle + V_3\lvert10\rangle + V_4\lvert11\rangle $ and $V_1^2+V_2^2+V_3^2+V_4^2 = 1$ results from the sum of  2 qubits like $\lvert \phi_1\rangle =   a_1\lvert0\rangle + b_1\lvert1\rangle$ and $\lvert \phi_2\rangle = a_2\lvert0\rangle + b_2\lvert1\rangle $. 
Their composition $\lvert \psi\rangle = \lvert \phi_1\rangle \otimes \lvert \phi_2\rangle = a_1 a_2 \lvert00\rangle + a_1 b_2\lvert01\rangle + a_2 b_1\lvert10\rangle + b_1 b_2\lvert11\rangle$. Then the coordinates $V_n$ are $V_1 = a_1 a_2$ , $V_2 = a_1 b_2$ , $V_3 = a_2 b_1$ and $V_4 = b_1 b_2$ .
To apply $\left( \begin{matrix} 0 & 1 \\ 1 & 0\end{matrix} \right)$ on the first qubit, $a_1$ and $b_1$ must be swapped, giving the transformation of $( V_1,V_2,V_3,V_4 )$ in $( V_3,V_4,V_1,V_2 )$. It's easy to verify starting from the above values or the $V_n$. The matrix doing this transformation is merely :
$\left( \begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{matrix} \right)$
To apply on the 2nd, $a_2$ and $b_2$ must be swapped, giving the transformation of $( V_1,V_2,V_3,V_4 )$ in $( V_2,V_1,V_4,V_3 )$ :
$\left( \begin{matrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{matrix} \right)$
