Vector space of $\mathbb{C}^4$ and its basis, the Pauli matrices How do I write an arbitrary $2\times 2$ matrix as a linear combination of the three Pauli Matrices and the $2\times 2$ unit matrix? 
Any example for the same might help ? 
 A: A slow construction would go...
$$$$
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}
=
a\begin{pmatrix}1&0\\0&0\end{pmatrix}
+b\begin{pmatrix}0&1\\0&0\end{pmatrix}
+c\begin{pmatrix}0&0\\1&0\end{pmatrix}
+d\begin{pmatrix}0&0\\0&1\end{pmatrix}
$$

$$
\begin{pmatrix}1&0\\0&0\end{pmatrix}
=\frac{1}{2}
\begin{pmatrix}1&0\\0&1\end{pmatrix}
+
\frac{1}{2}
\begin{pmatrix}1&0\\0&-1\end{pmatrix}
=\frac{1}{2}1_2+\frac{1}{2}\sigma_3
$$
$$
\begin{pmatrix}0&1\\0&0\end{pmatrix}
=\ ...
$$

$$
\Longrightarrow  \begin{pmatrix}a&b\\c&d\end{pmatrix}
=
\frac{a}{2}1_2+\frac{a}{2}\sigma_3+\ ...\ (\text{other combintations of the four matrices})
$$
A: I like to put it this way:
$$\left(\begin{array}{cc}
w+z&x-iy\\
x+iy&w-z\end{array}\right)$$
So, for example:
$$\left(\begin{array}{cc}
1&5\\1&2\end{array}\right) = 
\left(\begin{array}{cc}
(1.5)+(-0.5)&(3)-i(2i)\\
(3)+i(2i)&(1.5)-(-0.5)\end{array}\right)$$
So $w=1.5, x=3, y=2i, z=-0.5$ and
$$\left(\begin{array}{cc}
1&5\\1&2\end{array}\right) = 
1.5 + 3\sigma_x + 2i\sigma_y -0.5\sigma_z.$$
You can solve for $w,x,y,z$ from the entries in the array easily. I.e. $x$ is the average of the top right and bottom left entries, etc.
A: The matrices $\sigma_0\equiv \boldsymbol{1}_2$, $\sigma_x$, $\sigma_y$ and $\sigma_z$ form an orthonormal basis of your vector space w.r.t. the scalar product
$$
 (X,Y) \equiv \frac{1}{2}\operatorname{tr}(X\cdot Y),
$$
where $X$ and $Y$ label any two complex $2\times 2$ matrices. The factor $1/2$ is just for convenience, you may as well normalise your Pauli matrices by dividing them by $2$.
All you want to do now is to decompose an arbitrary element
$$
 M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
$$
of your vector space into the above basis and figure out the coefficients. As usual, this is done by projecting onto that basis by means of the scalar product
$$
 M = (\sigma_0,M)\cdot\sigma_0 + (\sigma_x,M)\cdot\sigma_x + (\sigma_y,M)\cdot\sigma_y + (\sigma_z,M)\cdot\sigma_z\ .
$$
This has essentially been said in the above comments, particularly in the link posted by Kostya.
