Construction of Pauli Matrices How can we construct the Pauli matrices starting from $$\sigma_i=\begin{bmatrix}
a & b\\ 
 c& d
\end{bmatrix}$$ by using the conditions $$\sigma^2_i=1,$$$$\left [ \sigma_x,\sigma_y \right ]=2i\sigma_z,$$ and so on?
 A: Since each $\sigma_i$ is a scalar multiple of a Lie bracket of other finite matrices, each $\sigma_i$ must be traceless. So straight away we know:
$$\sigma_i=\left(\begin{array}{cc}a&b\\c&-a\end{array}\right)\tag{1}$$
and $\sigma_i^2=\mathrm{id}$ then yields $a^2 + b\,c=1$.
The eigenvalues of any matrix of the form in (1) with $a^2 + b\,c=1$ are $\pm\sqrt{a^2+b\,c} = \pm1$. Therefore, for any set of matrices we find fulfilling all the given relationships, we can do a similarity transformation on the whole set and thus (1) diagonalize any member of the set we choose whilst (2) keeping all the required relationships intact. Exercise: Prove that the given relationships (Lie brackets and $\sigma_i^2=\mathrm{id}$) are indeed invariant under any similarity transformation.
Thus, without loss of generalness, we can always choose one of the set to be:
$$\sigma_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\tag{2}$$
So now work out the Lie bracket of $\sigma_z$ and $\sigma_x = \left(\begin{array}{cc}a_x&b_x\\c_x&-a_x\end{array}\right)$: result must be $2\,i\,\sigma_y$ and so we get:
$$\sigma_y = \left(\begin{array}{cc}0 & -i\,b_x \\i\,c_x & 0 \\\end{array}\right)\tag{3}$$
But given $\sigma_y^2=1$ we get $b_x\,c_x=1$ whence $a_x=0$ (since $a_x^2 + b_x\,c_x=1$). So our remaining two matrices are of the forms:
$$\sigma_x = \left(\begin{array}{cc}0 & b_x \\\frac{1}{b_x} & 0 \\\end{array}\right)$$
$$\sigma_y = \left(\begin{array}{cc}0 &  -i\,b_x \\\frac{i}{b_x} & 0\\\end{array}\right)\tag{4}$$
and the remaining commutation relationships then give you the unknown constant $b_x$.
Once you have found $b_x$, we know from our comments above that any set of matrices fulfilling the required commutation relationships and $\sigma_i^2=\mathrm{id}$ is  gotten from this particular set (the "standard" Pauli matrices) by a similarity transformation.
A: There are very good answers here already, however, for heuristic purposes, I will derive the Pauli matrices just from the 3 postulates presented, without any further assumptions, using only matrix multiplication and solving a few equations.
The 3 postulates presented by you can be written as:
$(1)$ $\sigma_i$ are $2 \times 2$ matrices;
$(2)$ $\sigma^2_i=I$;
$(3)$ $\left [ \sigma_i,\sigma_j \right ]=2i\sigma_k,$, where $i≠j≠k$ are indexes;
The general form of a $2 \times 2$ matrix is
$$\sigma_i=\begin{bmatrix}
a_i & b_i\\ 
c_i & d_i
\end{bmatrix} \tag{4}$$
Applying $(3)$
$$\begin{bmatrix}
a_i & b_i\\ 
c_i & d_i
\end{bmatrix}
\begin{bmatrix}
a_j & b_j\\ 
c_j & d_j
\end{bmatrix}
-\begin{bmatrix}
a_j & b_j\\ 
c_j & d_j
\end{bmatrix}
\begin{bmatrix}
a_i & b_i\\ 
c_i & d_i
\end{bmatrix}
=
2i\begin{bmatrix}
a_k & b_k\\ 
c_k & d_k
\end{bmatrix}
$$
Doing the math on the left-hand side and equating the terms occupying the same position at the right-hand side, we get a set of 4 equations, one of which is $d_k=-a_k$. Since, $k$ is a free index any of the particular indexes set there (${1,2,3};{x,y,z}$) must give the same result, meaning that the general form of the matrices is
$$\sigma_i=\begin{bmatrix}
a_i & b_i\\ 
c_i & -a_i
\end{bmatrix} \tag{4'}$$
$(4')$ indicates that any $2 \times 2$ matrices satisfying condition $(3)$ must be traceless.
Squaring $(4')$ and applying condition $(2)$, we get
$$\begin{bmatrix}
a_i^2+b_ic_i & 0\\ 
0 & a_i^2+b_ic_i
\end{bmatrix}
=
\begin{bmatrix}
1 & 0\\ 
0 & 1
\end{bmatrix}$$
or
$$a_i^2+b_ic_i=1 \tag{5}$$
If one imagines $a,b$ and $c$ as points in a 3D space, it is easy to check that once a particular choice for them is made, such a choice creates a basis vector into that particular direction, thus, every other choices that has a component in that direction, can be written as a linear combination of the former, consequently, individual choices do not reduce generality.
With this in mind, the solutions to $(5)$ are:
a) $a=±1$ (basis for all solutions that $a$ do not vanishes)
This choice demands every possible value for $b$ and $c$ to be in the $(a,b)$ or in the $(a,c)$ planes, meaning $b=0,\,c=\lambda_1$ or $b=\lambda_2,\,c=0$ with $\lambda$ any real or complex scalar.
The 2 choices above give us 2 different matrices. However, it is easy to show (you can try by applying $(3)$ into those matrices) that any choice for $\lambda_1$ and $\lambda_2$ not simultaneously $0$ won't satisfy $(3)$, hence, for $a=±1$ there is only one possible solution, which is
$$\sigma_3=\begin{bmatrix}
1 & 0\\ 
0 & -1
\end{bmatrix} \tag{6}$$
The solutions $a=-1$ are not needed as it will be a scalar multiple of $(6)$.
b) $a=0$ (solutions that vanishes the right-hand side of $(5)$’s sum)
$$bc=1 \tag{7}$$
This places the solutions into the $(b,c)$ plane. Considering that for the sake of generality $a$ and $b$ could be complex numbers, all possible points in the $(b,c)$ plane can just be written as linear combinations of a pure real and a pure imaginary basis vectors. The simplest real basis vector in the $(b,c)$ plane satisfying $(7)$ is $b=1,\,c=1$, that gives us
$$\sigma_1=\begin{bmatrix}
0 & 1\\ 
1 & 0
\end{bmatrix} \tag{8}$$
The pure imaginary basis vector to $(7)$ is $b=-i,\,c=i$ (or the scalar multiple of this $b=i,\,c=-i$, which is obviously the same choice).
$$\sigma_2=\begin{bmatrix}
0 & -i\\ 
i & 0
\end{bmatrix} \tag{9}$$
The 3 matrices $(6),(8)$ and $(9)$ are the Pauli matrices and they spans the set of all matrices that satisfy the proposed postulates. They form the full set of basis vector for the $2 \times 2$ matrix representation $\sigma_i$s.
