I want construct the Pauli matrices starting from $$\sigma_i=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$$ by using only the anti-commutation relation $$ \sigma^i\sigma^h+\sigma^h\sigma^i=\{\sigma^i,\sigma^h\}=2\delta^{ih} $$
(Construction of Pauli Matrices Is a similar question, but use the commutators that I want to avoid).
From $(\sigma^i)^2=I$ I found: $$ \begin{cases} a^2+bc=1\\ d^2+bc=1\\ b(a+d)=0\\ c(a+d)=0 \end{cases} $$ and from this:
if $d=a\ne 0$ we have a scalar matrix that commutes with all other matrices, so it cannot be a Pauli matrix.
if $d=-a\ne 0$, $a=\pm1$ and $b=c=0$ and the matrix has the form $$ \begin{bmatrix} 1 & 0\\ 0& -1 \end{bmatrix} \qquad \mbox{or}\qquad \begin{bmatrix} -1 & 0\\ 0& 1 \end{bmatrix} $$ which matches $\sigma^3$.
if $d=a=0$ we have $bc=1$, so a matrix of the form $$ \begin{bmatrix} 0 & b\\ 1/b& 0 \end{bmatrix} $$ but now, how can I prove that the possible values for $b$ are just $-i$ and $1$ ?
