Suppose we map each 4-vector $V \in M^4$, where $M^4$ is the Minkowski space, to a $2 \times 2$ Hermitian matrix $$ V= V^μ σ_μ $$ where $σ^0$ is the identity $2 \times 2$ matrix and $σ_i$ are the Pauli matrices. We see that the determinant of $V$ is $η_{μν} V^μ V^ν$, where $η_{μν}=diag(1,-1,-1,-1)$ is the Minkowski metric. It is quite easy to find the inverse map: $$V^μ = \frac12 Tr (V σ_μ).$$
Any transformation \begin{equation} {V} \rightarrow {V}^{\prime}=\alpha {V} \alpha^{*} \tag{1} \label{eq:1} \end{equation} with $a$ a complex $2 \times 2$ matrix with determinant $1$, and $a^*$ its Hermitian adjoint, preserves the determinant, ie $ det(V^{'})=det(V)$. Hence it corresponds to a Lorentz transormation $Λ^μ _ν$ (an element of $SO(3,1)$) : $$V^{\prime \mu}=\Lambda_{\nu}^{\mu} V^{\nu} $$
What I am trying to prove is the form of the Lorentz transformation, given the transformation of the matrix $V$ by $a$. I know that the result is:
$$\Lambda_{\nu}^{\mu}(\alpha)=\frac{1}{2} \operatorname{Tr} \alpha \sigma_{\nu} \alpha^{*} \sigma_{\mu} \tag{2} $$
I tried taking equation (1) to get:
$$Λ^μ_ν V^ν σ_μ = α V^μ σ_μ α^* $$
which looks promising. Now this has to holds for all $V^μ$, but I'm not sure what this allows me to do.
