Let $\varphi$ be the map from $\rm{SL}(2,\mathbb{C})\to SO(1,3)$ defined over $\widehat{\varphi(A)x}= A\hat{x}A^\dagger$, where $\hat{x}=\sum_{\mu=0}^3 x_\mu \sigma_\mu$ for all $x\in\mathbb{R}^4$ and $\{\sigma_i\}_{i=0}^3$ the set of Pauli matrices. I am trying to prove that $\varphi(A)_{ij} = \frac{1}{2}{\rm{tr}}(\sigma_i A\sigma_j A^\dagger)$.
For this I started with $$\widehat{\varphi(A)x}=[\varphi(A)x]_j\sigma_j = \varphi(A)_{jk}x_k\sigma_j=Ax_k\sigma_kA^\dagger = A\hat{x}A^\dagger.$$ Since this is true for all $x\in\mathbb{R}^4$ we find $\varphi(A)_{jk}\sigma_j=A\sigma_kA^\dagger$. For the next step I just multiplied both sides with $\sigma_p$, took the trace of it and multiplied the result with $1/2$, which gives me $$\frac{1}{2}{\rm tr}(\sigma_p\varphi(A)_{jk}\sigma_j)= \frac{1}{2}{\rm tr}(\sigma_pA\sigma_kA^\dagger).$$ The right-hand side is exactly what I'm looking for, but I don't understand why the left-hand side should equal to $\varphi(A)_{pk}$. Is this maybe the wrong approach to prove the statement, or what exactly am I missing?
The problem itself comes from a prove I'm trying to do in physics, but this seems to me more like a mathematical problem, so I hope asking this here is fine.
It looks good so far in my opinion. The missing step is the last one. Namely, $ \frac{1}{2} Tr(\sigma_p \phi(A)_{jk} \sigma_j) = \frac{1}{2} \phi(A)_{jk} Tr(\sigma_p \sigma_j)$.
Please note that $Tr(\sigma_p \sigma_j) = 2 \delta_{pj}$ with $\delta_{pj}$ the Kronecker delta symbol.
This gives that $ \frac{1}{2} \phi(A)_{jk} Tr(\sigma_p \sigma_j) = \phi(A)_{jk} \delta_{pj} = \phi(A)_{pk}$,
which equals $ \frac{1}{2} Tr(\sigma_p A \sigma_k A^\dagger)$ which you had to show.