homework and exercises - Representation $(1/2,1/2)$ of Lorentz group

I want to show that the Lorentz representation $(1/2,1/2)$ corresponds to the normal vectorial representation $A^\mu$. For this I need to show that the double spinors $A_{ij}=(A_\mu\sigma^\mu\sigma^2)_{ij}$ that transforms like $$A_{ij}\rightarrow(\Lambda_R)_{ik}(\Lambda_L)_{jl}A_{kl}$$ implies that the $A_\mu$ transforms like $$A^\mu\rightarrow\Lambda^\mu_{\>\>\nu} A^\nu.$$

But I really don't know how to even start this. Maybe I have issues understanding what is this representation. Here is what I did: $$A'_\mu\sigma^\mu\sigma^2=\Lambda_R\Lambda_LA_\mu\sigma^\mu\sigma^2$$ Then multiply by $\sigma^2\sigma_\nu$ on the left and taking the trace $$Tr(\sigma^2\sigma_\nu A'_\mu{\sigma^\mu\sigma^2)=Tr(\sigma^2\sigma_\nu\Lambda_R\Lambda_LA_\mu\sigma^\mu\sigma^2)\\ 2A'_\mu\delta_\nu^\mu=Tr(\sigma^2\Lambda_R\sigma^2\Lambda_LA_\mu\delta^\mu_\nu)\\ 2A'_\nu=Tr(\Lambda_L^\dagger\Lambda_LA_\nu)\\ 2A'_\nu=Tr(\sigma^2\Lambda_L^{-1}\sigma^2\Lambda_LA_\nu)\\ 2A'_\nu=A_\nu}$$ which can't be good.

EDIT:

I came up with a solution: $$A_{ij}'=(\Lambda_R)_{ik}(\Lambda_L)_{jl}A_{kl}\\ (A_\mu'\sigma^\mu\sigma^2)_{ij}=(\Lambda_R)_{ik}(A_\mu\sigma^\mu\sigma^2)_{kl}(\Lambda_L)_{lj}^T\\ \text{Tr}(\sigma^2\sigma_\nu A_\mu'\sigma^\mu\sigma^2)=\text{Tr}(\sigma^2\sigma_\nu\Lambda_RA_\mu\sigma^\mu\sigma^2\Lambda_L^T)\nonumber\\ \text{Tr}(\sigma_\nu A_\mu'\sigma^\mu)=\text{Tr}(\sigma_\nu\Lambda_RA_\mu\sigma^\mu\sigma^2\Lambda_L^T\sigma^2)\nonumber\\ A_\mu'\text{Tr}(\sigma_\nu\sigma^\mu)=\text{Tr}(\sigma_\nu\Lambda_R\sigma^\mu\Lambda_R^*)A_\mu\nonumber\\ 2A_\mu'{\delta}{_\nu^\mu}=\text{Tr}(\sigma_\nu\Lambda_R\sigma^\mu\Lambda_R^*)A_\mu\nonumber\\ A_\nu'=\frac{1}{2}\text{Tr}(\sigma_\nu\Lambda_R\sigma^\mu\Lambda_R^*)A_\mu\nonumber\\ A_\nu'={\Lambda}{_\nu^{\>\>\mu}}A_\mu.$$ but I'm not sure why my indices are not good (compared to what I'm supposed to get ${A^{\mu}}'=\Lambda^\mu_{\>\>\nu}A^\nu$).