Fierz Identity calculation

While reading an article, it’s said that to simplify the following Dirac structure $$left(P_Lv_j^dbar{v}^s_kP_Rright)_{alphabeta}tag{1}label{1}$$ where $j,k$ are color indices and $d,s$ indicate which quarks the spinor refer to, we’re using Fierz identities projecting onto the set of complete operators (we are in dimension $4$) for objects containing two Lorentz indices $$P_L, P_R, gamma^mu P_L, gamma^mu P_R, sigma^{munu}$$The $P_{L/R}$ operators are the Weyl projection operators. The only relevant structure for $(ref{1})$ is the $gamma^mu P_R$ structure, so we project on it using $frac{1}{2}text{Tr},gamma^mu P_Lcdots$. Now the article says that $$frac{1}{2}text{Tr},gamma^mu P_L P_Lv^d_jbar{v}^s_k P_R = -frac{1}{2}bar{v}^s_kgamma^mu P_L v^d_j$$ I think that what is going on is just that $P_L^2=P_L$ so that $$frac{1}{2}text{Tr},gamma^mu P_L P_Lv^d_jbar{v}^s_k P_R = frac{1}{2}text{Tr},gamma^mu P_Lv^d_jbar{v}^s_k P_R = frac{1}{2}text{Tr},P_Rgamma^mu P_L P_Lv^d_jbar{v}^s_k = frac{1}{2}text{Tr},gamma^mu P_L P_Lv^d_jbar{v}^s_k == frac{1}{2}text{Tr},gamma^mu P_Lv^d_jbar{v}^s_k$$ where I used the cyclicity of the trace, the fact that $P_Lgamma^mu = gamma^mu P_R$ since $gamma_5$ anticommutes with all the $gamma^mu$, and again the projector property. But then I don’t know how to go to the last result. I imagine that the minus sign comes from the anticommutation of the $v$ spinors but then I don’t know how he got rid of the trace.