Trace II

We have $T_AT_B(X) \; = \; ABX - AXB - BXA + XBA$ so, using the summation convention $\big(T_AT_B(X)\big)_{uv} \; = \; A_{ur}B_{rs}X_{sv} - A_{ur}X_{rs}B_{sv} - B_{ur}X_{rs}A_{sv} + X_{ur}B_{rs}A_{sv}$ If $X = E(i,j)$ is the matrix with $1$ in the $(i,j)$ th position, and $0$ everywhere else, then $X_{rs} = \delta_{ri}\delta_{sj}$ , and hence $\big(T_AT_BE(i,j)\big)_{u,v} \; = \; A_{ur}B_{ri}\delta_{vj} - A_{ui}B_{jv} - B_{ui}A_{jv} + \delta_{ui}B_{js}A_{sv}$ Thus, still using the summation convention, $\begin{aligned} B_2(A,B) & = \; \big(T_AT_BE(i,j)\big)_{i,j} \\ & = \; A_{ir}B_{ri}\delta_{jj} - A_{ii}B_{jj} - B_{ii}A_{jj} + \delta_{ii}B_{js}A_{sj} \\ & = \; 2nA_{ij}B_{ji} - 2A_{ii}B_{jj} \; = \; 2n\mathrm{Tr}(AB) - 2\mathrm{Tr}(A)\mathrm{Tr}(B) \\ & = \; 2B_1(A,B) - 2\mathrm{Tr}(A)\mathrm{Tr}(B) \; = \; 2B_1(A,B) \end{aligned}$ for all $A,B \in \mathcal{V}$ , so that $\lambda = \boxed{\tfrac12}$ .