Prove that for all $u,v\in\Hom(E)$: $\tr(uv)=\tr(vu)$ and $\tr u^*=\tr u$.
1. Suppose $(a_{jk})$ and $(b_{jk})$ are the matrices of $u$ and $v$ with respect to a fixed basis. Then the matrix of $uv$ is $(c_{jk})=AB$, i.e.
$$
\tr(uv)
=\sum_jc_{jj}
=\sum_{j,k}a_{jk}b_{kj}
=\sum_{j,k}a_{kj}b_{jk}
=\sum_{j,k}b_{jk}a_{kj}
=\tr(vu)~.
$$
2. Since $Je_j$ is the dual basis to $e_j^*$, we get by definition of the trace:
$$
\tr(u^*)
=\sum_jJe_j(u^*(e_j^*))
=\sum_ju^*(e_j^*)(e_j)
=\sum_je_j^*(u(e_j))
=\tr(u)~.
$$