Let $C \in M_n(\mathbb C)$ and suppose that $$Tr((m+m^*)C) = 0 $$ for every $m \in \mathfrak sl_n$ (i.e for every $n \times n$ matrix with null trace), with $*$ the transposition and conjugation. Then $C = c\cdot I$, i.e. $C$ is a scalar matrix.
The paper says that it's easy, but I can't see it.
Thanks!
Let $C \in \mathfrak{gl}(n, \mathbb{C})$. Then it follows from the fact that $\mathfrak{gl}(n, \mathbb{C})\cong \mathfrak{sl}(n, \mathbb{C})\oplus \mathbb{C}$, we see that \begin{align} C = cI+N \end{align} where $N$ is traceless. Next, observe \begin{align} (m+m^\ast)C = cm+cm^\ast + mN+m^\ast N \end{align} which means \begin{align} \operatorname{Tr}[(m+m^\ast)C] = \operatorname{Tr}[(m+m^\ast)N] =\operatorname{Tr}[mN]+\operatorname{Tr}[m^\ast N]= 0 \end{align} for all $m\in \mathfrak{sl}(n, \mathbb{C})$. In particular, if $m=N$, we see that \begin{align} \operatorname{Tr}[N^2]+\operatorname{Tr}[N^\ast N]=0 \end{align} if and only if $N =0$.
Edit: Set $m = N$ then we have \begin{align} \operatorname{Tr}[N^2]+\operatorname{Tr}[N^\ast N]= 0. \end{align} Next, set $m = iN$, then we have that \begin{align} i\operatorname{Tr}[NN]-i\operatorname{Tr}[N^\ast N]= 0. \end{align} Hence $\operatorname{Tr}[N^\ast N]= \pm \operatorname{Tr}[N^2]$ which means $\operatorname{Tr}[N^\ast N]=0$.