$ tr(C_1 C_2 \cdots C_{n+1}) = \frac{1}{(n+1)!} \sum_{\sigma \in S_{n+1}} tr(C_{\sigma_1} \cdots C_{\sigma_{n+1}}) $

Question

Let $C_i$ be a matrix. How can I prove the following?

$$\mbox{tr} \left( C_1 C_2 \cdots C_{n+1} \right) = \frac{1}{(n+1)!} \sum_{\sigma \in S_{n+1}} \mbox{tr} \left( C_{\sigma_1} \cdots C_{\sigma_{n+1}} \right)$$

For $n=1$, I can see this actually holds. However, I am having problems with general $n > 1$. I try to prove this by induction, but having trouble because when $n \to n+1$, symmetric group $S_n \to S_{n+1}$, and I have no idea how to handle this.

Is there any way to prove this? If there are better way than induction, please let me know.

Answer 1

For $n=2$ and using the invariance of the trace under cyclic permutations, this can be reduced to $$ tr(ABC)=\frac12(tr(ABC)+tr(BAC)). $$ As all matrices are free to chose, this now implies $AB=BA$. As most matrices do not commute, your claim is wrong already in this first non-trivial case.