Summing the traces of matrix powers

Question

Let $G=\langle h\rangle_n\subset{\rm GL}(m,\mathbb{C})$ be a cyclic group of order $n$. I wonder if there is a good formula for calculating the sum

$\sum_{g\in G}{\rm Tr}(g)$

via ${\rm Tr}(h)$, for example?

Thanks!

Answer 1

If $h$ has eigenvalues $\lambda_1, \ldots, \lambda_m$, then then $h^i$ has eigenvalues $\lambda_1^i, \ldots, \lambda_m^i$.

Since $h$ has order $n$, each $\lambda_j$ satisfies $\lambda_j^n=1$.

We have two cases:

• $\lambda_j=1$, in which case $\sum_{i=0}^{n-1} \lambda_j^i = n$, or
• $\lambda_j \neq 1$, in which case $0=(\lambda_j^n-1)/(\lambda-1) = \sum_{i=0}^{n-1} \lambda_j^i$

Hence $$\sum_{g\in G} \operatorname{Tr}(g) = \sum_{i=0}^{n-1} \operatorname{Tr}(h^i) = \sum_{i=0}^{n-1} \sum_{j=1}^m \lambda_j^i = \sum_{j=1}^m \sum_{i=0}^{n-1} \lambda_j^i = \sum_{j=1}^m \left\{\begin{matrix} n & \lambda_j=1 \\ 0 &\lambda_j \neq 1\end{matrix}\right\} = n\cdot \#\{ j : \lambda_j = 1\}$$

Unfortunately this quantity, “how many eigenvalues of $h$ are equal to 1?” cannot be read from the trace of $h$, since both $h_1 = \begin{bmatrix} i & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$ and $h_2 = \begin{bmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -i \end{bmatrix}$ have trace $i$ and order $4$, but $\sum_{g\in G} \operatorname{Tr}(g)$ equal to $4$ and $0$ respectively.