Infinite sum of matrix traces I am trying to work out the following sum where C is an n by n matrix:
$$S=\sum_{k=1}^\infty{\ln \left|e^{C^k} \right|}$$
The matrix $C$ has the property that $\operatorname{tr}(C^{2m-1})=0 $ when $m\in \mathbb{N}$ therefore I figured the sum should be equal to:
$$S=\sum_{k=1}^\infty{\ln \left|e^{C^{2k}} \right|}$$
Another property is that all the values in C are natural numbers and their sum is less than 2n.
From here I dont know where to go, but I tried the following:
$$S={\ln \left|\prod_{k=1}^{\infty} e^{C^{2k}} \right|}={\ln \left| e^{\sum_{k=1}^\infty{C^{2k}}} \right|}$$ Can I now treat the sum as a geometric series?
You will obtain
$$
S=\sum_{k=1}^\infty \ln|e^{C^k}|=\sum_{k=1}^\infty {\rm Tr}(C^k)
=\sum_{k=1}^\infty {\rm Tr}(C^{2k})=\sum_{k=1}^\infty \sum_{i=1}^N \lambda_i^{2k}
=\sum_{i=1}^N\frac{\lambda^2_i}{1-\lambda^2_i}
$$
and a property
$$\sum_{i=1}^N \lambda_i^{2k+1}=0$$
where $N$ is the dimension of the square matrix $C$ , $\lambda_i$ are its eigenvalues. Without any more conditions, you cannot simplify it further.
You will obtain
$$
S=\sum_{k=1}^\infty \ln|e^{C^k}|=\sum_{k=1}^\infty {\rm Tr}(C^k)
=\sum_{k=1}^\infty {\rm Tr}(C^{2k})=\sum_{k=1}^\infty \sum_{i=1}^N \lambda_i^{2k}
=\sum_{i=1}^N\frac{\lambda^2_i}{1-\lambda^2_i}
$$
and a property
$$\sum_{i=1}^N \lambda_i^{2k+1}=0$$
where $N$ is the dimension of the square matrix $C$ , $\lambda_i$ are its eigenvalues. Without any more conditions, you cannot simplify it further.