For some fixed square matrix $X$ for which $\lim_{n\rightarrow\infty} X^n = 0$, it is apparent that the trace of $X^n$ must also converge to zero as $n\rightarrow\infty$. Is there anything I can do to upper-bound the trace of $X^k$ for some positive integer $k$? It would be nice if I could do this just in terms of the trace and determinant of $X$, but I can use other properties of $X$ if they are necessary.
Does anyone have any such bounds which can be helpful?
If $X^n$ tends to $0$ then all eigenvalues must be $|\lambda_k| < 1$. This follows (for example) from Jordan's decomposition of $X$. Since $tr(X^n) = \sum \lambda_k^n$ you get $|tr(X^n)| <= M*max|\lambda_k|^n$. Where $M$ is the dimension of space. So, basically, trace has to tend to zero like power of the largest eigenvalue.