Let $M$ be a $mxm$ positive semi-definite matrix with $tr(M)\le1$. Is there some non-trivial inequality of the type $tr(M^n)\ge f(tr(M^i), tr(M^j))$ with $f$ some function and $i,j\le n$ ?
And what if M is not positive semi-definite but just satisfies $tr(M)\le1$?
Just to give some context, I am trying to compute a sequence $F_n = tr(M^n) \in [0,1] $ and I expect it to be monotonically decreasing with n, but the matrices are too big to find exact results (either by hand or numerically), so I'm trying to find out some bound on how this scales with n. An inequality of that type would be a good start, since I can compute $F_n$ for small n by other means (boring physics stuff :P).
Any hint would be helpful. Thanks!
the trace is the sum of the eigenvalues, and the eigenvalues of $M^n$ are the $n$-th power of those in $M$. If $M$ is positive semidefinite, and you use the inequalities between generalized means then you get $$ \sqrt[n]{tr(M^n)/m} \ge \sqrt[p]{tr(M^p)/m} $$ for every $n\ge p$.
If $M$ is not definite, then the situation is a lot more grim, since $tr(M)$, alone, does not provide enough information. For example, if $$ M = \begin{pmatrix} -1 & 0\\ 0 & 2 \end{pmatrix} $$ then its trace is 1, but $tr(M^n) = 2^n \pm 1$ that diverges to infinity.