TITLE: Pair of square matrices related by traces formulas
QUESTION [8 upvotes]: Let $A$ and $B$ be two $n\times n$ matrices over $\mathbb{C}$. Assume that for every $k\geq 1$ it holds  $tr(A^k) = tr(B^{2k-1})$. What can we say about the possible eigenvalues of $A$ and of $B$? How limited is the set of possible eigenvalues? Must it be, for example, finite?

REPLY [4 votes]: Suppose that all of the nonzero eigenvalues of $A$ are the same, so $\text{tr}(A^k) = m \lambda^k$ for some $\lambda$ and some $m \le n$. If $\lambda = \frac{1}{m^2}$ then $\text{tr}(A^k) = \frac{1}{m^{2k-1}}$ and so $B$ can have eigenvalue $\frac{1}{m}$ and $\pm$ pairs as in Mike Jury's comment.
Similarly, suppose that all of the eigenvalues of $B$ that don't come in $\pm$ pairs are the same, so $\text{tr}(B^{2k-1}) = m \lambda^{2k-1}$ for some $\lambda$ and some $m \le n$. If $\lambda = m$ then $\text{tr}(B^{2k-1}) = m^{2k}$ and so $A$ can have eigenvalue $m^2$and all other eigenvalues $0$ as in Mike Jury's comment.
On the other hand, if you require that all of the nonzero eigenvalues of $A$ and $B$ have distinct absolute values then I think you can show that $A$ has no nonzero eigenvalues by a comparison of the growth rates of the LHS and the RHS as $k \to \infty$.