Does there exist a symmetric matrix $A$ such that $2^{\sqrt{n}}\le |\operatorname{Tr}(A^n)|\le2020 \ \cdot 2^{\sqrt{n}}$ for all $n$ Does there exist a symmetric matrix $A$  such that $2^{\sqrt{n}}\le |\operatorname{Tr}(A^n)|\le2020 \cdot2^{\sqrt{n}}$ for all $n$?
I think no. The trace of $A^n$ equals $\sum\limits_{i=1}^n\lambda_i^n$ where $\lambda_i$ are the eigenvalues of $A$. Now, if the absolute value of trace of $A$ is bounded below by $2$, then I think the trace of $A^n$ will grow infinitely. Am I right? Thanks beforehand.
 A: As you said, for a symmetric matrix $A$ we have $Tr(A^n) = \sum_i \lambda_i^n$. Now consider two cases for $\lambda_\max = \max_i |\lambda_i|$:

*

*$\lambda_\max \le 1$. Then $|Tr(A^n)| \le n$, and which is smaller than $2^{\sqrt n}$ for a sufficiently large $n$.

*$\lambda_\max > 1$. Then for even $n$ we have $Tr(A^n) \ge \lambda_\max^n$, which is greater than $2020 \cdot 2^{\sqrt n}$ for a sufficiently large $n$.

A: Let $A\in\mathbb{R}^{d\times d}$ and $A$ is symmetric. The eigenvalues $\lambda_1,\ldots,\lambda_d$ of $A$ are real.
For all $n\in\mathbb{N}$
$$Tr(A^n)=\sum_{i=1}^n\lambda_i^n$$
If we assume that  for all $n\in\mathbb{N}$
$$2^{\sqrt{n}}\leq Tr(A^n)\leq 2020 \times2^{\sqrt{n}}$$
There exists necessarily $i\in\{1,\ldots,d\}$ such that $|\lambda_i|>1$, because otherwise the sequence $Tr(A^n)$ is bounded.
Then for all $n\in\mathbb{N}$
$$\lambda_i^{2n}\leq 2020\times 2^{\sqrt{2n}}$$
So, for all $n\in\mathbb{N}$
$$ 2n\ln(|\lambda_i|)-\sqrt{2n}\ln (2)\leq\ln(2020)$$
which is impossible since $\lim(2n\ln(|\lambda_i|)-\sqrt{2n}\ln (2))=+\infty$.
