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.

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 $n$.