# 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 $$n$$.
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 $$\lim(2n\ln(|\lambda_i|)-\sqrt{2n}\ln (2))=+\infty$$.