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.


2 Answers 2


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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .