$ 2^{\sqrt n} \leq |\operatorname{ tr}( A^n)| \leq 2020\cdot2^{\sqrt n} $

Question

I am trying to prove the following claim
There does not exist an integer $r\geq1 $ and a symmetric matrix $ A \in M_r(\Bbb R) $ Such that for all $ n \in \Bbb N $ we have
$$ 2^{\sqrt n} \leq |\operatorname{tr}( A^n)| \leq 2020\cdot2^{\sqrt n} $$ Please give me some idea how to start this problem?

Answer 1

I'm going to prove it by contradiction using the fact that real symmetric matrices have real eigenvalues and are diagonalizable. Let $\lambda_1,\lambda_2,\cdots,\lambda_r$ be the real eigenvalues of the matrix $A$. Then $|tr(A^n)|=|\lambda_1^n+\lambda_2^n+\cdots+\lambda_r^n|$.

Dividing both sides of your equation by $2^\sqrt{n}$. We get, \begin{align*} 1\leq \biggl|{\sum_{i=1}^{r} \frac{\lambda_i^n}{2^\sqrt{n}}}\biggr| \leq2020 \end{align*} Call the middle expression $S_n$

$\mathbf{Case 1}$: When $|\lambda_i| \leq 1 \: \forall i=1,2,\cdots, r$. Then taking limit $ n \rightarrow \infty $ in the above inequality we get $1\leq \lim_{n\rightarrow \infty}S_n = 0$. A contradiction

$\mathbf{Case 2}$: When at least one $|\lambda_i| > 1$. Wlog assume that to be $\lambda_1$. Then taking limit $n\rightarrow \infty$ in the above inequality we get $\lim_{n\rightarrow \infty}S_{2n} = +\infty \leq 2020$. Again a contradiction

$\mathbf{Edit}$: As pointed out in the comments the cases should be split accordingly whether $\lambda_i \geq 1$ and $\lambda_i < 1$.