Prove norm of matrix greater than spectral radius

Question

Given an $n \times n$ matrix $A$ with spectral radius $\sigma(A) = 1$, prove that there is an $m_0$ so that $\|A^m\| > 1$ for every $m \geq m_0$.

Solution attempt:

I think the question is in error, and I think the question should state $||A^m||\geq1$ instead of $||A^m||>1$. I found a proposition which states for any $k\geq 1$, $\sigma (A) \leq||A^k||^{\frac{1}{k}}$

By the proposition, $$||A^k||^{\frac{1}{k}}\geq\sigma(A)=1\\ (||A^k||^{\frac{1}{k}})^k \geq 1^k=1\\ \text{thus,}\,||A^k||\geq 1$$

Does anyone think this solution is incorrect? If so, what have I done wrong?

Answer 1

Yes, there is an error in the question. Your solution is correct.

Answer 2

For example, $\|I^m\| = 1 = \sigma(I)$ for all $m$, so $\|A^m\|>1$ is incorrect.

Answer 3

Remark: one can show that

$$\sigma(A)= \inf \{||A^k||^{\frac{1}{k}}: k \in \mathbb N\}= \lim_{k \to \infty}||A^k||^{\frac{1}{k}}.$$