I'm asked to prove that if two matrices $A,B$ are similar and $||\cdot||$ is a norm in $\mathbb{C}^n$, then we would have $\lim_{m\to \infty}||A^m||^{\frac{1}{m}} = \lim_{m\to \infty}||B^m||^{\frac{1}{m}}$.
What I got:
Since $A,B$ are similar, we have $B= P^{-1}AP$ for some invertible $P$. Then $\lim_{m\to \infty}||B^m||^{\frac{1}{m}} = \lim_{m\to \infty}||P^{-1}A^mP||^{\frac{1}{m}}$, but what next?
Thank you
On
Gelfand's formula https://en.wikipedia.org/wiki/Spectral_radius#Gelfand's_formula gives $\rho(A)= \lim_n \sqrt[n]{\|A^n\|}$ for any matrix norm.
If $A,B$ are similar they have the same eigenvalues and so $\rho(A) = \rho(B)$.
$\|AB\|\leq \|A\| \|B\|$. So $\|P^{-1}A^{m}P\|^{1/m}\leq \|P^{-1}\|^{1/m} \|A^{m}\|^{1/m} \|P\|^{1/m}$. Can you finish?