Gelfand's formula, different field Gelfand's formula says that for a complex matrix $A \in \mathbb{C}^{n \times n}$, $$\rho(A) = \lim_{m \rightarrow \infty} \|A^m\|^{1/m},$$ where $\rho$ is the spectral radius (norm of maximal eigenvalue / maximal norm of eigenvalues) and $\|\cdot \|$ is any matrix norm.
I looked at the proof at http://en.wikipedia.org/wiki/Spectral_radius and to me it seems like this should be valid (maybe with slight changes to the proof) over any algebraically closed field $K = \overline{K}$ with a nontrivial valuation (in particular, over $\overline{\mathbb{Q}_p}$).
Can anyone confirm this?
 A: It seems fine to me: the only point to be careful of is to note that since you've assumed the normed field $K$ to be algebraically closed, its value group must be dense (it cannot be discrete since if it were, there would be a uniformizer and it couldn't have $n$th roots). So for any $\epsilon > 0$, we can find some $a \in K^\times$ such that $\rho(A) < |a| < \rho(A) + \epsilon$. Then
$\rho(a^{-1}A) = |a|^{-1} \rho(A) < 1$
again, so we do the same thing as on Wikipedia to get that for sufficiently large $k$, we have
$\left\lvert A^k \right\lvert ^{1/k} < |a| < (\rho(A) + \epsilon)$.
Then we use density again to find some $b \in K^\times$ such that $\rho(A) - \epsilon < |b| < \rho(A)$ and then we note that
$\rho(b^{-1}A) = |b|^{-1} \rho(A) > 1$
so then, for sufficiently large $k$ again, we have
$\left\lvert A^k \right\lvert^{1/k} > |b| > (\rho(A) - \epsilon)$.
Note also that the proof that given on Wikipedia that if $J_m(\lambda)$ is a Jordan block with $|\lambda| < 1$ then $\lim_{k \to \infty} J_m(\lambda)^k = 0$ is even easier when $K$ is nonarchimedean, since all of the binomial coefficients in the expression for $J_m(\lambda)^k$ are then integers which must have norm at most 1 (in the archimedean case, one needs to note that these binomial coefficients are polynomials in $k$ and then compare their convergence to exponentials, which is also easy but not quite as easy). 
