Characteristic polynomial and $p$-adic valuation Suppose I had a linear operator $L$ whose characteristic polynomial was $f(x) = x^{n} + a_{1}x^{n-1} + \cdots + a_{n-1}x + a_{n}$. Furthermore, I also know that the eigenvalues of $L$ have $p$-adic valuation $\geq 1$. Why does it follow that $a_{i} \in p^{i}\mathbb{Z}_{p}$ for all $i$? What if all the eigenvalues have $p$-adic valuation equal to 1?
The eigenvalues are the roots of the characteristic polynomial. Now you just need to recall the relation between the roots of a (monic) polynomial and its coefficients.
I can't comment at the moment - insufficient points. You need to go back to what the p-adic valuation means and basic facts about divisibility of integers by primes. I think it is a lot simpler than you realise.