Let $ f(x)= x^{n}-nx+1 $ and let $A$ be an $ n \times n $ matrix with characteristic polynomial $f$. I am going to prove that if $n> 2$ then $A$ is diagonalizable over the complex numbers.
If we show that $f$ and f' do not have common zeros can we say that $f$ has no multiple zeros namely if $ \alpha $ is a zero of f $ (x-\alpha)^{2} $ is not a divisor of $f$? So we have $n$ different eigenvalues of $f$. Hence $A$ is diagonalizable.
Also $A$ is diagonalizable if $n=2$.