Over a perfect field, every linear map splits uniquely into commuting semisimple and nilpotent parts.Suggest a slogan
Let \(k\) be a perfect field, let \(V\) be a finite-dimensional \(k\)-vector space, and let \(T\colon V\to V\) be \(k\)-linear. Then there are unique \(k\)-linear maps \(S,N\colon V\to V\) with \[ T=S+N,\qquad SN=NS, \] where \(S\) is semisimple, that is, diagonalizable over an algebraic closure \(\kbar\), and \(N\) is nilpotent. If \(T\) is invertible, there are unique \(k\)-linear maps \(S,U\colon V\to V\) with \(T=SU\) and \(SU=US\), where \(S\) is semisimple and \(U\) is unipotent.