If \(N\) is a nilpotent linear operator on \(V\), show that for any polynomial \(f\) the semi-simple part of \(f(N)\) is a scalar multiple of the identity operator (\(F\) a subfield of \(C\)).
Solution
Write \[ f(x)=f(0)+xq(x). \] Then \[ f(N)=f(0)I+Nq(N). \] The scalar operator \(f(0)I\) is semisimple. Also \(Nq(N)\) is nilpotent: if \(N^r=0\), then, since \(N\) commutes with \(q(N)\), \[ (Nq(N))^r=N^rq(N)^r=0. \] The two summands commute. Hence this is a semisimple–nilpotent decomposition of \(f(N)\). By uniqueness of that decomposition [@HK71], the semisimple part of \(f(N)\) is exactly \[ \boxed{f(0)I}. \]