Let \(T\) be a linear operator on a finite-dimensional space over a subfield of \(C\). Prove that \(T\) is semi-simple if and only if the following is true: If \(f\) is a polynomial and \(f(T)\) is nilpotent, then \(f(T) = 0\).
Solution
1. If \(T\) is semisimple, every nilpotent polynomial in \(T\) is zero.
By Exercise 2, \(f(T)\) is semisimple for every polynomial \(f\) [@HK71]. If in addition \(f(T)\) is nilpotent, then its minimal polynomial divides both a squarefree polynomial and a power of \(x\). Hence its minimal polynomial divides \(x\), so \(f(T)=0\).
2. Conversely, the stated property forces \(T\) to be semisimple.
Let \[ T=D+N, \qquad DN=ND, \] be the semisimple–nilpotent decomposition. Hoffman–Kunze’s decomposition theorem gives \(D=g(T)\) for some polynomial \(g\) [@HK71]. Therefore \[ N=T-D=(x-g(x))(T) \] is itself a polynomial in \(T\). By construction \(N\) is nilpotent. The assumed property therefore implies \(N=0\). Hence \(T=D\) is semisimple.
Thus the two conditions are equivalent.