Let \(V\) be a finite-dimensional vector space over the field of complex numbers. Let \(T\) be a linear operator on \(V\) and let \(D\) be the diagonalizable part of \(T\). Prove that if \(g\) is any polynomial with complex coefficients, then the diagonalizable part of \(g(T)\) is \(g(D)\).
Solution
Write the Jordan decomposition of \(T\) as \[ T=D+N, \] where \(D\) is diagonalizable, \(N\) is nilpotent, and \(DN=ND\) [@HK71].
1. The operator \(g(D)\) is diagonalizable.
Choose a basis in which \[ D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). \] Then in the same basis \[ g(D)=\operatorname{diag}(g(\lambda_1),\ldots,g(\lambda_n)), \] so \(g(D)\) is diagonalizable.
2. The difference \(g(T)-g(D)\) is nilpotent and commutes with \(g(D)\).
Let \[ g(x)=\sum_{r=0}^m a_rx^r. \] Since \(D\) and \(N\) commute, the binomial theorem gives \[ (D+N)^r-D^r =\sum_{j=1}^r\binom rjD^{r-j}N^j =N\sum_{j=1}^r\binom rjD^{r-j}N^{j-1}. \] Hence \[ M:=g(T)-g(D)=NQ(D,N) \] for a polynomial \(Q\) in the commuting operators \(D,N\). Thus \(M\) commutes with \(D\), \(N\), and in particular with \(g(D)\).
If \(N^s=0\), then \[ M^s=N^sQ(D,N)^s=0, \] because all factors commute. Therefore \(M\) is nilpotent.
3. Uniqueness identifies the diagonalizable part.
We have written \[ g(T)=g(D)+M \] as a sum of a diagonalizable operator and a commuting nilpotent operator. The uniqueness part of the diagonalizable–nilpotent decomposition theorem therefore implies that the diagonalizable part of \(g(T)\) is exactly \[ \boxed{g(D)}. \]