Let \(T\) be a normal operator on a finite-dimensional complex inner product space. Prove that there is a polynomial \(f\), with complex coefficients, such that \(T^* = f(T)\). (Represent \(T\) by a diagonal matrix, and see what \(f\) must be.)
Solution
Let the distinct characteristic values of \(T\) be \[ \lambda_1,\ldots,\lambda_r. \] Since \(T\) is normal, the spectral theorem gives an orthonormal basis in which \[ [T]=\operatorname{diag}(\lambda_{i_1},\ldots,\lambda_{i_n}), \] where each diagonal entry is one of the distinct \(\lambda_j\). In the same basis, \[ [T^*]=\operatorname{diag}(\overline{\lambda_{i_1}},\ldots, \overline{\lambda_{i_n}}). \]
Choose the Lagrange interpolation polynomial \[ f(x)= \sum_{j=1}^r \overline{\lambda_j} \prod_{k\ne j}\frac{x-\lambda_k}{\lambda_j-\lambda_k}. \] The eigenvalues are distinct, so the denominators are nonzero, and \[ f(\lambda_j)=\overline{\lambda_j} \] for every \(j\).
Therefore, in the chosen orthonormal eigenbasis, \[ [f(T)] =\operatorname{diag}(f(\lambda_{i_1}),\ldots,f(\lambda_{i_n})) =\operatorname{diag}(\overline{\lambda_{i_1}},\ldots, \overline{\lambda_{i_n}}) =[T^*]. \] Hence \[ \boxed{T^*=f(T)}. \]