$T$ is normal if and only if exist polynomial $p$ s.t $T^{*}=p(T)$

Question

Let $V$ be an inner product space over $\mathbb{C}$, and let $T:v\to V$ be a linear transformation. Need proving that $T$ is normal if and only if exist polynomial $p\in\mathbb{C}[x]$ s.t $T^{*}=p(T)$.

Thanks!

Answer 1

Commutativity of $T$ and $T^*$ is trivial if $ T^* = p(T) $. To prove the converse, take $A : \mathbb{C}^n \rightarrow \mathbb{C}^n $ to be the corresponding normal matrix of $T$. Then let $ \lambda_1, \lambda_2,..., \lambda_k \in \sigma(A) $ be distinct eigenvalues, with $ E_{\lambda_j}(A) $ being orthogonal eigenspaces. Spectral theorem guarantees that $$ \mathbb{C}^n = E_{\lambda_1}(A)\oplus E_{\lambda_1}(A)\oplus...\oplus E_{\lambda_k}(A) $$ and hence if $ P_j $ is the projection to $E_{\lambda_j}(A) $ then we have the resolution of identity as $ P_1 + P_2 + ...+P_k = I $. Now we define polynomials $ \phi_j \in \mathbb{R}[X] $ for $ 1 \leq j\leq k $ as the following $$ \phi_j(X) = \prod_{ j \neq l = 1}^k \Bigl(\frac{X-\lambda_l}{\lambda_j-\lambda_l}\Bigr) $$ Now observe that $ \phi_j(\lambda_i) = \delta_{ij} $, thus for all $x = x_1 + x_2 + ...+ x_k \in \mathbb{C}^n $ with $ P_jx = x_j $ hence $ Ax_j = \lambda_jx_j $, we find that $ \phi_j(A)x_i = x_j\delta_{ij} $, in other words $ \phi_j(A)x = P_jx $. So $ P_j = \phi_j(A) $. Now recalling that $ E_\lambda(A) = E_{\bar{\lambda}}(A^*) $, we see that $ A^*P_j x = \bar{\lambda_j}P_j x$ and with resolution of identity we find $ A^* = A^*P_1 + A^*P_2 + ... + A^*P_k = \bar{\lambda_1}P_1 + ...+ \bar{\lambda_k}P_k $ which using previous results imply $$ A^* = \bar{\lambda_1}\phi_1(A) + \bar{\lambda_2}\phi_2(A) + ... + \bar{\lambda_k}\phi_k(A) $$