Originally Posted by

**skyking** Thinking about this again, I am not so sure that I can say that if T is normal then P(T) is normal. Can I say this?

Let $\displaystyle \displaystyle P(T)=\sum_{j=0}^{n}\alpha_j T^j$ we then have that $\displaystyle \displaystyle P(T)^\ast=\sum_{j=0}^{n}\overline{\alpha_j}\left(T ^j\right)^\ast$. Our first claim that $\displaystyle T^k\left(T^j\right)^\ast=\left(T^j\right)^\ast T^k$ for every $\displaystyle k,j\in\mathbb{N}\cup\{0\}$. We do this by letting $\displaystyle k$ be arbitrary and proceeding by induction on $\displaystyle j$. Evidently this is true when $\displaystyle j=0$ and so assume it's true for $\displaystyle j$ then

$\displaystyle \begin{aligned}T^k\left(T^{j+1}\right)^\ast &=\left(T^k \left(T^j\right)^\ast T^\ast\\ &=\left(T^j\right)^ast T^k T^\ast\\ &=\left(T^j\right)^\ast T^{k-1}TT^\ast\\ &= \left(T^j\right)^\ast T^{k-1}T^\ast T\\ &= \vdots\\ &= \left(T^j\right)^\ast T^\ast T^k\\ &= \left(T^{j+1}\right)^\ast T^k\end{aligned}$

then with this we see that

$\displaystyle \displaystyle \begin{aligned}P(T)P(T)^\ast &= \left(\sum_{k=0}^{n}\alpha_k T^k\right)\left(\sum_{j=0}^{n}\overline{\alpha_j}\ left(T^j\right)^\ast\right)\\ &= \sum_{k=0}^{n}\sum_{j=0}^{n}\left(\alpha_j T^k\right)\left(\overline{\alpha_j}\left(T^\ast\ri ght)^j\right)\\ &= \sum_{k=0}^{n}\sum_{j=0}^{n}\left(\overline{\alpha _j}\left(T^j\right)^\ast\right)\left(\alpha_k T^k\right)\\ &= \left(\sum_{j=0}^{n}\overline{\alpha_j}\left(T^j\r ight)^\ast\right)\left(\sum_{k=0}^{n}\alpha_k T^k\right)\\ &= P(T)^\ast P(T)\end{aligned}$