characteristic vectors for $Τ$ which are associated with distinct characteristic values are orthogonal

Question

I am having problem in doing the following problems:

$1)$ If $T$ is a normal operator, prove that characteristic vectors for $Τ$ which are associated with distinct characteristic values are orthogonal.

$2)$ Let $Τ$ be a normal operator on a finite-dimensional complex inner product space. Prove that there is a polynomial $f$, with complex coefficients, such that $T^* = f(T)$.

Answer 1

The following Lemma 1 and Theorem 1 will suffice for a solution to (1), and Theorem 2 for a solution to (2).

Let $A\in M_{n\times n}(\mathbb{C})$ and $v\in \mathbb{C}^n$. Define $A^*= \overline{A}^T$, $v^*=\overline{v}^T$, and $\langle v, w \rangle = v^* w$.

Lemma 1 [Common Eigenvector for Commuting Matrices]

Let $A, B \in M_{n\times n}(\mathbb{C})$ with $AB=BA$. Then there is a common eigenvector of $A$ and $B$.

Proof) Let $\lambda$ be an eigenvalue of $B$ and let $V_{\lambda}$ be its corresponding eigenspace. Then for any $v\in V_{\lambda}$, \begin{align*} BAv = ABv=\lambda Av. \end{align*} This gives $Av\in V_{\lambda}$. Thus, $V_{\lambda}$ is $A$-invariant. Let $w$ be an eigenvector for the restriction $A|_{V_{\lambda}}$, we see that $w$ is an eigenvector for both $A$ and $B$.

Theorem 1 [Spectral Theorem for Normal Matrices]

Let $A\in M_{n\times n}(\mathbb{C})$ be a normal matrix, i.e. $A$ satisfies $AA^*= A^* A$. Then there is a unitary matrix $U$ such that $U^{-1}AU$ is diagonal.

Proof) Let $v$ be a common eigenvector of $A$ and $A^*$. Let $w\in \mathbb{C}^n$ be a nonzero vector such that $\langle v, w\rangle =0$. Then $$ \langle v, Aw \rangle = \langle A^* v, w \rangle=0. $$ This shows that $(\mathbb{C} v)^{\perp}$ is $A$-invariant. Repeat this argument until we find $n$ orthogonal eigenvectors of $A$. Normalize them to obtain $U$.

Theorem 2 [Adjoint Operator Written as a Polynomial]

Let $A$ be a normal matrix. Then there is $p(x)\in \mathbb{C}[x]$ such that $A^*=p(A)$.

Proof) By the spectral theorem, there is a unitary matrix $U$ and a diagonal matrix $D=Diag(\lambda_1,\ldots, \lambda_n)$ such that $A=UDU^*$. Then $A^*=UD^* U^*$. It follows that $D^*=Diag(\overline{\lambda_1}, \ldots , \overline{\lambda_n})$. By Lagrange interpolation, there is $p(x)\in \mathbb{C}[x]$ such that $$ p(\lambda_1)=\overline{\lambda_1}, \ \ldots , \ p(\lambda_n)=\overline{\lambda_n}. $$ Then we have $$ p(A)=Up(D)U^*=UD^*U^*= A^*. $$