How to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable? I have showed that normal matrix is diagonalizable, but how can I know that the normal matrix is unitarily diagonalizable using inner products?
 A: Let $A$ be a normal matrix of $\mathbb C(n)$.
For $x \in \mathbb C^n$, you have $$\Vert Ax \Vert^2 = \langle Ax,Ax \rangle = \langle x,A^*Ax \rangle = \langle x,AA^*x \rangle = \langle A^*x,A^*x \rangle=\Vert A^*x \Vert^2$$ As for $\lambda \in \mathbb C$, $A - \lambda I_n$ commutes with $A^* - \bar{\lambda} I_n$ you have for all $(\lambda,x) \in \mathbb C \times \mathbb C^n$ $$\Vert Ax - \lambda x \Vert = \Vert A^*x - \bar{\lambda} x \Vert$$ Based on that we can conclude that if $\lambda \in \mathbb C$ is an eigen value for $A$, then $\bar{\lambda}$ is an eigen value for $A^*$ and all eigen vector of $A$ associated to the eigen value $\lambda$ is an eigen vector of $A^*$ associated to the eigen value $\bar{\lambda}$.
Then if $x$ and $y$ are two eigen vectors associated to the different eigen values $\lambda$ and $\mu$ you have $$\langle Ax,Ay \rangle = \lambda \bar{\mu} \langle x , y \rangle = \lambda \bar{\lambda} \langle x , y \rangle = \mu \bar{\mu} \langle x , y \rangle$$ From there, we can conclude that $\langle x , y \rangle = 0$ as desired.
