Normal operator in Hilbert Complex space share an eigenvalue. Everyone, I get stuck in an exercise of Functional Analysis.
Let $T \in B(H)$ (H a complex Hilbert space) and $T^*$ adjoint of $T$. Supose $T$ is a normal operator.
1) Prove that $Ker(T)= Ker(T^*) = R(T)^\perp$ - I've finished this.
2) Using previous proof. If $\alpha$ is an eigenvalue of $T$ then the conjugate $\bar\alpha$ is an eigenvalue of $T^*$.
3) If $\alpha \neq \beta $ are eigenvectors of T, then the associated eigenspaces are orthogonal between them.
My try:
2) Let $x \in H$ such that $Tx = \alpha x $. 
$$<x,Tx> = <T^*x,x> = \alpha<x,x> = <\bar\alpha x,x>$$ 
Then we have that 
$$<T^*x - \bar \alpha x,x>=0$$
Here I don't know if above implies that $T^*x - \bar \alpha x =0$.
3) I didn't make a great progress here. 

$T$ is normal means $T^*T=TT^*$, which is equivalent to
$$
    \|Tx\| = \|T^*x\|,\;\;\; x\in H.
$$
So $\mathcal{N}(T)=\mathcal{N}(T^*)$ follows. The sum of normal operators is normal, and any scalar times a normal operator is normal. And the identity $I$ is normal. So, if $T$ is normal, then so is $\alpha I-T$ for any scalar $\alpha$. Therefore $\mathcal{N}(T-\alpha I)=\mathcal{N}(T^*-\overline{\alpha}I)$, and any eigenvector of a normal $T$ with eigenvalue $\alpha$ is an eigenvector of $T^*$ with eigenvalue $\overline{\alpha}$. If $Tx=\alpha x$ and $Ty=\beta y$, then
\begin{align}
    (\alpha-\beta)\langle x,y\rangle
  & = \langle \alpha x,y\rangle-\langle x,\overline{\beta}y\rangle \\
  & = \langle Tx,y\rangle-\langle x,T^*y\rangle \\
  & = \langle Tx,y\rangle-\langle Tx,y\rangle =0.
\end{align}
Therefore, if $\alpha\ne \beta$, it follows that $x\perp y$.
