Normal operator in Hilbert Complex space share an eigenvalue.

Question

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.

Answer 1

For part 2, using the proof of part 1 (that $\langle Tx, Tx\rangle = \langle T^*x, T^*x \rangle$), we see$$||T^*x-\overline{\alpha}x||^2 = \langle T^*x-\overline{\alpha}x, T^*x-\overline{\alpha}x\rangle = \langle T^*x, T^*x\rangle - \langle \overline{\alpha}x, T^*x\rangle - \langle T^*x, \overline{\alpha}x\rangle+\langle \overline{\alpha} x, \overline{\alpha} x\rangle$$ $$ = \langle Tx, Tx \rangle-\langle Tx, \alpha x\rangle-\langle \alpha x, Tx\rangle + \langle \alpha x, \alpha x\rangle = 0.$$

For part 3, if $Tx = \alpha x, Ty = \beta y$, then $$\alpha\langle x,y \rangle = \langle Tx,y \rangle = \langle x,T^*y \rangle = \langle x, \overline{\beta} y \rangle = \beta \langle x,y \rangle$$ implies $\langle x,y \rangle = 0$. This means that the eigenspaces are orthogonal.

Answer 2

$T$ 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 $x\perp y$.