# 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 $$Tx = \alpha x$$.

$$ = = \alpha = <\bar\alpha x,x>$$

Then we have that

$$=0$$

Here I don't know if above implies that $$T^*x - \bar \alpha x =0$$.

3) I didn't make a great progress here.

• Use T^* to get $T^*$. Your notation without the * in the exponent can be confusing. – Disintegrating By Parts Nov 29 '18 at 19:05

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.

• Thank you! for this part. – Kutz Nov 29 '18 at 19:10
• @Kutz I added part 3 – mathworker21 Nov 29 '18 at 19:13
• Thank you for answer. – Kutz Nov 29 '18 at 19:22

$$T$$ 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 $$x\perp y$$.