Can someone tell me if this proof is correct?
Claim:V is a vector space over the Complex field. $T:V\rightarrow V$ is a normal operator. Then if $v\in V$ is an eigenvector with the eigenvalue $\lambda \in C$ then , v is eigenvector of $T^*$ and $\overline{\lambda}$ is its eigenvalue.
Proof: $\langle Tv,v\rangle=\langle \lambda v,v\rangle=\overline{\lambda}\langle v,v\rangle=\langle v,\overline{\lambda}v\rangle$
On the other hand
$\langle Tv,v\rangle=\langle v,T^*v\rangle$ and therefore we have $\langle v,T^*v\rangle=\langle v,\overline{\lambda}v\rangle$ for every eigenvector v that belongs to the eigenvalue $\lambda$. We can notate $V_{\lambda}$ as the space of eigenvectors of $\lambda$.
If we reduce the domain from V to $V_{\lambda}$ we will get that
$\langle v,T_{|V_{\lambda}}^{*}v\rangle=\langle v,\overline{\lambda}v\rangle$ is true for every $v \in V$ and then we will get
$\langle v,(T_{|V_{\lambda}}^{*}-\overline\lambda Id)v\rangle=0$ for every $v\in V_{\lambda}$ and therefore, $T_{|V_{\lambda}}^{*}=\overline\lambda Id$ for every $v\in V_{\lambda}$ as desired.
Best Answer
If $T$ is normal on a Complex space $X$, then $$ \|Tx\|^{2}=\|T^{\star}x\|^{2},\;\;\; x \in X. $$ Because $\lambda I$ is normal with adjoint $\overline{\lambda}I$, then $\|(T-\lambda I)x\|=\|(T^{\star}-\overline{\lambda}I)x\|$ also holds for all $x$ if $T$ is normal. That's the missing piece.