Eigenvectors associated to distinct eigenvalues are orthogonal (with T a normal operator) Let be V a inner product space over $\mathbb{C}$, $T$ a normal operator in $V$ and $u,v \in V$ two eigenvectors of T corresponding to different eigenvalues. Prove that $u$ and $v$ are ortogonal.
I was trying to prove this fact, until I found this proof in a Lineal Algebra Book (Friedberg):
Proof:
\begin{align*}
\lambda_1 \left \langle u,v \right \rangle &=\left \langle \lambda _1 u,v \right \rangle\\
&=\left \langle Tu,v \right \rangle\\
&=\left \langle u,T^{*}v \right \rangle\\
&=\left \langle u,\overline{\lambda}_{2} v \right \rangle\\
&=\lambda_{2}\left \langle u,v \right \rangle
\end{align*}
and since, $\lambda_{1}\neq \lambda_{2}$ (both eigenvalues) $\Rightarrow \left \langle u,v \right \rangle=0$
Nerverthless I still have a doubt in one step. I don't understand why there appears $\lambda_{2}$. I think that the correct step should be $\left \langle u,T^{*}v \right \rangle=\left \langle u,\overline{\lambda}_{1}v \right \rangle=\lambda_{1}\left \langle u,v \right \rangle$. I'm conscious that if I'm correct then the proof is incorrect, but I still don't understand why is $\lambda_{2}$. Can you help me to understand this please?

Extra note:
I also know that with $V$ a inner product space over $\mathbb{C}$ and T a normal operator it satisfies:

*

*If $\lambda$ is an eigenvalue of $T$ $\Rightarrow$ $\overline{\lambda}$ is a eigenvalue of $T^{*}$.

And I know $\lambda \neq \overline{\lambda}$, but this (unless I'm not understanding well) supports what I've exposed. I hope I have made myself understood well.
 A: The answer came to my mind! The reason is because $u,v$ are distinct eigenvectors linked to a eigenvalue respectively. So,
\begin{align*}
\left \langle \lambda_{1} u,v \right \rangle &= \left \langle Tu,v \right \rangle
\end{align*}
Then by properties of $T^{*}$ we have:
\begin{align*}
\left \langle Tu,v \right \rangle &= \left \langle u, T^{*}v \right \rangle
\end{align*}
And here is the key step! $T^{*}$ is applied in the vector $v$ (and no in the vector $u$, so this was what I wasn't understanding well in my original question). And as $v$ and $u$ are two vectors with two different eigenvalues, in this step necessarily has to be another eigenvalue:
\begin{align*}
\left \langle u, T^{*}v \right \rangle &=\left \langle u ,\overline{\lambda}_{2} v\right \rangle \\
&=\lambda_{2}\left \langle u,v \right \rangle
\end{align*}
And, to finish, since $\lambda_1 \neq \lambda_{2}$ $\Rightarrow$ $\left \langle u,v \right \rangle=0$
\begin{align*}
\therefore u \text{ and } v \text{ are orthogonal}
\end{align*}
