Normal operators intuition I wanted some more intuition of normal operators defined as $\| \textsf{T}v \| = \| \textsf{T}^∗ v \|$.
Here is a theorem from Linear Algebra Done Right :

Corollary 7.7 : Suppose $\textsf T \in \mathcal{L}(\textsf V)$ is normal. If $v\in \textsf V$ is an eigenvector of $\textsf T$ with eigenvalue $\lambda \in F$, then $v$ is also an eigenvector of $\textsf{T}^∗$ with eigenvalue $\overline{ \lambda }$.

The proof is the following :

Proof : Suppose $v\in \textsf V$ is an eigenvector of $\textsf T$ with eigenvalue $\lambda$. Thus $(\textsf{T}-\lambda \textsf{I})v=0$ because $\textsf T$ is normal, so is $\textsf{T}-\lambda \textsf{I}$.
By definition of normal operator we have :
$$0=\| (\textsf{T}−\lambda \textsf{I})v \| = \| (\textsf{T}-\lambda \textsf{I})^∗ v \| = \| (\textsf{T}^∗ -\overline{λ} \textsf{I})v\|$$
and hence $v$ is an eigenvector of $\textsf{T}^∗$ with eigenvalue $\overline \lambda$, as desired.

The proof is ok to follow along but I want some intuition on this. Why does this special case happen to normal operators? What's the deal with them?
 A: The author mentions right before this proof that $T$ being a normal operator is equivalent to $T$ and its adjoint, $T^*$, having the property $\forall v \in V, ||Tv|| = ||T^*v||$, as you posted as well. The statement about eigenvectors you posted should intuitively follow from that, plus some properties of adjoints. And the norm preserving property, follows from the definition of a normal operator; $T$ and $T^*$ commute. 
To further clarify, intuitively, from the norm preserving property, if $v$ were an eigenvector of $T$ then $T^*$ must also map $v$ with eigenvalue $\nu$, to some $\lambda v, \lambda \in \mathbb{C}$. The only option is a $\lambda$ such that $|\lambda| = |\nu|$. By the properties of the adjoint operation, $\lambda$ and $\nu$ must be complex conjugates. 
Edit : To clarify, in the statement that I mentioned about $T^*$ must map $v$ to some scalar multiple of $v$ doesn't directly follow from the norm preserving property for $T$, but comes from the fact that $T-\nu I$ is normal as well. This is stated in the proof in the book. This justifies why it can't be $T^*v = u, ||u|| = ||\nu v||$.
