If $A$ is normal and $AA^* x = \lambda x$, then is $x$ an eigenvector of $A$?

Question

Let $H$ be a Hilbert space and let $A$ be a normal linear transform. Let $A^*$ be the adjoint of $A$. If $(\theta, x)$ is an eigenpair of $A$, then it is easy to prove that $(\lvert \theta \rvert^2, x)$ is an eigenpair of $AA^*$, because $$ A x = \theta x \rightarrow A^*A x = \theta A^*x=\theta \bar{\theta} x = \lvert \theta \rvert^2 x $$. However, I am interested to know if the converse holds true, that is, if $AA^* x = \lambda x$ for some $x \in H$, then is $x$ an eigenvector of $A$?

Answer 1

This is not true, consider for example a unitary operator on a finite dimensional space.