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

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 $A$?

• Already there's a simple counter-example in two dimensions: let $Ae_1=\mu\cdot e_1$ and $Ae_2=\overline{\mu}\cdot e_2$. Then $AA^*$ is the identity... – paul garrett Jul 24 '17 at 19:17

• Would it be true in infinite dimensions ($e.g. H = L^2(R)$, for example)? – user38397 Jul 24 '17 at 19:25
• Yes, take the shift on $l^2$. – Tsemo Aristide Jul 24 '17 at 19:27