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 $A$?

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    $\begingroup$ 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... $\endgroup$ – paul garrett Jul 24 '17 at 19:17

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

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

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