3
$\begingroup$

Possible Duplicate:
Prove that $\sigma(AB) \backslash \{0\} = \sigma(BA)\backslash \{0\} $

Let $A$ be a bounded linear operator on a (complex) Hilbert space. I want to prove that $\sigma(AA^*)=\sigma(A^*A)$, where $\sigma(\cdot)$ is the spectrum. The finite dimensional case is (almost) trivial. How could we extend to infinite dimensional case? (Maybe a reference is appreciated.) (I tried to find the answer, but search engine cannot handle formula.)

  • 0
    Yes it is valid (sorry for the late response): $A$ is invertible if and only if $A^\ast$ is invertible, thus $A^\ast A$ and $AA^\ast$ are both invertible, so $0$ doesn't belong to neither $\sigma(AA^\ast)$ nor to $\sigma(A^\ast A)$ and hence the spectra are equal.2012-07-11

0 Answers 0