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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.)

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vesszabo
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    That's wrong. Conside $\ell^2$ and $A\colon (x_n) \mapsto (0, x_1, \ldots)$. Then $A^\colon (x_n) \mapsto (x_2, \ldots)$. So $AA^\colon (x_n) \mapsto (0, x_2, \ldots)$ and $A^A = \mathrm{id}$. Hence $0 \in \sigma(AA^) \setminus \sigma(A^A)$. But you will allways have $\sigma(AA^)\setminus {0} = \sigma(A^*A) \setminus{0}$. – martini Jul 10 '12 at 09:11
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    More generally you have $\sigma(AB) \smallsetminus {0} = \sigma(BA) \smallsetminus {0}$ whenever $A,B$ are elements of a unital $\mathbb{C}$-algebra. See e.g. this thread – t.b. Jul 10 '12 at 09:28
  • Thanks the link and the counterexample. I'm awfully sorry, I forgot to write that A is invertible. In this case the above equality is valid, isn't it? – vesszabo Jul 10 '12 at 12:23
  • 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. – t.b. Jul 11 '12 at 16:56

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