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Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.

Let $T,S\in\mathcal{B}(F)$, be two self-adjoint operators. Why $$\sigma (TS)\subseteq\mathbb{R}?$$

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This follows from combining the next two facts: $$ \sigma( T \, S ) \cup \{0\} = \sigma( S \, T ) \cup \{0\}, $$ this is sometimes called "Jacobson's lemma", and it can be proved by using, e.g., https://math.stackexchange.com/a/1928728/58577

The second fact is $$ \sigma( U ) = \overline{\sigma( U^\star) }.$$

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  • $\begingroup$ Thank you for your answer. If $T,S\in\mathcal{B}(F)^+$, it is true that $$\sigma (TS)\subseteq\mathbb{R}_+?$$ $\endgroup$ – Schüler Mar 1 '18 at 9:39
  • $\begingroup$ Yes. By introducing the square-root $\sqrt{T}$, you have $\sigma( T \, S) \cup\{0\} = \sigma(\sqrt{T} \, S \, \sqrt{T}) \cup \{0\} \subset [0,\infty)$. $\endgroup$ – gerw Mar 1 '18 at 9:42
  • $\begingroup$ $$ \sigma( T \, S ) \cup \{0\} = \sigma( S \, T ) \cup \{0\}, $$ is true even if $T$ and $S$ aren't self-adjoint? $\endgroup$ – Schüler Mar 1 '18 at 10:08
  • $\begingroup$ Yes, see the link above $\endgroup$ – gerw Mar 1 '18 at 10:45
  • $\begingroup$ Thank you but the link is for matrice and here $F$ is an infinite dimentional Hilbert space. $\endgroup$ – Schüler Mar 1 '18 at 10:47

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