# Spectrum of product of self adjoint operators [closed]

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 $$\sigma (TS)\subseteq\mathbb{R}?$$

## closed as off-topic by user223391, Namaste, The Phenotype, JonMark Perry, Claude LeiboviciMar 3 '18 at 6:28

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## 1 Answer

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 $$\sigma( U ) = \overline{\sigma( U^\star) }.$$

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