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

Martin Argerami
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Schüler
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1 Answers1

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Edit: You need additionally that one of the operators is positive, see the comment by Martin Argerami below.

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

gerw
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  • Thank you for your answer. If $T,S\in\mathcal{B}(F)^+$, it is true that $$\sigma (TS)\subseteq\mathbb{R}_+?$$ – Schüler Mar 01 '18 at 09:39
  • 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 01 '18 at 09:42
  • $$ \sigma( T , S ) \cup {0} = \sigma( S , T ) \cup {0}, $$ is true even if $T$ and $S$ aren't self-adjoint? – Schüler Mar 01 '18 at 10:08
  • Yes, see the link above – gerw Mar 01 '18 at 10:45
  • Thank you but the link is for matrice and here $F$ is an infinite dimentional Hilbert space. – Schüler Mar 01 '18 at 10:47
  • The arguments are the same. It is never used that those are matrices. – gerw Mar 01 '18 at 11:20
  • -1: This answer is wrong. Consider for instance $$ S=\begin{bmatrix} 1&0\0&-1\end{bmatrix},\ \ \ \ T=\begin{bmatrix} 0&1\1&0\end{bmatrix}. $$ – Martin Argerami Mar 08 '20 at 00:48
  • @MartinArgerami: Thank you. Indeed, my answer only shows that $\overline{\sigma(S T)} \cup {0} = \sigma(S T) \cup {0}$. – gerw Mar 08 '20 at 07:45