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

Let $T,S\in\mathcal{B}(F)^+$. Assume that there exists $\lambda\in \mathbb{C}$ such that $T=\lambda S$. Why $$\lambda\in\mathbb{R}_+?$$

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We denote the inner product on $F$ by $(*|*)$.

For $x \in F$ we have $Tx=\lambda Sx$, hence

$$(Tx|x)= \lambda (Sx|x).$$

$(Tx|x)$ and $(Sx|x)$ are $ \ge 0$, conclusion .... ?

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  • $\begingroup$ I think that since $\sigma(T)\subseteq \mathbb{R}_+$ and $\sigma(S)\subseteq \mathbb{R}_+$, then $\lambda\in \mathbb{R}_+$ . Right? Thank you $\endgroup$
    – Student
    Mar 8 '18 at 12:20
  • $\begingroup$ If $(Tx|x), (Sx|x) \ge 0$ and $(Tx|x)= \lambda (Sx|x)$, then we must have $ \lambda \ge 0$. $\endgroup$
    – Fred
    Mar 8 '18 at 12:24

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