# If $T=\lambda S$ with $T,S\geq 0$. why $\lambda\in\mathbb{R}_+?$

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 $$\lambda\in\mathbb{R}_+?$$

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 .... ?

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