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Let $A$ be a non-unital sub-C*-algebra of $B(H)$. I want to show that if $T\in A$ and $\lambda \in \mathbb{C}$ are such that $T+\lambda I_H\geq 0$ then $T$ is self-adjoint and $\lambda \geq 0$.

Let $\lambda =a+ib$ with $a, b\in \mathbb{R}$. Since $(T+\lambda I_H)^*=(T+\lambda I_H)$ we have $T^*-T=2ibI_H$. That implies $b=0$ because $A$ is not unital. Thus $T^*=T$ and $\lambda \in \mathbb{R}$. It only remains to show that $\lambda \geq 0$. If $T$ is not injective it's obvious by using the characterization $A\geq 0$ iff for all $h\in H$ $\langle Ah,h \rangle \geq 0$ (we can choose $h\neq 0$ such that $Th=0$). But in the general case I don't see how to use the positivity of $T+\lambda I_H$ in order to show that $\lambda \geq 0$.

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Without loss of generality $A$ acts non-degenerately on $H$, i.e. $$\cap_{X\in A}\ker X=\{0\},$$ otherwise consider the restriction of $A$ onto the orthogonal complement of $\cap_{X\in A}\ker X$.

Let $\tilde A$ be the unitization of $A$. Then $\tilde A$ is naturally isomorphic to $A+\mathbb C I_H\subseteq B(H)$. The notion of spectrum of an element $X\in A$ coincide in $A\subseteq\tilde A\subseteq B(H)$.

Since $A$ is non-unital, we have $0\in\sigma(X)$ for all $X\in A$. Hence $\lambda\in\sigma(T+\lambda I_H)$. Since $T+\lambda I_H\geq 0$ we have $\sigma(T+\lambda I_H)\subseteq[0,+\infty)$ i.e. $\lambda \geq 0$.

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