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I am reading Sec33 of Conway's A Course in Operator Theory, according to his definition,

An operator system is a linear manifold $\mathcal{S}$ in a $C^*$-algebra such that $1\in\mathcal{S}$ and $\mathcal{S}=\mathcal{S}^*$.

Then he makes the comment that operator systems have an abundance of positive elements. His argument is that for every hermitian element $a\in\mathcal{S}$, $|a|$ also lies in $\mathcal{S}$ and hence the positive parts and negative parts lie in $\mathcal{S}$ and hence $\mathcal{S}$ is spanned by its positive elements.

However, I do not know why $|a|$ lies in $\mathcal{S}$ since $\mathcal{S}$ is only assumed to be a linear manifold, not an algebra.

Can somebody give a hint? Thanks!

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On page 187, on the same page and very soon after the definition, he shows why $S$ is spanned by its positive elements, with nothing about positive and negative parts. (He uses $\|a\|$ as shorthand for $\|a\|1$, if that is what is confusing you.) – Jonas Meyer Jul 14 '12 at 21:11
@JonasMeyer Yes. I mixed $\|a\|$ with $|a|$. That is what confused me. – Hui Yu Jul 14 '12 at 21:31
up vote 3 down vote accepted

I assume "linear manifold" means "subspace." I also do not understand the given argument (if $|a|$ denotes the absolute value), but here's another one: the second condition implies that $S$ is spanned by its self-adjoint elements. Let $a$ be such an element. If $\lambda \in \mathbb{R}$ is greater than the spectral radius of $a$ then $a + \lambda$ is positive, so $a = (a + \lambda) - \lambda$ is a difference of positive elements.

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Thanks! Your argument works well. Maybe you already know this but the difference between a linear manifold and a subspace (at least in the literature of operator theory) is that a subspace is closed in $\|\cdot\|$ while there is no such restriction on manifolds. – Hui Yu Jul 14 '12 at 21:12
@HuiYu: Of course, in other areas of functional analysis, when we say "(linear) subspace" we mean "linear subspace", and when we want to refer to one which is closed, we say "closed (linear) subspace". I think one should be aware that not everyone will follow the terminology of one's particular sample of books – user16299 Jul 14 '12 at 23:37
@Hui: I respectfully disagree. If one wants to talk about a densely defined operator, its domain is certainly not going to be a closed subspace, and there are many situations where one wants to talk about densely defined operators. – Qiaochu Yuan Jul 15 '12 at 0:50
@QiaochuYuan All right. I deleted that comment. I should have said that 'most of the time' we talk about closed subspaces, and we might gain a little efficiency if we could drop the 'closed' before each 'subspace'. – Hui Yu Jul 15 '12 at 1:18

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