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You ask what is the definition of a positive element in $E$; the element $a \in E$ is positive if $a$ is positive when we think of it as an element in $A$. But an operator system is more than that. It also allows us to define an order structure on $M_n(E)$ for each $n$ and to say whether each matrix $[a_{ij}]\in M_n(E)$ is positive or not. The statement "an ...

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If you do not need any control over the norm of the vector $\alpha$, then, yes, such a vector exists. Take any vector with norm less than $\min (\frac{\varepsilon}{2}, \frac{ \varepsilon}{ 2 \Vert T \Vert})$. If you add the assumption that $\Vert \alpha \Vert=1$, such a vector $\alpha$ need not exist. Consider the operator $T(x)=-x$. Its restriction to any ...

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If you're talking linear algebra (finite or infinite dimensional), any element in the subspace $$\;\sum_{i\in I}S_i\;,\;\;S_i\;\;\text{a vector subspace}\;\;\forall\,i\in I$$ is a finite expression of the form $$s_{i_1}+…+s_{i_k}\;,\;\;s_{i_m}∈S_{i_m}$$ If you allow infinite sums then it is because you have some kind of analytic structure in your linear ...

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This result is true for any bounded operator with $\|T\|\leq1$. Fix $x\in X$ with $\|x\|\leq1$. Put $$C=\overline{\{t\,T^nx:\ t\in[0,\tfrac1{2}],\ n\in\mathbb N\}}\subsetneq B_X$$ (every element in $C$ has norm at most $1/2$). This is the most general possible result: let $T=(1+\delta)I$ for some $\delta>0$. A subset $C$ as desired satisfies ...

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