Let $A$ be a unital $C^*$- algebra and let $\pi: A\to B(H)$ be a representation. Then we know that $\pi$ is irreducible if and only if $\pi(A)'$ are scalars. Is it true that $\{T \in \pi(A)': 0\leq T \leq I\}$ are scalars implies the full $\pi(A)'$ are scalars?
Since $\pi(A)'$ is a von Neumann algebra, it is spanned by projections. Projections are positive contractions and so we can conclude that $\pi(A)'$ are scalars if $\{T \in \pi(A)': 0\leq T \leq I\}$ are scalars. Is the argument correct?
What you say is correct, but you don't even need to talk about von Neumann algebras. For any C$^*$-algebra $A$, you have $$ A=\overline{\operatorname{span}}\{a\in A:\ 0\leq a\leq 1\}. $$ This a straightforward consequence of the continuous functional calculus, since you always have $$ a=a_1-a_2+i(a_3-a_4) $$ with $a_1,a_2,a_3,a_4$ positive.