A proposition about cyclic representation in C*-algebra

Question

Let $A$ be a C*-algebra, if for an arbitrary cyclic representation $\pi: A \rightarrow B(H)$, we have $\pi(a) \geq 0$, $a\in A$, then can we conclude that $a \geq 0$?

Answer 1

The state space $S_A$ of $A$ separates points. Doing GNS for each $\phi\in S_A$ we obtain cyclic representations $\pi_\phi$. Then the direct sum $\pi_S=\oplus_{\phi\in S_A}\pi_\phi$ is a faithful representation. So if $\pi_\phi(a)\geq0$ for all $\phi\in S$, then $\pi_S(a)\geq0$ and so $a\geq0$.

Answer 2

Every C$^*$-algebra has a faithful representation that is a direct sum of cyclic representations. So, yes, if $\pi(a)>0$ for every cyclic representation $\pi$, then the element $a$ must be positive.

Take a look at the wikipedia page for the Gelfand-Naimark Theorem for further details.