Let $A$ be a $C^*$-algebra with an identity. If $A$ has a unique tracial state $\varphi:A \rightarrow \mathbb{C}$ i.e. $\varphi(ab) = \varphi(ba)$, $\varphi(x^*x) \geq 0$, and $\varphi(1) = 1$. I am trying to show that the double commutant of the image of GNS representation for $\varphi$ is a factor i.e. $Z((\pi_\varphi(A))'') = \mathbb{C}I$.
Suppose that $Z((\pi_\varphi(A))'')$ is not a factor. There exists a nontrivial projection in $Z((\pi_\varphi(A))'')$ since $\mathbb{C}I$ is proper to $Z((\pi_\varphi(A))'')$ and using the fact the double commutant of projections in $Z(M)$ equals $Z(M)$. I am having trouble constructing tracial state to contradict the uniqueness. I know if I find a cyclic vector $\xi$ (maybe the trivial i.e. $1 + N_\varphi$), then I can construct at least a state $$\phi(a) = \langle \pi_\varphi(a)\xi, \xi \rangle$$ then $\phi$ is a state and maybe I can just it is tracial.
Let $P\in Z(\pi_\varphi(A)'')$ be a non-trivial projection. Write $\tilde\varphi(X)=\langle X\xi_\varphi,\xi_\varphi\rangle$ the normal tracial state induced by $\varphi$. We have $\tilde\varphi(P)>0$. Choose $s\in(0,1)$ with $s\ne \tilde\varphi(P) $ and define a state $\psi$ on $A$ by $$ \psi(a)=\frac s {\tilde\varphi(P)}\,\tilde\varphi(P \pi_\varphi(a)) +\frac {(1-s)} {\tilde\varphi(I-P)}\,\tilde\varphi((I -P) \pi_\varphi(a)). $$ Because $\tilde\varphi$ is a trace and $P$ is central, $\psi$ is a trace. Suppose that $\psi=\varphi$. Let $\{a_n\}$ be a sequence such that $\pi_\varphi(a_n)\to P$. Then $$ \tilde\varphi(P)=\lim_n\varphi(a_n)=\lim_n\psi(a_n)=s, $$ a contradiction. So $\psi$ would be a different tracial state on $A$. The uniqueness thus implies that $Z(\pi_\varphi(A)'')$ is trivial.