I'm reading Davidson's book C* Algebras by Example.in the proof of the irrational rotational algebra has unique trace. he define a automorphism:
$$\rho_{\lambda,\mu}(U)=\lambda U, \rho_{\lambda,\mu}(V)=\mu V, $$ where $U,V$ are the generator of the irrational rotational algebra. then define two maps: $$\Phi_{1}(A)=\int_{0}^{1}{\rho_{1,e^{2\pi it}}(A)}dt,\Phi_{2}(A)=\int_{0}^{1}{\rho_{e^{2\pi it,1}}(A)}dt$$ then we can prove that $$\Phi_{1} (\sum_{k,l}{a_{k,l}U^{k}V^{l}})=\sum_{k}{a_{k,0} U^{k}} $$ in the book it says ''If $A$ is positive and non-zero,then $\rho_{1,e^{2\pi it}}(A)$ is positive and nonzero for all $t$. thus the integral $\Phi(A)$ is positive and nonzero. Hence $\Phi_{1}$ is positive and faithful.'' I agree that $\Phi_{1}$ is positive, but how can it be faithful since $\Phi_{1}(V)=0$ by the above formula?
Faithful means "nonzero on positive elements". Here $V$ is a unitary distinct from the identity, so it is not positive.
The situation is no different than that for the trace in $M_2(\mathbb C)$, which is faithful, while $\operatorname{Tr}(V)=0$ when $V$ is, for instance, $$ \begin{bmatrix} 1&0\\0&-1\end{bmatrix},\ \ \ \begin{bmatrix} 0&1\\ 1&0\end{bmatrix},\ \ \text{ or }\ \ \begin{bmatrix} 1/\sqrt2&1/\sqrt2\\ 1/\sqrt2&-1\sqrt2\end{bmatrix}. $$