# GNS representation

Suppose $$A$$ has a tracial state $$\psi$$, I want to prove $$A/ker(\pi_{\psi})$$ has a faithful tracial state,where $$\pi_{\psi}$$ is the $$GNS$$ respresentation with respect to $$\psi$$. My thought: define $$\tau(a+ker(\pi_{\psi}))=\psi(a)$$,it is well defined, in order to prove it is faithful, let $$a\in A/ker(\pi_{\psi})$$.

Then $$\psi(a*a)=\langle\pi_{\psi}（a^*a）e_{\psi},e_{\psi}\rangle=0$$, where $$e_{\psi}$$ is the cyclic vector of $$\psi$$. How to show that $$\pi_{\psi}(a）=0$$?

As you said, assuming you take $$b=a+ker(\pi_\psi)\in A/ker(\pi_\psi)$$ such that $$\tau(b)=0$$, then $$0 = \tau(bb^*) = \tau(b^*b)=\psi(a^*a) =\langle\pi_{\psi}(a^*a)e_{\psi},e_{\psi}\rangle=\langle\pi_{\psi}(a)\pi_\psi(a)^*e_{\psi},e_{\psi}\rangle=\langle\pi_\psi(a)^*e_\psi,\pi_\psi(a)^*e_\psi\rangle$$, so $$\pi_\psi(a)^*e_\psi = 0$$, using the symmetry of the trace and the fact that $$\pi_\psi$$ is a *-morphism. I claim that $$a\in ker_(\pi)$$ automatically follows.
Indeed, you want to show that $$\pi_\psi(a)\pi(A)e_\psi = 0$$, and it will imply (because $$e_\psi$$ is cyclic for $$\pi$$) that $$\pi_\psi(a)H_\pi = 0$$, hence $$\pi_\psi(a)=0$$, so $$a\in ker(\pi_\psi)$$, hence $$b=0$$.
But this is immediate, since $$\vert\vert\pi_\psi(a)\pi(c)e_\psi\vert\vert^2=\langle\pi_\psi(a)\pi_\psi(c)e_\psi,\pi_\psi(a)\pi_\psi(c)e_\psi\rangle=\langle\pi_\psi(c^*a^*ac)e_\psi,e_\psi\rangle = \psi((c^*a^*)(ac))=\psi((ac)(c^*a^*))=\langle\pi_\psi(a)\pi_\psi(c)\pi_\psi(c^*)\pi_\psi(a^*)e_\psi,e_\psi\rangle = 0$$.