This question is from Pieter Naaijken's "Quantum Spin Systems on Infinite Lattices": Let $\mathcal{U}$ be a C*-algebra and $\omega$ a faithful state on $\mathcal{U}$, meaning that $\omega(A^*A)>0$ if $A\neq 0$. Show that the corresponding GNS representation is faithful (in the sense that the kernel is trivial).
My attempt: Let $A\neq 0$. We wish to show that $\pi_\omega(A)$ is not the zero operator on $\mathcal{H}_\omega$. To that end, let $B$ be any nonzero element of $B(\mathcal{H}_\omega)$. By construction, $\pi_\omega(A)[B] = [AB]$, and we have $\langle [AB],[AB] \rangle = \omega((AB)^*AB) > 0$ by faithfulness of $\omega$; hence $\pi_\omega(A)$ is not the zero operator for any nonzero $A$, which implies that the kernel of $\pi_\omega$ is the trivial kernel $\{0\}$ (i.e. $\pi_\omega$ is injective).
Is it this simple? I am very new to this so I'm doubting myself a little.
Usually, you want to reason the other way around. If $\pi_\omega(A)=0$, then for any $B\in\mathfrak A$ you have that $$ 0=\langle \pi_\omega(A)B,\pi_\omega(A)B\rangle=\omega(B^*A^*AB). $$ Because $\omega$ is faithful, you get that $AB=0$. As you are free to choose $B$, you can let it be $A^*$, so you get that $AA^*=0$, and so $A=0$ and then $\pi_\omega$ is injective (in case it is not obvious: $\|A\|^2=\|A^*\|^2=\|AA^*\|$).