A state $\tau$ on a C*-algebra $A$ is called faithful if $\tau(a^{*}a)=0$ implies $a=0$ for all $a\in A$.
Since $a=0$ if and only if $a^{*}a=0$ (by the C*-identity) and since the positive elements of $A$ are characterized by elements of the form $a^{*}a$, I interpreted the defintion of faithful as $$\ker(\tau)^{+}=\ker(\tau)\cap A^{+}=\{0\}.$$
But since the elements of (the closed ideal, hence C*-subalgebra) $\ker(\tau)$ are spanned by the positive elements in $\ker(\tau)$, this implies that $\ker(\tau)=\{0\}$. Clearly, the converse is also true: If $\ker(\tau)=\{0\}$, then $$\{0\}\subset\ker(\tau)^{+}\subset\ker(\tau)=\{0\}.$$ But then $\tau$ is faithful if and only if $\tau$ is injective. This seems very weird, because this implies that if $A$ has a faithful state, then $\dim(A)\leq\dim(\mathbb{C})=1$, i.e. $A=\{0\}$ or $A=\mathbb{C}$.
I must be doing something wrong. Any help will be greatly appreciated!
Something that in my opinion is not encouraged enough, is to try basic examples. You can produce a faithful state on $\mathbb C\oplus\mathbb C$ by $\tau(a,b)=\frac{a+b}2$.
Then you can easily calculate $$\ker\tau=\{(a,-a):\ a\in\mathbb C\},$$ while $(\ker\tau)^+=\{0\}$.
Also, note that linear functionals always have big kernels, as $\dim A/\ker\tau=1$. Indeed, if $a\in A$ is such that $a\not\in\ker\tau$, then for any $b\in A$ we have $b-\frac{\tau(b)}{\tau(a)}\,a\in\ker\tau$, so $b+\ker\tau=\frac{\tau(b)}{\tau(a)}\,a+\ker\tau$. That is, $A/\ker\tau=\mathbb C\,a+\ker\tau$.