I am currently reading through the proof of the GNS Construction, and there are two parts I simply do not understand. Below is the entire proof, the parts in red are what I do not understand. It would be great if someone could clear these up for me.
Let $A$ be a $C^\ast$-algebra with $f\colon A\to\mathbb{C}$ a positive linear functional. Define
$$
N=\{a\in A: f(a^\ast a)=0\}.
$$
Note that $N$ is not only a vector subspace of $A$, but a left ideal as well. In particular, $A/N$ is a vector space. We let $\xi\colon A\to A/N$ be the quotient map. We have chosen $N$ so that
$$
\langle\xi(a),\xi(b)\rangle:=f(b^\ast a)
$$
is a well-define inner product on $A/N$. Therefore we can complete $A/N$ to a Hilbert space $\mathcal{H}$.
Since $N$ is a left-ideal in $A$, for each $a\in A$ we can define an operator $\pi_0(a)$ on $A/N$ by
$$
\pi_0(a)\xi(b):=\xi(ab).
$$
Furthermore, $\pi_0$ is a homomorphism of $A$ into the linear operators on $A/N$.
Since $a^\ast a\leq\|a\|^2\mathbf{1}_{\tilde{A}}$, we have $b^\ast a^\ast ab\leq\|a^2\|b^\ast b$ for any $a,b\in A$.
Therefore
$$
\|\pi_0(a)\xi(b)\|^2=\|\xi(ab)\|^2=f(b^\ast a^\ast ab)\leq\|a\|^2f(b^\ast b)=\|a\|^2\|\xi(b)\|^2.
$$
It follows that $\pi_0(a)$ is bounded and extends to an operator $\pi(a)$ on $\mathcal{H}$. Clearly, $\pi\colon A\to B(\mathcal{H})$ is an algebra homeomorphism. Since
$$
\langle\pi_0(a)\xi(b),\xi(c)\rangle=f(c^\ast ab)=\langle\xi(b),\pi_0(a^\ast)\xi(c)\rangle,
$$
we must have $\pi(a)^\ast=\pi(a^\ast)$. Thus $\pi$ is a representation.
Now let $\{e_\lambda\}$ be an approximate identity in $A$. If $\mu\leq\lambda$, then $e_\lambda-e_\mu\geq0$ and $\left(\mathbf{1}_{\tilde{A}}-(e_\lambda-e_\mu)\right)\geq0$. This implies that $\|e_\lambda-e_\mu\|\leq1$. Thus, $(e_\lambda-e_\mu)^2\leq(e_\lambda-e_\mu)$ and
$$
\|\xi(e_\lambda)-\xi(e_\mu)\|^2=f\left((e_\lambda-e_\mu)^2\right)\leq f\left((e_\lambda-e_\mu)\right).
$$
Since $\lim_\lambda f(e_\lambda)=\|f\|=\lim_\mu f(e_\mu)$, $\{\xi(e_\lambda)\}$ is a Cauchy net in $\mathcal{H}$. Hence, there is a $h\in\mathcal{H}$ such that $\xi(e_\lambda)\to h$.
This means that
$$
\pi(a)h=\lim_\lambda\pi_0(a)\xi(e_\lambda)=\lim_\lambda\xi(ae_\lambda).
$$
But $a\mapsto\xi(a)$ is continous because $f$ is: $\|\xi(a)-\xi(b)\|^2=f\left((b-a)^\ast(b-a)\right)$. So we can conclude that $\pi(a)h=\xi(a)$. $\color{red}{\text{Therefore }h\text{ is a cyclic vector for }\pi}$ and
$$
\langle\pi(a^\ast a)h,h\rangle=\langle\pi(a)h,\pi(a)h\rangle=\langle\xi(a),\xi(a)\rangle=f(a^\ast a).
$$
$\color{red}{\text{By linearity, } f(a)=\langle\pi(a)h,h\rangle\text{ for all }a\in A}$.
Since $\pi$ has a cyclic vector, it is definitely nondegenerate, so $\pi(e_\lambda)h\to h$. Thus
$$
\|f\|=\lim_\lambda f(e_\lambda)=\lim_\lambda\langle\pi(e_\lambda)h,h\rangle=\|h\|^2.
$$
Best Answer
First red part:
You have $\xi(a)=\pi(a)h$. By construction, the image of $\xi$ is dense in $\mathcal{H}$. In other words, $\pi(A)h$ is dense in $\mathcal{H}$ which means exactly that $h$ is cyclic for $\pi$.
Second red part:
The elements of the form $a^*a$ are exactly the positive elements. Every element in a $C^*$-algebra is a linear combination of four positive elements, so indeed you can use linearity on the line above the claim.