 Solution

A state on a C-star algebra is a positive linear functional of norm one. In the unital case this is equivalent to positivity and $x(1)=1$. A unital star-homomorphism $\pi :A\to \mathcal{B}(K)$ is a cyclic representation of a C-star algebra if it has a vector $\xi$ with $\overline{\pi (A)\xi }=K$. For the representation associated with a state, the cyclic vector is taken to have norm one.

Use inner products linear in the first argument, as in the displayed formula for the projection in the source. In the Gelfand–Naimark–Segal construction, setThe Cauchy–Schwarz inequality for positive C-star functionals makes this inner product well-defined on $A/N_x$. Indeed a null vector pairs to zero with every vector. The set $N_x$ is a left ideal: $c^{\ast }c\le \Vert c{\Vert }^{2}1$ impliesso $a\in N_x$ implies $ca\in N_x$. The same estimate shows that left multiplication induces a bounded operatorComplete the quotient pre-Hilbert space to $H_x$, and extend these operators by continuity. The formula for the inner product gives ${\pi }_x(c{)}^{\ast }={\pi }_x(c^{\ast })$, while multiplication and the unit are preserved. Put ${\xi }_x=[1]$. Then $\Vert {\xi }_x\Vert =1$, the vectors ${\pi }_x(a){\xi }_x=[a]$ are dense in $H_x$, andThis gives the requested cyclic representation for an arbitrary state.

For the specified unit vector $\eta$, the functional $x_{\eta }$ is complex-linear and obeys $|x_{\eta }(a)|\le \Vert a\eta \Vert \Vert \eta \Vert \le \Vert a\Vert$. MoreoverThus it is positive and has norm exactly one, proving that $x_{\eta }$ is a vector state on a C-star algebra.

The rank-one projection satisfiesIf $a\eta =0$, this is zero for every $\zeta$, so $a{p}_{\eta }=0$. Conversely, evaluation at $\zeta =\eta$ gives $(a{p}_{\eta })\eta =a\eta$. HenceIn particular $N_{x_{\eta }}=\{a:a\eta =0\}=\{a:a{p}_{\eta }=0\}$.

The concrete GNS space is the completion of $A/N_{x_{\eta }}$ withDefine $U_0[a]=a\eta$. The description of the null ideal proves it is well-defined, and the displayed equality proves it is a linear isometry with range $A\eta$. It therefore extends to an isometry $U:H_{x_{\eta }}\to \overline{A\eta }$. The extended range is closed, since its domain is complete, and contains the dense set $A\eta$. Thus $U$ is onto the required closure.

In fact here $A=\mathcal{B}(H)$, so $A\eta =H$: given any $\zeta \in H$, the bounded rank-one operator $b(v)=⟨v,\eta ⟩\zeta$ sends $\eta$ to $\zeta$. Thus $U$ is a unitary map onto all of $H$, and $U{\xi }_{x_{\eta }}=\eta$.

Finally, on the dense quotient vectors,Both sides are bounded operators, so this identity extends to all of $H_{x_{\eta }}$. Multiplication by the inverse unitary map givesThis is the GNS representation of a vector state, with the vector state, null ideal, cyclic vector and intertwining isometry all made explicit.