 Solution

For a bounded linear functional on a unital C*-algebra, the state criterion isEquivalently it is positive and takes the unit to one.

In the Gelfand–Naimark–Segal construction, set $N_x=\{b:x(b^{\ast }b)=0\}$. This is a left ideal. Give $A/N_x$ the inner product $⟨[b],[c]⟩=x(c^{\ast }b)$, linear in the first variable, and complete it to a Hilbert space $H_x$. Left multiplication gives the bounded *-representation ${\pi }_x(a)[b]=[ab]$. The vector $\xi =[1_A]$ is cyclic and satisfies $x(a)=⟨{\pi }_x(a)\xi ,\xi ⟩$.

The Hermitian matrix $\rho$ has eigenvalues ${\lambda }_{\pm }=1/2\pm |\alpha |$ and trace one. In an orthonormal eigenbasis $v_{+},v_{-}$,If $|\alpha |\le 1/2$, both weights are nonnegative and sum to one, so this functional is positive and normalized, hence a state. If $|\alpha |>1/2$, evaluating on the positive rank-one projection onto the negative-eigenvalue eigenspace gives a negative value, so it is not a state. ThereforeTo obtain the representation dimension, cyclicity of trace givesThus $A/N_{x_{\rho }}$ identifies isometrically with the matrix space $\{b{\rho }^{1/2}:b\in M_2(\mathbb{C})\}$, with left multiplication as the representation. If $r=\mathrm{rank}\rho$, this space consists, after a unitary change of basis, of arbitrary first $r$ columns and zero remaining columns. Its complex dimension is $2r$. It is already complete because it is finite-dimensional. Rank is one at $|\alpha |=1/2$ and two at $|\alpha |<1/2$, proving