State Space of Finite-Dimensional Abelian C*-Algebras I'm studying some properties of C*-algebras and I found the following statement.
In the case of $\mathbb{C}$ the state space $\mathcal{S}(\mathbb{C})$ is a point and in the case of $\mathbb{C} \bigoplus \mathbb{C}$ the state space $\mathcal{S}(\mathbb{C} \bigoplus \mathbb{C})$ is the interval $[0,1]$. 
Well, I don't understand how I can get these results. 
The state space $\mathcal{S}(\mathbb{C})$ is the identity map?
I'm using the following definition: a state on a unital $C^*-algebra$, $\mathfrak{A}$, is map $\omega: \mathfrak{A} \to \mathbb{C}$ s.t. $\omega(A)\ge 0$, $\forall A \in \mathfrak{A}^+$ and $\omega(\mathbb{I})=1$ (the map is normalized). 
In the previous definition we have $\mathfrak{A}^+=\{A \in \mathfrak{A}_{\mathbb{R}}:\sigma(z)\subset\mathbb{R}^+ \}$ with $\mathfrak{A}^+_{\mathbb{R}}$ the set of self-adjoints elements of $\mathfrak{A}$ and $\sigma(A)$ the spectrum of $A$.
 A: Note that a linear map $\Bbb C\to \Bbb C$ is uniquely determined by its image of $1$. In this case the condition $\omega(1)=1$ uniquely determines a map and it is also easy to see that this is a state.
A linear map $\Bbb C\oplus \Bbb C\to\Bbb C$ is uniquely determined by its image of $(1,0)$ and $(0,1)$. Note that $(1,0)+(0,1)=(1,1)=\Bbb1$ so $\omega(1,0)+\omega(0,1)=1$ must hold for states and
$$\omega(1,0)=1-\omega(0,1)\tag{1}$$
where both terms must be positive since $(1,0)$ and $(0,1)$ are positive elements of $\Bbb C\oplus\Bbb C$.
Denote $\omega_0,\omega_1$ the linear maps so that $\omega_0(1,0)=0,\omega_0(0,1)=1$ and $\omega_1(1,0)=1,\omega_1(0,1)=0$. You can verify these are states and that any state is a convex combination of these two by equation $(1)$. It follows that the state space is
$$\{t\omega_0+(1-t)\omega_1\mid t\in[0,1]\}$$
A: s.harp has answered your exact question. I just wanted to mention that "$[0,1]$" is a poor way of saying what $\mathcal S(\mathbb C\oplus \mathbb C)$ is. Concretely, $$\mathcal S(\mathbb C^n)=\{(t_1,\ldots,t_n)\in[0,1]^n:\ t_1+\cdots+t_n=1\},$$
under the natural duality $$ \langle (t_1,\ldots,t_n),(z_1,\ldots,z_n)\rangle=t_1z_1+\cdots+t_nz_n.$$
Saying that $\{(t,1-t):\ t\in[0,1]\}$ is $[0,1]$ is true to a certain extent, but definitely not very enlightening, as it doesn't express at all how $t\in[0,1]$ is a functional on $\mathbb C\oplus\mathbb C$.  
