Question about special $C^*$-algebra I have a question about a $C^*$ algebra $A$ namely $M_2(\mathbb{C})$. I want to prove that every state $\alpha$ of $M_2(\mathbb{C})$ (thus a positive linear functional with norm $1$) is of the form $\alpha_r(a):=Tr(ra)$ for $Tr$ the trace, $a\in A$ and $r$ a density matrix (thus matrix which is self-adjoint with positive eigenvalues). For example $$r=\frac{1}{2}\begin{pmatrix}1+z&x+iy\\x-iy&1-z\end{pmatrix}$$ is a density matrix iff $x^2+y^2+z^2\leq1$. But why can every state be writen as such a special thing? Can you use the fact that $\alpha$ is positive iff $\alpha(1)=||\alpha||$?
When i can prove that i can also prove that $B^3$ (the $3$-ball) and the state space of $M_2\mathbb{C}$ are "the same convex sets" (of homeomorphic to each other).
Can someone help me with this problem?
Thank you very much :)
 A: That states are of the form you mention is true for $B(H)$ for any Hilbert space. In this case, you are asking about the dimension 2 case, and in such case one can give a very explicit argument. 
Let $f:M_2(\mathbb C)\to\mathbb C$ be a state. For any $x\in M_2(\mathbb C)$, we have $x=\sum_{k,j}x_{kj}e_{kj}$ for the matrix units $e_{11},e_{12},e_{21},e_{22}$. Then
$$
f(x)=\sum_{k,j}x_{kj}f(e_{kj})=\text{Tr}(xh),
$$
where $h=\sum_{k,j}f(e_{jk})e_{kj}$. One can obtain the properties for $h$ from this last formula, but it is also possible to proceed this way: as $f$ is positive, we know that $\text{Tr}(xh)\geq0$ for all positive  $x$. Now fix any norm-one vector $\xi\in\mathbb C^2$, and let $p=\xi\xi^*$ be the rank-one projection onto the range of $\xi$. Then $p$ is positive, so 
$$
\langle h\xi,\xi\rangle=\langle hp\xi,p\xi\rangle=\text{Tr}(php)=\text{Tr}(ph)=f(p)\geq0.
$$
So $h\geq0$. And clearly $\text{Tr}(h)=f(I)=1$. These two conditions are also sufficient: if $h\geq0$ with $\text{Tr}(h)=1$, then $f(x)=\text{Tr}(xh)$ defines a state. 
To move from $M_2(\mathbb C)$ to arbitrary $B(H)$, the only issue is whether the definition $$h=\sum_{k,j}f(e_{jk})e_{kj}$$ makes sense. 
If $\{\xi_j\}$ is the orthonomal basis corresponding to the matrix units $\{e_{kj}\}$, let $\xi=\sum_{j=1}^nc_j\xi_j$. The key step below is Kadison's inequality $|g(x)|^2\leq g(x^*x)$ that is satisfied by any state $g$. We have
\begin{align}
\left\|\sum_{k=1}^m\sum_{j=1}^rf(e_{jk})e_{kj}\xi \right\|^2
&=\left\|\sum_{k=1}^m\sum_{j=1}^nf(e_{jk})c_j\xi_k \right\|^2
=\left\|\sum_{k=1}^mf\left(\sum_{j=1}^nc_je_{jk})\right)\xi_k \right\|^2\\ \ \\
&=\sum_{k=1}^m\left|f\left(\sum_{j=1}^nc_je_{jk})\right)\right|^2
\leq\sum_{k=1}^mf\left(\sum_{j,h=1}^n\bar{c_j}c_he_{kj}e_{hk}\right)\\ \ \\
&=\sum_{k=1}^mf\left(\sum_{j=1}^n|c_j|^2 e_{kk}\right)\\ \ \\
&=\|\xi\|^2\,f\left(\sum_{k=1}^me_{kk}\right)\leq\|\xi\|^2\,f(I)\\ \ \\
&=\|\xi\|^2.
\end{align}
Thus $\|h\xi\|\leq\|\xi\|$, which shows that $h$ is bounded on a dense subset of $H$, and thus extends as a bounded operator to all of $H$. 
