Bloch representation. Why Pauli operators? Why do I know that a general qubit state can be written as
$$
\rho = \frac 1 2 \big(\mathbb 1 +\vec r \vec \sigma\big)\;\text ?
$$
It is clear that the factor of $1/2$ comes from $\text{tr}\rho=1$. But is the set $\{\mathbb 1 , \sigma_x,\sigma_y,\sigma_z\}$ a unique basis for positive hermitian operators? If so, how can I see that there is no other?
 A: Let me elaborate on the answers in the comments and answer this question completely:
First note that the space of Hermitian matrices is a vector space, but a real one. Real linear combinations of Hermitian matrices $A$ and $B$ are also Hermitian, i.e. $\alpha A+B$ is Hermitian for $\alpha \in \mathbb{R}$. However, it is not a complex vector space, since $iA$ is anti-Hermitian for Hermitian $A$ as $(iA)^{\dagger}=-iA^{\dagger}=-iA$. 
Now, we need to find a basis for the Hermitian operators. General considerations tell us that the real dimension of the vector space is four, i.e. our basis must consist of four linear independent Hermitian matrices. There are infinitely many bases of such kind, but since the Pauli matrices are very familiar to us, we'll stick to those. Adding unity, we have four matrices that are clearly linear independent and Hermitian: $\{1,\sigma_x,\sigma_y,\sigma_z\}$ hence any Hermitian $2\times 2$ matrix $A$ can be represented as a linear combination $A=x_11+x_2\sigma_x+x_3\sigma_y+x_4\sigma_z$ for some $x_i\in \mathbb{R}$. 
Since we are talking about states, as you say, the normalization $\operatorname{tr}(A)=1$ forces us to fix $x_1$ and therfore, we obtain the formula you quote. The last constraint on $\vec{r}$ which is not mentioned in the question is obtained when we make sure that the state is a positive semi-definite matrix.
