Why do the density operators span the whole operator space $\mathcal{B}(H)$? The convex set of density operators on a finite-dimensional Hilbert space $H$ defined by
$$\mathcal{D}(H):=\{\rho\in\mathcal{B}(H)\,|\,\rho\geq 0,\, \operatorname{tr}\rho =1\},$$
This set is said to span the entire space of operators $\mathcal{B}(H)$. Why is that so?
I guess there is an easy explanation, I just do not see it.
 A: You want to prove that given an arbitrary matrix $A$, we can write $A$ as a linear combination of positive, unit-trace matrices.
To do this, you start by writing $A$ in terms of its Hermitian and skew-Hermitian components (see also this post about this decomposition):
$$A=\underbrace{\frac{A+A^\dagger}{2}}_{\equiv A_1}+i\underbrace{\frac{A-A^\dagger}{2i}}_{
\equiv A_2}\equiv A_1+i A_2,$$
where $A_1,A_2$ are Hermitian (one can also easily show that this decomposition is unique).
Then, one can use the fact that for any Hermitian matrix $H$, there are positive matrices $H_+$ and $H_-$ such that $H=H_+-H_-$. An easy way to construct these is by having $H_+$ contain only the terms of the spectral decomposition of $H$ corresponding to positive eigenvalues, and similarly for $H_-$. Equivalently, we just define $H_+\equiv (H+|H|)/2$ and $H_-\equiv (|H|-H)/2$.
In conclusion, we managed to write
$$A=\frac{1}{2}[(A_{1,+}-A_{1,-})+i(A_{2,+}-A_{2,-})],$$
which tells you that any operator is a linear combination of positive ones. To show that it is also a linear combination of positive, unit-trace ones, you simply need to rescale each element in the sum to get operators with unit trace. For example, $A_{1,+}$ might not have unit-trace, but $A_{1,+}=\lambda(A_{1,+}/\lambda)$ for any $\lambda\in\mathbb R$, and we can choose $\lambda$ such that $\operatorname{tr}(A_{1,+}/\lambda)=1$.
A: Choose your favourite operator $\,x\in\mathcal{B}(H)\,$ and write it as $\,x=a+ib\,$ where both
$$a={x+x^*\over 2}\quad\text{and}\quad b={x-x^*\over 2i}$$
are self-adjoint (or 'hermitian', synonymously). Here $x^*$ denotes the adjoint of $\,x$; it corresponds to transposition and complex conjugation if $\,x\,$ is represented as a matrix.
The linear combinations $\,a,b\,$ generalise the real and imaginary parts of a complex number.
To get the desired conclusion note that the subset of self-adjoint operators 
in $\mathcal{B}(H)$ equals the $\mathbb{R}$-linear span of $\,\mathcal{D}(H)\,$.
