How to recognize a tomographically complete set of operators? What I understand...
A tomographically complete set of operators is a set of operators such that measuring the probability distributions of an unknown quantum state over the spectrum of each of these operators allows you to write down the pre-measurement quantum state (of course, one needs infinitely many copies of the identically prepared unknown quantum state to carry this out). For example, for a two-dimensional Hilbert space of spin half particles, the spin operators in $x$, $y$, and $z$ directions form such a set which I can verify by explicitly writing a unique state consistent with a given set of probability distributions over the spectrum of these three operators.
What I'm looking for...
I'm not sure if I understand the mathematical conditions that I can write down for a set of operators to tell me if it's a tomographically complete set of operators or not. Intuitively, I expect it to be something like "a largest set of non-commuting operators" because such a set would give me all the information about the phases which would be hidden if I perform measurements over a commuting set of operators. But what is the precise mathematical definition/criterion for such a largest set of non-commuting operators?
Wikipedia says that a tomographically complete set of operators forms an "operator basis on the Hilbert space". I don't think I understand this statement, for example, the three spin operators constitute a tomographically complete set of operators but I can't write down $S^2$ as a linear combination of $S_x, S_y, S_z$ which is something I should be able to do if $S_x, S_y, S_z$ formed a basis for all operators in the Hilbert space, or so do I think. Or, is this not what an operator basis on the Hilbert space supposed to mean?
 A: The space of operators is a complex Hilbert space with scalar product $\langle X,Y\rangle = \mathrm{tr}[X^\dagger Y]$. Even further, the space of hermitian matrices is a real Hilbert space with the same scalar product.
In order to completely reconstruct a vector $\rho$ in a Hilbert space, you thus want the scalar product with a set of hermitian operators $\{B_i\}$ which span the full space of hermitian matrices (as a real vector space). Then, you can write
$$
\rho = \sum C_i \, \langle B_i,\rho\rangle = 
\sum C_i\,\mathrm{tr}[B_i^\dagger\rho]\ ,
$$
where the $\{C_i\}$ and the $\{B_i\}$ form a bi-orthogonal set, $\langle B_i,C_j\rangle = \delta_{ij}$.
You might wonder whether there are constraints from positivity which simplify the problem, but you can see that this will not be the case by considering $\rho+\lambda I$ in case of a non-positive $\rho$, which is positive for sufficiently large $\lambda$ and requires the same amount of information to be reconstructed (since $\langle B_i,I\rangle=\mathrm{tr}\,B_i$ contains no information).
On the other hand, we also know that $\mathrm{tr}\,\rho=1$, and this does provide a constraint: Specifically, if you choose your basis that it contains the identity $I$, then you know $\langle I,\rho\rangle = 1$.
Now to your example above: What you forget (in case you talk about spin 1/2) is the identity operator.  Only with the identity operator, the Paulis form a basis for the space of operators, and allow you to express e.g. $S^2=3 I/4$.
