What does the rank of a two-qubit density matrix represent? I understand that the rank of a matrix is the number of linearly independent rows or columns, but what does the matrix rank say about a two qubit density matrix? 
As an example: I understand that for a Bell diagonal state, if we rotate the density matrix to the Bell basis, then the rank of the state becomes the number of nonzero diagonal terms, and hence says something about how many Bell states (and therefore pure states) the Bell diagonal state is composed of.
How does this generalize?  Does the rank always say something about the minimum number of pure states that need to be summed to create a given density matrix?  Does this say something about the physical meaning of diagonalizing a density matrix in general?
 A: The rank of a density matrix (that is, the number of nonzero eigenvalues) does not hold information about the entanglement structure of the state: $\rho=\lvert00\rangle\langle00\rvert$ and
$$
\rho'=\frac{1}{2}\big(\lvert00\rangle+\lvert11\rangle\big)
\big(\langle00\rvert+\langle11\rvert\big)
$$
have both rank $1$.
By definition, the rank of a state $\rho$ is the minimum number of pure states that you need to obtain the state: $\rho=\sum_{i=1}^{\text{rank}} p_i \rho_i$, with $\sum_i p_i = 1$ and $\rho_i^2=\rho_i$.
The rank of a density matrix gives on the other hand information about the purity of the state, which is defined as $\operatorname{Tr}(\rho^2)$.
The purity is a measure of the non-mixedness of a state.
It is unital if and only if the state is of the form $\lvert\psi\rangle\langle\psi\rvert$. A state with purity $1$ is said to be a pure state, and has always rank $1$.
When on the other hand the purity is less than $1$, then the state must be written as a mixture of some number of states, and its rank is greater than $1$.
However, there is some relation between entanglement and rank.
In particular, the rank of the reduced density matrix gives information about the entanglement of the full state.
Consider a pure bipartite state $\rho\in\mathcal H_A\otimes\mathcal H_B$.
Then, $\rho$ is entangled if and only if the reduced state $\rho_A\equiv\mathrm{Tr}_B(\rho)$ is not pure (that is, have rank greater than one).
Equivalently, $\rho$ is separable if and only if $\rho_A$ is pure (that is, has unitary rank).
Note that you can equivalently replace every occurrence of $\rho_A$ with $\rho_B$ in the above.
Note also that this equivalency fails for more general states: a mixed separable state can have mixed partial traces, a simple example being $\rho=\frac{1}{4}(I\otimes I)$.
