This is part of a larger problem. I have the expression
$$E = A^{-1}B(C^{-1} + D)^{-1}A,$$
where all matrices are of appropriate dimensions, in my notes. I am told that E is of rank n. Given that all matrices are of full rank. The only remaining part of the puzzle to to see why the summation of these matrices also results in a rank $n$ matrix.
Let $\mathcal{R}_A,\mathcal{R}_B,\mathcal{C}_A,\mathcal{C}_B$ be the row spaces for $A,B$ and the column spaces for $A,B$ respectively. Then $\text{rank}(A+B) = \text{rank}(A)+\text{rank}(B)$ if and only if $\text{dim}(\mathcal{R}_A\cap \mathcal{R}_B) = \text{dim}(\mathcal{C}_A\cap \mathcal{C}_B) = 0$.
The proof for this is given here: http://www.dtic.mil/dtic/tr/fulltext/u2/600471.pdf