Assume that $A\in\mathbb{R}^{n\times \nu}$, $B\in\mathbb{R}^{\nu\times \nu}$ and $C\in\mathbb{R}^{\nu\times n}$. I'm interested in the rank of the matrix $$AB^{-1}C.$$ $A$ and $C$ are full rank ($n$), and of course, $B$ is as well ($\nu$). In light of this operation I'm wondering if:
- Is there any explicit condition so that $\operatorname{rank}(M)=n$?
- Can I say something about $\operatorname{rank}(AB^{-1}C)$ using $\operatorname{rank}(ABC)$ instead? (that would be much easier for my problem).
Let me answer you second question first: unfortunately no, the conditions are not equivalent. Take $$A=\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}, B = \begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\ 1 & 0 & 0\end{bmatrix}, C = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}.$$ Then $A^TBC = 0$ but $A^T B^{-1}C = 1.$ Here the matrix $B$ was carefully crafted to rotate the image of $C$ into the kernel of $A^T$, but not vice-versa.
For the first question, I know of no simple condition (simpler than performing the multiplication and probing the product's rank). You can take the SVD of $A$ and $C$ and try to reason how B maps the different subspaces to each other, but at that point you're already better off taking the determinant...