Suppose $A$ and $B$ are $(n \times m)$ and $(m \times n)$ matrices respectively, with $n<m$ and $\operatorname{rank}(A)=\operatorname{rank}(B)=n$. Consider the matrix $M$ given by
$$ M = B[AB]^{-1}A. $$
I want to prove that all the principal minors of $M$, of order greater than $n$, are equal to zero.
It follows from the fact that $$\operatorname{rank}(AB) \leq \min(\operatorname{rank}(A), \operatorname{rank}(B))$$ that the rank of $\operatorname{rank}(M)=n$. So, I think there is just one step missing to prove this, but I haven't been able to figure it out.