Let $A \in GL(n)$ and $B \in GL(m)$, i.e. $A$ and $B$ are invertible. Let $F$ be an arbitrary $m\times n$ matrix. Suppose $$ BF(A-I_n) = FA(A-I_n), $$ where $I_n$ is the $n \times n$ identity.
What can be said about $A$ in terms of $B$ and $F$? Is there a simplification of this formula or a different way to express the relationship?
I would particularly like to know if there is a nicer description of $A$ than the expression above.
We assume that $m>n$ and $rank(F)=n$.
Then there is $G\in M_{n,m}$ s.t. $GF=I_n$.
Thus $F(A-I)=B^{-1}FA(A-I)$ and $A-I=K(A^2-A)$, where $K=GB^{-1}F\in M_n$ is invertible.
Finally $A^2-(K^{-1}+I)A+K^{-1}=0$ is a Riccati matricial equation (in the unknown $A$) over $M_n$.
$A=I_n$ is always a particular solution but there exists an infinity of other solutions.