How does the identity A(I+BA)=(I+AB)A connect the inverses of I+BA and I+AB? Prove that they are both invertible or both singular.
Here is an idea, if we know something about if $\displaystyle A$. If $\displaystyle A$ is invertible$\displaystyle (I+BA)=A^{-1}(I+AB)A$ then
This would say that $\displaystyle I+BA$ is similar to $\displaystyle I+AB$. Since similar Matrices have the same determinant They are both either singular or invertible.
In the general case, that is $\displaystyle A$ singular or not, suppose ( without loss of generality ) $\displaystyle I+BA$ singular and $\displaystyle I+AB$ invertible, then:
$\displaystyle (i)\quad \left |{I+BA}\right |=0 \Leftrightarrow \left |{BA-(-1)I}\right |=0 \Leftrightarrow -1\in\textrm{spec}(BA)$
$\displaystyle (ii)\quad \left |{I+AB}\right |\neq 0 \Leftrightarrow \left |{AB-(-1)I}\right |\neq 0 \Leftrightarrow -1 \notin \textrm{spec}(AB)$
This is a contradiction because
$\displaystyle \textrm{spec}(BA)=\textrm{spec}(AB)$
Fernando Revilla