
Originally Posted by
AMI
$\displaystyle \bullet$ $\displaystyle (I+BCA)(I-BA)=I-BA+BCA-BCABA$
$\displaystyle C=(I-AB)^{-1}\Rightarrow C(I-AB)=I$$\displaystyle \Rightarrow C-CAB=I\Rightarrow B(C-CAB)A=BIA$$\displaystyle \Rightarrow BCA-BCABA=BA\Rightarrow(I+BCA)(I-BA)=I$.
$\displaystyle \bullet$ $\displaystyle (I-BA)(I+BCA)=I+BCA-BA-BABCA$
$\displaystyle C=(I-AB)^{-1}\Rightarrow (I-AB)C=I$$\displaystyle \Rightarrow C-ABC=I\Rightarrow B(C-ABC)A=BIA$$\displaystyle \Rightarrow BCA-BABCA=BA\Rightarrow(I-BA)(I+BCA)=I$.
$\displaystyle \Longrightarrow(\exists)\text{ }(I-BA)^{-1}=I+BCA$.