# Invertible matrices proofs

• Jun 13th 2009, 02:37 AM
profmq
Invertible matrices proofs
Hey, i am having with this question. Would appreciate any help:

A and B are matrices, with (I-AB) an invertible matrix.

Show that the inverse of (I-BA) = I+BCA

Where C is the inverse of (I-AB)
• Jun 13th 2009, 03:36 AM
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$.

$\displaystyle \Longrightarrow(\exists)\text{ }(I-BA)^{-1}=I+BCA$.
• Jun 13th 2009, 04:40 AM
profmq
Quote:

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$.

$\displaystyle \Longrightarrow(\exists)\text{ }(I-BA)^{-1}=I+BCA$.

thanks :)
• Jun 13th 2009, 10:44 PM
Isomorphism
Just for additional information: It is a simple application of the more general Woodbury formula. Trying to prove that identity is fun :D