Invertible matrices proofs

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June 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)
June 13th 2009, 03:36 AM
AMI

$\bullet$ $(I+BCA)(I-BA)=I-BA+BCA-BCABA$
$C=(I-AB)^{-1}\Rightarrow C(I-AB)=I$ $\Rightarrow C-CAB=I\Rightarrow B(C-CAB)A=BIA$ $\Rightarrow BCA-BCABA=BA\Rightarrow(I+BCA)(I-BA)=I$ .
$\bullet$ $(I-BA)(I+BCA)=I+BCA-BA-BABCA$
$C=(I-AB)^{-1}\Rightarrow (I-AB)C=I$ $\Rightarrow C-ABC=I\Rightarrow B(C-ABC)A=BIA$ $\Rightarrow BCA-BABCA=BA\Rightarrow(I-BA)(I+BCA)=I$ .

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

Quote:

Originally Posted by AMI

$\bullet$ $(I+BCA)(I-BA)=I-BA+BCA-BCABA$
$C=(I-AB)^{-1}\Rightarrow C(I-AB)=I$ $\Rightarrow C-CAB=I\Rightarrow B(C-CAB)A=BIA$ $\Rightarrow BCA-BCABA=BA\Rightarrow(I+BCA)(I-BA)=I$ .
$\bullet$ $(I-BA)(I+BCA)=I+BCA-BA-BABCA$
$C=(I-AB)^{-1}\Rightarrow (I-AB)C=I$ $\Rightarrow C-ABC=I\Rightarrow B(C-ABC)A=BIA$ $\Rightarrow BCA-BABCA=BA\Rightarrow(I-BA)(I+BCA)=I$ .

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

thanks :)
June 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

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