if I-AB is invertible , is I-BA invertible ?

**poorbuttryagin** · February 4th 2013, 05:39 PM

Hi, everyone ~

I read Linear Algebra by Hoffman & Kunze.

At 190pg #8,

A, B := n*n matrices . Prove that if I-AB is invertible, then I-BA is invertible and (I-BA)^(-1) = I +B(I-AB)^(-1) A.

Any hint or comment are welcomed !

Please help !

Thanks.

**johng** · February 4th 2013, 08:02 PM

Let $X=(I-AB)^{-1}$ . Then $I=X(I-AB)=X-XAB$ or $XAB=X-I$ . Now use this fact and the distributive law to show

$(I+BXA)(I-BA)=I$ . (You know this proves the statement, don't you?)

**jakncoke** · February 4th 2013, 08:11 PM

well assume $I - AB$ is invertible

then $(I-AB)*A = (A - ABA) = A(I - BA)$ so $(I-AB)*A = A*(I-BA)$ using the determinant rule $det(I-AB)A = det(I-AB)det(A) = det(I-BA)*A) = det(I-BA)*det(A)$ since determinants are just numbers, cancelling det(A) on both sides. $det(I-AB) = det(I-BA) = 1$ thus I-BA is invertible.

**johng** · February 4th 2013, 08:17 PM

Hi Jakncoke,

What if det(A) = 0?. I think your argument fails.

**jakncoke** · February 4th 2013, 08:24 PM

Originally Posted by johng

Hi Jakncoke,

What if det(A) = 0?. I think your argument fails.

yes you are right! what was i thinking.

**Deveno** · February 4th 2013, 09:01 PM

since they actually give the answer, i don't see what the fuss is about:

$(I + B(I - AB)^{-1}A)(I - BA) =$

$[I + B(I - AB)^{-1}A] - [BA + B(I - AB)^{-1}ABA] =$

$I - BA + B[(I - AB)^{-1} - (I - AB)^{-1}AB]A =$

$I - BA + B[(I - AB)^{-1}I - (I - AB)^{-1}AB]A =$

$I - BA + B[(I - AB)^{-1}(I - AB)]A = I - BA + BA = I$ .

**poorbuttryagin** · February 5th 2013, 03:42 AM

Yes, i see.
Thanks everyone !!

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