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

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.

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

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

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

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

well assume $\displaystyle I - AB $ is invertible

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

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

Hi Jakncoke,

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

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

Quote:

Originally Posted by

**johng** Hi Jakncoke,

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

yes you are right! what was i thinking.

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

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

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

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

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

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

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

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

Yes, i see.

Thanks everyone !!