I-AB is invertible if and only if I-BA is invertible

In this entry $A$ and $B$ are endomorphisms of a vector space $V$. If $V$ is finite dimensional, we may choose a basis and regard $A$ and $B$ as square matrices of equal dimension.

Theorem - Let $A$ and $B$ be endomorphisms of a vector space $V$. We have that

1. $I-AB$ is invertible (http://planetmath.org/LinearIsomorphism) if and only if $I-BA$ is invertible, and moreover

2. $I-AB$ is injective if and only if $I-BA$ is injective.

Proof :

1. Suppose that $I-AB$ is invertible. We shall prove that $B{(I-AB)}^{-1}A+I$ is the inverse of $I-BA$. In fact

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

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

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

   $=$ $BA+I-BA$

   $=$ $I$

   A similar computation shows that $\left(B{(I-AB)}^{-1}A+I\right)\left(I-BA\right)=I$, i.e. $I-BA$ is invertible.

   Exchanging the roles of $A$ and $B$ we can prove the ”if” part. So $I-AB$ is invertible if and only if $I-BA$ is invertible.

2. Let us first recall that a linear map between vector spaces is invertible if and only if its kernel $\ker$ is the zero vector (see this page (http://planetmath.org/KernelOfALinearTransformation)).

   Suppose $I-AB$ is not injective, i.e. there exists $u\ne 0$ such that $(I-AB)u=0$. Then

   $$(I-BA)Bu=B(I-AB)u=0$$

   i.e. $Bu\in \ker (I-BA)$. Notice that $Bu\ne 0$ because $u=ABu$ (by definition of $u$), so $I-BA$ is also not injective.

   Similarly, if $I-BA$ is not injective then $I-AB$ is not injective. $\mathrm{\square }$

Remark - It is known that for finite dimensional vector spaces a linear endomorphism is invertible if and only if it is injective. This does not remain true for infinite dimensional spaces, hence 1 and 2 are two different statements.

0.1 Comments

The result stated in 1 can be proven in a more general context — If $A$ and $B$ are elements of a ring with unity, then $I-AB$ is invertible if and only if $I-BA$ is invertible. See the entry on techniques in mathematical proofs, in which this result is proven using several different techniques.

This entry is based on http://planetmath.org/?op=getmsg&id=5088this discussion on PM.

Title	I-AB is invertible if and only if I-BA is invertible
Canonical name	IABIsInvertibleIfAndOnlyIfIBAIsInvertible
Date of creation	2013-03-22 14:44:43
Last modified on	2013-03-22 14:44:43
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	16
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 16B99
Classification	msc 15A04
Classification	msc 47A10
Related topic	TechniquesInMathematicalProofs