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 :

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

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

A similar computation shows that $\Big{(}B(I-AB)^{-1}A+I\Big{)}\Big{(}I-BA\Big{)}=I$ , 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.

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

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

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

i.e. $Bu\in\operatorname{ker}(I-BA)$ . Notice that $Bu\neq 0$ 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. $\square$

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