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

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

1. 1.
2. 2.

Proof :

1. 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

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

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.

2. 2.

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.

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.