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[parent] I-AB is invertible if and only if I-BA is invertible (Theorem)

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

    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. 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).

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

    $\displaystyle (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.

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 this discussion on PM.



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See Also: techniques in mathematical proofs


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$AB$ and $BA$ are almost isospectral (Corollary) by asteroid
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Cross-references: techniques in mathematical proofs, ring with unity, infinite dimensional, zero vector, kernel, linear map, similar, injective, dimension, square matrices, basis, finite dimensional, vector space, endomorphisms

This is version 13 of I-AB is invertible if and only if I-BA is invertible, born on 2004-10-17, modified 2007-12-07.
Object id is 6383, canonical name is IABIsInvertibleIfAndOnlyIfIBAIsInvertible.
Accessed 3409 times total.

Classification:
AMS MSC: 15A04 (Linear and multilinear algebra; matrix theory :: Linear transformations, semilinear transformations)
 16B99 (Associative rings and algebras :: General and miscellaneous :: Miscellaneous)
 47A10 (Operator theory :: General theory of linear operators :: Spectrum, resolvent)
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