I-AB is invertible if and only if I-BA is invertible
Theorem - Let and be endomorphisms of a vector space . We have that
Suppose that is invertible. We shall prove that is the inverse of . In fact
A similar computation shows that , i.e. is invertible.
Exchanging the roles of and we can prove the ”if” part. So is invertible if and only if is invertible.
Suppose is not injective, i.e. there exists such that . Then
i.e. . Notice that because (by definition of ), so is also not injective.
Similarly, if is not injective then is not injective.
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 and are elements of a ring with unity, then is invertible if and only if 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|
|Date of creation||2013-03-22 14:44:43|
|Last modified on||2013-03-22 14:44:43|
|Last modified by||asteroid (17536)|