Drazin Invertibility of Products

If a and b are elements of an algebra, then we show that ab is Drazin invertible if and only if ba is Drazin invertible. With this result we investigate products of bounded linear operators on Banach spaces. 1. Drazin inverses Throughout this section A is a real or complex algebra with identity e 6= 0. We denote the group of invertible elements of A by A. We call an element a ∈ A relatively regular if there is b ∈ A for which a = aba. In this case b is called a generalized inverse of a. An element a ∈ A is said to be Drazin invertible if there is c ∈ A and k ∈ N ∪ {0} such that aca = a, cac = c and ac = ca . In this case c is called a Drazin inverse of a and the least non-negative integer k satisfying aca = a is called the Drazin index i(a) of a. We write D(A) for the set of all Drazin invertible elements of A. With the convention a = e we have a ∈ A ⇐⇒ a ∈ D(A) and i(a) = 0 . Proposition 1. If a ∈ D(A), then a has a unique Drazin inverse. Proof. [3]. Proposition 2. For a ∈ A the following assertions are equivalent: (1) a ∈ D(A) and i(a) ≤ 1; (2) there is b ∈ A such that aba = a and e− ab− ba ∈ A. Proof. [5, Proposition 3.9]. The main result of this section reads as follows: Theorem 1. Let a, b ∈ A. Then: ab ∈ D(A) ⇐⇒ ba ∈ D(A) . In this case we have: (1) |i(ab)− i(ba)| ≤ 1; (2) if c is the Drazin inverse of ab, then bca is the Drazin inverse of ba. 2000 Mathematics Subject Classification. 47A10, 15A09.