 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-10/4/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Let

c = (1 - ab)^{- 1} \in A

. From the two inverse identities we have

c ab = c - 1 = ab c

. The candidate inverse is

(1 - ba)^{- 1} = 1 + b c a .

Indeed

(1 - ba) (1 + b c a) = 1 + b (c - 1 - ab c) a = 1,

and

(1 + b c a) (1 - ba) = 1 + b (c - 1 - c ab) a = 1.

Both products have been checked, so this is valid in a noncommutative Banach algebra. It is the identity underlying the nonzero spectra of products in opposite orders result.

 New to topics? Read the docs here!