Question

Let a, b be elements in a ring R. If 1 − ba is left-invertible, show that 1 − ab is left-invertible, and construct a left inverse for it explicitly

Answer 1

The left ideal R(1 − ab) contains Rb(1 − ab) = R(1 − ba)b = Rb, so it also contains (1 − ab) + ab = 1. This shows that 1 − ab is left-invertible. This proof lends itself easily to an explicit construction: if u(1 − ba) = 1,then
b = u(1 − ba)b = ub(1 − ab), so 1 = 1− ab + ab = 1− ab + aub(1 − ab) = (1+aub)(1 − ab).
Hence, (1 − ab)−1 = 1+a(1 − ba)−1b, where x−1 denotes “a left inverse” of x. The case when 1 − ba is invertible follows by combining the “left-invertible” and “right-invertible”
cases