Abstract
As is well known, the generalized inverse, Moore–Penrose inverse and group inverse are not continuous, i.e. for θ = {1, 2}, {1, 2, 3, 4} and {1, 2, 5}, a linear bounded operator T has a θ-inverse , the perturbed operator
is not necessary θ-invertible and even if it is θ-invertible,
may not be true. In this paper, we prove that
is θ-invertible and its θ-inverse
has the simplest possible expression, which satisfies
. Thus, we have found a continuous orbit for the generalized inverse, Moore–Penrose inverse and group inverse.
Acknowledgments
This research is supported by the National Natural Science Foundation of China (11771378, 11871064, 11971419).
Disclosure statement
No potential conflict of interest was reported by the author(s).