Assume $T$ is compact operator and $S(I- T) = I $.Is this true that $(I- T)S =I$?

Question

Suppose $S,T \in {\rm B}(X)$ and assume $T$ is compact operator and $S(I- T) = I $.Is this true that $(I- T)S =I$?

Welcome to Mathematics Stack Exchange! I would suggest you to explain a little bit your thoughts about it, so other people could help you better. Good luck! — iadvd, Jun 15 '15 at 06:30

score 5 · Answer 1 · answered Jun 15 '15 at 06:44

5

Yes, your hypothesis implies that $(I-T)$ is injective, which, by the Fredholm alternative means that it is surjective as well. By the bounded inverse theorem, this means $(I-T)$ is invertible in $B(X)$, and so its right and left-inverses must coincide.

answered Jun 15 '15 at 06:44

Prahlad Vaidyanathan

31,554

Why does the assumption imply that $(I-T)$ is injective? – 1233dfv Jul 12 '15 at 14:02
1

Because $\ker(I-T) \subset \ker(S(I-T))$ – Prahlad Vaidyanathan Jul 12 '15 at 15:46

Assume $T$ is compact operator and $S(I- T) = I $.Is this true that $(I- T)S =I$?

1 Answers1

Linked