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

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

#### Solutions Collecting From Web of "Assume $T$ is compact operator and $S(I- T) = I$.Is this true that $(I- T)S =I$?"

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.