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In this question, quoted below, $T$ is a compact operator. Why is this condition necessary?

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

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  • Welcome to Math.SE! Since the original question you link to is very short, it would be helpful if you repeat that question here. – Hrodelbert Jun 15 '15 at 17:55
  • The word "necessary" is unfortunate here. The compactness of $T$ is sufficient for invertibility; if you remove that condition and don't replace it with anything else, then invertibility will no longer follow; but that condition is not necessary for invertibility, in that just because $I-T$ is invertible doesn't mean $T$ is compact. (For example, $I-\lambda I$ is invertible for most scalars $\lambda$.) –  Jun 16 '15 at 00:13

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Without compactness, $I-T$ could be, for example, the forward shift operator on $\ell^2$, with $S$ being the backward shift.