# For $T$ compact, $I-T$ left or right invertible implies $I-T$ invertible

Let $S\in B(X)$ be a bounded linear operator from $X$ onto $X$ and let $T\in K(X)$ be a compact linear operator from $X$ onto $X$. Then

$$S(I-T)=I \iff (I-T)S=I.$$

I don't know if we need the fact that $T$ is compact here, we might only need to know that $T\in B(X)$. I needed compactness for another part of this same problem.

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We do need that $T$ is compact. Let $S$ be the left shift given by $S(e_1) = 0$ and $Se_{n} = e_{n-1}$ for $n \geq 2$ and let $R$ be the right shift given by $R e_n = e_{n+1}$ on $\ell^2(\mathbb N)$. Then $SR = I$, so $T = I-R$ satisfies $S(I-T) = I$, but $RS e_1 = 0$, so $(I-T)S \neq I$. Using this idea you can show that both implications are false if $T$ isn't assumed to be compact. – t.b. May 23 '12 at 8:49

This is a simple consequence of Fredholm alternative: for a compact operator $T$, then a value $\lambda \neq 0$ is either an eigenvalue or is in the resolvent. In this case, you're interested in $\lambda = 1$. In fact the theorem is even stronger: it states that $\lambda I - T$ is injective iff it's surjective, which is exactly your question.