# 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.

• 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

## 1 Answer

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.

Since you might be interested in a proof (which isn't short), instead of copying all of it here, you can find it in "A simple proof of the Fredholm alternative " for example, proposition 1. The proof is only given in the case of an approximable operator (which is always the case in Hilbert spaces, for example), but it's true in general.