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Suppose we have a linear isomorphism $T_{0}: \mathbb{R}^{n} \mapsto \mathbb{R}^{n}$, and $T_{\epsilon}$ is a small linear perturbation of it, i.e.

$$T_{\epsilon} = T_{0} + \epsilon T_{1} + O(\epsilon^{2}),$$

where $T_{1}$ is also linear isomorphism and $0< | \epsilon | <<1$. Does it follow that $T_{\epsilon}$ is also an isomorphism? If so, I am wondering is it possible to use an implicit function theorem for this, or can some result from perturbation theory of linear operators be used? (Note that this question comes from an earlier one: How do I set up Implicit Function Theorem to verify this function is a $C^{r}$ diffeomorphism?)

Thanks!

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In any complete normed space, the set of invertible operators is open. This follows from the usual Neumann trick that shows that if $\|I-T\|<1$, then $T$ is invertible, with the inverse given by $$ \sum_{k=0}^\infty (I-T)^n. $$ One then extends this to show that if $S$ is invertible and $\|S-T\|<\|S^{-1}\|^{-1}$, then $T$ is invertible.

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The answer is yes and it does not depend on $T_1$ being an isomorphism. The reason is that the map $$\mathcal{L}(\Bbb R^n\to\Bbb R^n)\to\Bbb R: T\mapsto \det(T)$$ is continuous, and the set of isomorphisms is the preimage of $\Bbb R\setminus \{0\}$ which is an open set, hence the set of isomorphisms is open itself.

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  • $\begingroup$ Thanks. Does this mean, that if I have some perturbed map $F_{\epsilon} (x)= F_{0}(x) + \epsilon F_{1}(x, \epsilon)$ where each $F_{0}$ and $F_{1}$ are $C^{k}$ diffeomorphisms, then $F_{\epsilon}$ is, as well? (From applying inverse function theorem and looking at the linearisation $DF_{0} + \epsilon DF_{1}) $ ? @Arnaud Mortier $\endgroup$ – Alex May 22 '18 at 13:55
  • $\begingroup$ @Alex the linearisation will tell you that locally $F_\epsilon$ is still a diffeomorphism, but globally you cannot be certain that it is still one-to-one just by looking at $DF$. $\endgroup$ – Arnaud Mortier May 22 '18 at 14:00
  • $\begingroup$ Could you clarify a bit why locally $F_{\epsilon}$ would be a diffemorphism, but not always globally, if I assume that $F_{0}, F_{1}$ are $\textbf{global}$ diffeomorphisms and $\epsilon$ is small? Also, what happens if I remove the assumption that perturbation $F_{1}$ is a diffeomorphism - does then implicit function theorem apply?@Arnaud Mortier $\endgroup$ – Alex May 22 '18 at 14:04
  • $\begingroup$ @Alex I think this should be a separate question, where you define the settings regarding the maps $F$. Yes, the IFT applies as soon as the Jacobian is invertible, which will be the case when $\epsilon$ is small enough, no matter what $F_1$ is (although it has to be differentiable of course). $\endgroup$ – Arnaud Mortier May 22 '18 at 14:10
  • $\begingroup$ Thanks once again for your comments (yes indeed, I added the link to this separate question in my main post above) @Arnaud Mortier $\endgroup$ – Alex May 22 '18 at 14:41
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The set of linear isomorphisms is open ! Hence there is $ \delta>0$ such that if $||T_0-A|| < \delta$, then $A$ is a linear isomorphism.

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