Is a small linear invertible perturbation of a linear isomorphism also an isomorphism?

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.