$T_{\varepsilon} = Id + \varepsilon T$ is an isomorphism for some $\varepsilon$

Question

Let $T: V \to V$ be a linear map on some finite dimensional real vector space, and define $$ T_{\varepsilon} = Id + \varepsilon T $$ Is it true that for $\varepsilon$ small enough, $T_{\varepsilon}$ is an isomorphism? How can I go about proving this? I think it's something to do with the fact that $T$ only has finitely many eigenvalues, but I don't have an exact rigorous formulation of the proof.

Additionally, are any generalizations of this possible? Say I have a non-linear but $C^{\infty}$ function $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$ and I define $f_{\varepsilon}$ as $f_{\varepsilon}(x) = x + \varepsilon f(x)$; then when is this function bijective, if we get the choice of $\varepsilon$? If we can't make it bijective, can we at least make it injective? Even in the linear case, we're proving injectivity only, which implies bijectivity in the former case but not the latter.

Edit: I'm happy with answers on the part with the linear assumption, but still want to know whether the generalization is possible, just for injectivity? In the generalization, we can assume that $f$ is a $C^{\infty}$ function.

Answer 1

Since the determinant is a continuous function and since $\det\operatorname{Id}=1$, you have that $\det(\operatorname{Id}+\varepsilon T)\neq0$ if $\varepsilon$ is close enough to $0$, and therefore $\operatorname{Id}+\varepsilon T$ is an isomorphism then.

Answer 2

One way to prove it using your own intuition: Consider the null space. $T_\epsilon x=0$ if and only if $(\epsilon T + Id)x=0$ if and only if $(T-\frac{-1}{\epsilon}Id)x=0$. Since $V$ is finite dimensional, there exists a minimum eigenvalue, $\lambda_0$. For $\epsilon$ sufficiently small, we can guarantee $\frac{-1}{\epsilon}<\lambda_0$. In this case, the kernel of $T+\frac{1}{\epsilon}$ is trivial. By rank-nullity theorem, $T_\epsilon$ is an isomorphism.

Answer 3

Suppose that for any $y$ we can find unique $x$ such that $x=y-\epsilon Tx$. Then $id+\epsilon T$ has an inverse and we are done.

This will be the case if $x \mapsto y-\epsilon T x$ is a contraction mapping.

Similarly for the nonlinear function $f_{\epsilon}(x)=x+\epsilon f(x)$ we need for any $y$ that $x \mapsto y-\epsilon f(x)$ be a contraction. If $f$ is Lipschitz continuous with constant $k$, i.e. if $\lVert f(x)-f(x')\rVert \leq k \lVert x - x'\rVert$ for all $x,x'$, then this is satisfied for $\epsilon<1/k$.

Answer 4

For the your generalized case, the answer is no. Take $f(x) =x^2 \sin x$. Then $f$ is $C^\infty $, and $x+\varepsilon f(x) $ takes arbitrarily big positive values at $x=(4n+1)\pi/2$ and arbitrarily big negative values at $(4n+3)\pi/3$.

On another note, the answer to your main question is yes even in infinite dimension, if you have a complete normed space. The result is that if $T$ is bounded linear and $\|I-T\|<1$, then $T$ is invertible. The inverse is, explicitly, $$T^{-1}=\sum_{n=0}^\infty (I-T)^n.$$