# Linear application and différential

1. Aug 6, 2016

### Calabi

1. The problem statement, all variables and given/known data
Let be E a normded vectoriel space, $dim(E) = m \in \mathbb{N}^{*}$, I have to show that $\exists \rho_{1}, \rho_{2} > 0 | \forall u \in L(E), ||u^{m} - Id| \leq \rho_{1} \Rightarrow |u - Id| \leq \rho_{2}$.

2. Relevant equations
Nothing.

3. The attempt at a solution

I try to constrcut a $C^{1}$ application which will have by definition a continuous différential. But I don't see what the dimension m as to do in here and I'm really locked.

Could you give me a clue please?

Thank you in advance and have a nice afternoon.

2. Aug 6, 2016

### pasmith

Think about the eigenvalues of $u$. There are at most $m$ of them, and a finite set of real numbers always has a maximum and a minimum.

3. Aug 6, 2016

### Calabi

If I look in $\mathbb{R}$ it's possible ther's no eigenvalues and the matrice I ut is not necessarely diagonaloisabled.
Are you sure it's really exploitable?

I find this problem on differential calcul set of problem.

4. Aug 6, 2016

### Calabi

I have an idea.

5. Aug 6, 2016

### Calabi

Let put us around the identity $Id$, the application $\phi : u \in L(E) \rightarrow u^{m} \in L(E)$ is $C^{\infty}$, as the differential in $Id$ is an isomorphism, then the inverse function theorem show that $\exists \rho_{1}, \rho_{2} > 0 | \phi : B(Id, \rho_{1}) \rightarrow B(Id, \rho_{2})$ is a bijection. But as $\phi$ is not injective it doesn't work.

6. Aug 6, 2016

### Calabi

I corrrecte the ennoncee : $\exists \rho_{1}, \rho_{2} > 0 | \forall v \in L(E) \exists u \in L(E) | u^{m} = v \text{ and } ||v - Id|| \leq \rho_{1} \Rightarrow ||u - Id|| \leq \rho_{2}$

7. Aug 6, 2016

### Calabi

I think we use m is the dimension for the unicity but I don't see why, the rest is an application of the inverse function theorem.
Yeah what I wroght non works I think but I don't use $m = dim(E)$.