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Linear application and différential

  1. Aug 6, 2016 #1
    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:oldbiggrin:.
     
  2. jcsd
  3. Aug 6, 2016 #2

    pasmith

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    Homework Helper

    Think about the eigenvalues of [itex]u[/itex]. There are at most [itex]m[/itex] of them, and a finite set of real numbers always has a maximum and a minimum.
     
  4. Aug 6, 2016 #3
    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.
     
  5. Aug 6, 2016 #4
    I have an idea.
     
  6. Aug 6, 2016 #5
    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.
     
  7. Aug 6, 2016 #6
    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}##
     
  8. Aug 6, 2016 #7
    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.
    Thanks for your help.
     
  9. Aug 6, 2016 #8
    Yeah what I wroght non works I think but I don't use ##m = dim(E)##.
    What do you think please?
     
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