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Endomorphism sequence

  1. Jul 21, 2015 #1
    Hello let be a finish normed vectoriel space $$(E, ||.||)$$, and $$u \in L(E) / ||u|| \leq 1$$.
    1) Show that $$Ker(u - Id) = Ker((u - Id)^{2})$$.
    For $$\subset$$ it's obvious, but for $$\supset$$ I don't know. I suppose $$\exists x_{1} \in Ker((u - Id)^{2})\Ker(u - Id)$$. So I can say that $$u(x_{1}) - x_{1} \neq 0_{E}$$ But I don't know.

    2) Show that $$E = Im(u - Id) \oplus Ker(u - Id)$$
    If I have an $$x \in Im(u - Id) \cap Ker(u - Id)$$, I have a w in E and I have $$u(w) - w = x$$ so $$u(x) - x = 0_{E} = (u - Id)^{2}(w)$$, so $$w \in Ker((u - Id)^{2}) = Ker(u - Id)$$ so $$x = 0_{E}$$.
    And like I have $$dim(E) = rg(u - Id) + dim(Ker(u - Id))$$. It conclude.

    3) Show that $$u_{n} = \frac{1}{n}(Id + \sum_{i=2}^{n} u^{i-1})$$ converge in an projector on $$Ker(u - Id)$$.
    I don't know. By the previous question I can say that if I fix an x $$Ker(u - Id)$$ like u(x) = x,
    $$u_{n}(x) = x$$. But like with the image I find nothing I give up with this idea.

    Could you help me please?

    Thank you in advance and have a nice afternoon:oldbiggrin:.
  2. jcsd
  3. Jul 21, 2015 #2
    Let's start with this. Take ##x\in \text{Ker}((u-\text{Id})^2)\setminus \text{Ker}(u-\text{Id})##.
    1) Show that ##y=(u -\text{Id})(x)## is an eigenvector of ##u## with eigenvalue ##1##.
    2) Show that ##\{x,y\}## are linearly independent. Show that we can find ##x## and ##y## to be orthonormal.
    3) Calculate ##u(x)## in terms of the "basis" ##\{x,y\}##.
    4) Find a lower bound for ##\|u(x)\|## and use it to show that ##\|u\|>1##.
  4. Jul 21, 2015 #3
    For this, consider an element ##y= u(w) - w## in the image of ##u - \text{Id}##. Can you rewrite

    [tex]y + u(y) + ... + u^{n-1}(y)[/tex]

    in a suitable way? Think of telescoping sums.
  5. Jul 22, 2015 #4
    Hello and thanks for answer, for 1) : $$y = u(x) - x$$ so $$u(y) = y$$ because $$x \in \text{Ker}((u-\text{Id})^2)$$.
    Then $$ax + by = 0_{E} \Rightarrow a(u(x) - x) = 0_{E} \Rightarrow a = 0$$ because $$u(x) - x \neq 0_{E}$$ as $$
    x\in \text{Ker}((u-\text{Id})^2)\setminus \text{Ker}(u-\text{Id})(1)$$. And because of (1) $$y \neq 0_{E}$$ so $$b = 0$$. I don't see what you means by orthonormal. So in the subspace geberae by x and y, the matrix of $$u - Id$$ is $$\beginn{pmatrix}(0 1\\11 )\end{pmatrix}$$. For 4) I don't know. Could you help me more pkease?

    For 3) as $$y \in Im(u - Id) $$ as you say we get $$u_{n}(y) = \frac{1}{n}(u^{n}(w) - w)$$. Coukld you help me more please?

    Thank you in advance and have a nice morning:oldbiggrin:.
  6. Jul 22, 2015 #5
    So if I fix a $$y \in Im(u - Id)$$, I have $$\forall n \in \mathbb{N}, ||u_{n}(y)|| \leq \frac{1}{n}||u^{n}(w)|| + \frac{1}{n}||w|| = \frac{2}{n}||w||$$ because $$||u|| \leq 1$$ and $$||u^{n}|| \leq ||u||^{n}$$. So lim $$u_{n}(y) = 0_{E}$$.
    So for $$x \in E, x = a + b, a \in Ker(u - Id), b \in Im(u - Id), lim u_{n}(x) = a$$. But why does a such simple convergence make that my sequence of $$u_{n}$$ converge****** in a projector on $$Ker(u - Id)$$ please?

    ****** For the norme $$||u|| = sup_{x \neq 0_{E}} \frac{||u(x)||}{||x||}$$ in $$L(E)$$.

    Thank you in advance and have a nice afternoon:oldbiggrin:.
  7. Jul 22, 2015 #6
    What happened to ##b##?

    Try to prove that ##\|Ay\|>\|y\|##.
  8. Jul 22, 2015 #7
    So you see that the pointswise limit ##u(x) = \lim_n u_n(x)## is a projection. Try to find a simple expression for ##u(x) - u_n(x)## and use this to show that ##\|u - u_n\|\rightarrow 0##.
  9. Jul 22, 2015 #8
    I said after we have $$by = 0_{E}$$ and y is not nul otherwise $$x \in Ker(u - Id)$$. SO b is nul.

    OK but it's just said me that for a certain x(fixed.). the sequence $$u_{n}(x)$$ have the projection of x on $$Ker(u - Id)$$ I can call it $$p(x)$$ so p(x) define an application which is linear. I really don't see the link with $$||u - u _{n}|| =
    sup_{x \neq 0_{E}} \frac{||u(x) - u_{n}(x)||}{||x||}$$. Could you help me please?

    Oh by the way how to stop latex from go at the line each time I rigght somethnig please?

    Thank you in advance and have a nice afternoon:oldbiggrin:.
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