Solve Endomorphism Sequence Questions in a Vectoriel Space

[Calabi]

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.

 

[micromass]

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.

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 $\|u\|>1$.

 

[micromass]

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.

For this, consider an element $y= u(w) - w$ in the image of $u - \text{Id}$. Can you rewrite

$$y + u(y) + ... + u^{n-1}(y)$$

in a suitable way? Think of telescoping sums.

 

[Calabi]

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.

 

[Calabi]

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.

 

[micromass]

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 $$ax + by = 0_{E} \Rightarrow a(u(x) - x) = 0_{E} \Rightarrow a = 0$$

What happened to $b$?

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?

Try to prove that $\|Ay\|>\|y\|$.

 

[micromass]

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?

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 $\|u - u_n\|\rightarrow 0$.

 

[Calabi]

What happened to bb?

I said after we have $$by = 0_{E}$$ and y is not nul otherwise $$x \in Ker(u - Id)$$. SO b is nul.

So you see that the pointswise limit u(x)=limnun(x)u(x) = \lim_n u_n(x) is a projection. Try to find a simple expression for u(x)−un(x)u(x) - u_n(x) and use this to show that ∥u−un∥→0\|u - u_n\|\rightarrow 0.

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.