Solve Endomorphism Sequence Questions in a Vectoriel Space

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In summary, we have a conversation about a normed vector space with a linear operator u satisfying a specific condition. The conversation covers three questions: 1) showing that the kernel of u-Id is equal to the kernel of (u-Id)^2, 2) proving that E is equal to the image of u-Id and the kernel of u-Id, and 3) discussing the convergence of a sequence of operators u_n to a projection on the kernel of u-Id. The conversation also includes dialogue about finding eigenvectors and eigenvalues, using orthonormality, and proving convergence.
  • #1
Calabi
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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:.
 
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  • #2
Calabi said:
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##.
 
  • #3
Calabi said:
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

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

in a suitable way? Think of telescoping sums.
 
  • #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:.
 
  • #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:.
 
  • #6
Calabi said:
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\|##.
 
  • #7
Calabi said:
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##.
 
  • #8
micromass said:
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.
micromass said:
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 ∥uun∥→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:oldbiggrin:.
 

FAQ: Solve Endomorphism Sequence Questions in a Vectoriel Space

What is an endomorphism?

An endomorphism is a linear transformation from a vector space to itself. In other words, it maps vectors from a vector space back to the same vector space.

What is a sequence of endomorphisms?

A sequence of endomorphisms is a series of endomorphisms applied in a specific order. This can also be thought of as a chain of linear transformations, where the output of one endomorphism becomes the input for the next one.

What is a vector space?

A vector space is a mathematical structure that consists of a set of vectors and operations such as addition and scalar multiplication. It is used to represent and manipulate quantities that have both magnitude and direction.

How do you solve endomorphism sequence questions?

To solve endomorphism sequence questions in a vector space, you need to apply the individual endomorphisms in the given sequence to the vector in the correct order. This can be done by multiplying the vector by each endomorphism's corresponding matrix representation.

Why are endomorphism sequence questions important?

Endomorphism sequence questions are important because they allow us to study how multiple linear transformations affect a vector in a vector space. These questions also have practical applications in fields such as physics, computer science, and engineering.

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