Math Problem Statement

Soient E un K-espace vectoriel de dimension finie n et f ∈ L(E). 1. On suppose f nilpotent et on note p son indice de nilpotence. (a) Écrire avec des quantificateurs les propositions fp = 0 L(E) et fp−1 ≠ 0 L(E). (b) Montrer que pour un certain x ∈ E, la famille (x, f(x), f2(x), …, fp−1(x)) est libre. (c) En déduire que fn = 0 L(E). 2. On suppose dans cette question que p = n. D’après 1.b, E possède donc une base de la forme: (x, f(x), f2(x), …, fn−1(x)).

Solution

Let's analyze and solve the given problems step by step.


Problem 1:

We have a KK-vector space EE of finite dimension nn and fL(E)f \in \mathcal{L}(E). We assume ff is nilpotent with index of nilpotence pp.

(a) Write the propositions fp=0L(E)f^p = 0_{\mathcal{L}(E)} and fp10L(E)f^{p-1} \neq 0_{\mathcal{L}(E)} using quantifiers.

Solution:

  • fp=0L(E)f^p = 0_{\mathcal{L}(E)}: xE, fp(x)=0\forall x \in E, \ f^p(x) = 0
  • fp10L(E)f^{p-1} \neq 0_{\mathcal{L}(E)}: xE such that fp1(x)0\exists x \in E \ \text{such that} \ f^{p-1}(x) \neq 0

(b) Show that for a certain xEx \in E, the family (x,f(x),f2(x),,fp1(x))(x, f(x), f^2(x), \ldots, f^{p-1}(x)) is linearly independent.

Solution: We assume ff is nilpotent with index pp, i.e., fp=0f^p = 0. Let us consider xEx \in E such that fp1(x)0f^{p-1}(x) \neq 0. We need to show that the family (x,f(x),f2(x),,fp1(x))(x, f(x), f^2(x), \ldots, f^{p-1}(x)) is linearly independent.

Assume there exist scalars α0,α1,,αp1K\alpha_0, \alpha_1, \ldots, \alpha_{p-1} \in K such that: α0x+α1f(x)+α2f2(x)++αp1fp1(x)=0\alpha_0 x + \alpha_1 f(x) + \alpha_2 f^2(x) + \cdots + \alpha_{p-1} f^{p-1}(x) = 0 Applying ff repeatedly to this equation, we obtain:

\alpha_0 f(x) + \alpha_1 f^2(x) + \alpha_2 f^3(x) + \cdots + \alpha_{p-2} f^{p-1}(x) + \alpha_{p-1} f^p(x) = 0 \\ \alpha_0 f^2(x) + \alpha_1 f^3(x) + \cdots + \alpha_{p-2} f^p(x) = 0 \\ \vdots \\ \alpha_0 f^{p-1}(x) + \alpha_1 f^p(x) = 0 \\ \alpha_0 f^p(x) = 0 \end{cases}$$ Since $$f^p(x) = 0$$, the last equation is trivial. The second last equation $$\alpha_0 f^{p-1}(x) = 0$$ implies $$\alpha_0 = 0$$ because $$f^{p-1}(x) \neq 0$$. Repeating this argument, we conclude $$\alpha_1 = \alpha_2 = \cdots = \alpha_{p-1} = 0$$. Hence, the family $$(x, f(x), f^2(x), \ldots, f^{p-1}(x))$$ is linearly independent. **(c)** Deduce that $$f^n = 0_{\mathcal{L}(E)}$$. **Solution:** Since $$f$$ is nilpotent with index $$p$$ and $$p \leq n$$, we have: $$f^p = 0_{\mathcal{L}(E)}$$ Given that $$p \leq n$$, it follows that: $$f^n = 0_{\mathcal{L}(E)}$$ --- **Problem 2:** We suppose in this question that $$p = n$$. According to 1.b, $$E$$ therefore has a basis of the form: $$(x, f(x), f^2(x), \ldots, f^{n-1}(x))$$ **Solution:** Since $$p = n$$, we have shown in 1.b that the family $$(x, f(x), f^2(x), \ldots, f^{n-1}(x))$$ is linearly independent and spans $$E$$. Therefore, it forms a basis of $$E$$. --- Do you want details or have any questions about the solutions? Here are five related questions to further your understanding: 1. What is the definition of a nilpotent linear operator? 2. How do we determine the index of nilpotence for a given operator? 3. What are the properties of a vector space with a nilpotent operator? 4. Can you provide an example of a nilpotent operator in a 3-dimensional vector space? 5. How does the concept of nilpotent operators relate to the Jordan canonical form? **Tip:** When dealing with nilpotent operators, always remember to verify the highest power of the operator that results in the zero transformation to find the index of nilpotence.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Nilpotent linear operators
Linear algebra
Vector spaces
Linear independence

Formulas

fp = 0 L(E)
fp−1 ≠ 0 L(E)
Linear independence of (x, f(x), f^2(x), …, f^(p-1)(x))

Theorems

Nilpotent operator theorem
Basis of a vector space

Suitable Grade Level

University level