Math Problem Statement

Sei K ein Körper, und sei V = ⟨1, T², T⁴, T⁶⟩ ⊆ K[T]. 1. Definieren Sie eine lineare Abbildung f : V → V mit Kern(f) = Bild(f). 2. Wählen Sie eine Basis B von V, und bestimmen Sie BMB(f). 3. Beweisen Sie: Ist W ein endlich erzeugter K-Vektorraum und ist f : W → W eine lineare Abbildung mit Kern(f) = Bild(f), so ist die Dimension von W eine gerade Zahl.

Solution

Let's go through each part of this problem step-by-step:

  1. Defining a Linear Map f:VVf: V \to V with ker(f)=Bild(f)\ker(f) = \operatorname{Bild}(f):

    • To define a linear map ff from VV to VV with ker(f)=Bild(f)\ker(f) = \operatorname{Bild}(f), we need to think about an endomorphism of the vector space VV with this property.
    • Since VV is spanned by the elements {1,T2,T4,T6}\{1, T^2, T^4, T^6\}, any linear map ff can be represented by a matrix with respect to this basis.
    • A possible choice for such an ff is to create a map that "shifts" elements in a way that reduces dimension by mapping certain elements to others in the span, satisfying ker(f)=Bild(f)\ker(f) = \operatorname{Bild}(f).
  2. Choosing a Basis B\mathcal{B} of VV and Determining BMB(f)\mathcal{B} M_{\mathcal{B}}(f):

    • The standard basis B={1,T2,T4,T6}\mathcal{B} = \{1, T^2, T^4, T^6\} can be used.
    • Once we define ff, we can represent it as a matrix MB(f)M_{\mathcal{B}}(f) in this basis.
    • To find MB(f)M_{\mathcal{B}}(f), we apply ff to each basis vector and express the results as linear combinations of the basis vectors. This will help in writing out the matrix of ff.
  3. Proof: If WW is a Finite-Dimensional KK-Vector Space and f:WWf: W \to W is a Linear Map with ker(f)=Bild(f)\ker(f) = \operatorname{Bild}(f), then dim(W)\dim(W) is Even.

    • This result can be shown using the Rank-Nullity Theorem, which states dim(W)=rank(f)+nullity(f)\dim(W) = \operatorname{rank}(f) + \operatorname{nullity}(f).
    • Since ker(f)=Bild(f)\ker(f) = \operatorname{Bild}(f), we have rank(f)=nullity(f)\operatorname{rank}(f) = \operatorname{nullity}(f).
    • Let rank(f)=k\operatorname{rank}(f) = k. Then, dim(W)=k+k=2k\dim(W) = k + k = 2k, which is even.
    • This proves that dim(W)\dim(W) must indeed be an even number.

Would you like further details on each part, or do you have specific questions? Here are some related questions to explore:

  1. What are examples of linear maps with ker(f)=Bild(f)\ker(f) = \operatorname{Bild}(f)?
  2. How can we construct matrices for linear maps in polynomial vector spaces?
  3. How does the Rank-Nullity Theorem apply to this type of linear transformation?
  4. What does it mean for a space to have an even dimension?
  5. How do we verify the kernel and image equality for a given transformation?

Tip: When constructing a matrix for a linear map, always consider the action of the map on each basis vector and express the results in terms of the basis. This approach will guide you in writing the matrix representation correctly.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Linear Transformations
Vector Spaces
Kernel and Image of a Linear Map
Rank-Nullity Theorem

Formulas

dim(V) = rank(f) + nullity(f)

Theorems

Rank-Nullity Theorem

Suitable Grade Level

Undergraduate