Let's break down and address each part of this mathematical problem:

Given Information

Let $K$ be a field, and $V = \langle 1, T^2, T^4, T^6 \rangle \subseteq K[T]$, where $K[T]$ is the ring of polynomials over $K$.

Problem Analysis

1. Defining a Linear Mapping $f$:

   • We are asked to define a linear map $f: V \to V$ such that the kernel of $f$ is equal to the image of $f$, i.e., $\ker(f) = \text{Im}(f)$.
   • To construct such a map, we need to ensure that the dimension of the image and kernel are equal, and that the dimensions of both the kernel and image match. This is often achieved through an endomorphism on $V$ with specific properties.

2. Choosing a Basis $\mathcal{B}$ of $V$ and Determining $\mathcal{B} M_{\mathcal{B}}(f)$:

   • A basis for $V$ can be taken as $\mathcal{B} = \{ 1, T^2, T^4, T^6 \}$, which spans $V$ as given in the problem.
   • We need to determine the matrix $M_{\mathcal{B}}(f)$ of $f$ with respect to this basis $\mathcal{B}$. The entries of this matrix will depend on how $f$ acts on each basis element.

3. Proving a Dimension Condition on $W$:

   • The problem states that if $W$ is a finitely generated $K$-vector space and $f: W \to W$ is a linear mapping with $\ker(f) = \text{Im}(f)$, then the dimension of $W$ must be even.
   • This condition can be approached by considering the Rank-Nullity Theorem and the structure of vector spaces where $\ker(f) = \text{Im}(f)$.

Solution Outline

1. Defining $f$:

   • Define $f$ as a linear transformation with a specific action, such as mapping each basis element $T^{2i}$ to a linear combination of other basis elements in such a way that it satisfies $\ker(f) = \text{Im}(f)$.
   • A possible approach is to define $f$ with a matrix that has rank 2, for instance, with a form that guarantees equal dimensions of the kernel and image.

2. Calculating $M_{\mathcal{B}}(f)$:

   • Once $f$ is defined, express its action on each basis element in terms of the basis $\{1, T^2, T^4, T^6\}$ and construct the corresponding matrix.

3. Proof of the Dimension Condition for $W$:

   • Use the fact that if $\ker(f) = \text{Im}(f)$, then by the Rank-Nullity Theorem, the dimension of $W$ is twice the rank of $f$, which is an even number.

Would you like to go through each step in detail? Here are some related questions:

1. How can we ensure that a linear map has $\ker(f) = \text{Im}(f)$?
2. What are the implications of the Rank-Nullity Theorem in this context?
3. How can we construct the matrix representation of $f$ once the basis is chosen?
4. Why does having $\ker(f) = \text{Im}(f)$ imply that the dimension of $W$ is even?
5. What alternative mappings could satisfy $\ker(f) = \text{Im}(f)$?

Tip: When dealing with linear maps, remember that the dimensions of the image and kernel are interrelated, and specific transformations can give balanced dimensions for both.