This task involves defining a linear mapping $f : V \rightarrow V$ on the vector space $V = \langle 1, T^2, T^4, T^6 \rangle \subseteq K[T]$ over a field $K$ such that the kernel of $f$, $\text{Ker}(f)$, is equal to its image, $\text{Im}(f)$. It also requires choosing a basis $\mathcal{B}$ for $V$ and determining the matrix representation of $f$ with respect to $\mathcal{B}$.

Let's proceed step-by-step.

Step 1: Define the Linear Mapping $f$

Since $\text{Ker}(f) = \text{Im}(f)$, the mapping $f$ should have a property where the dimension of $\text{Im}(f)$ is half the dimension of $V$. The vector space $V$ is spanned by $\{1, T^2, T^4, T^6\}$, so $\dim(V) = 4$.

To satisfy $\text{Ker}(f) = \text{Im}(f)$, $\dim(\text{Im}(f))$ should be 2. This suggests that $f$ is a mapping that reduces the dimension by half.

One possible choice is to define $f$ in a way that maps higher powers of $T$ to lower powers, for example: $f(1) = 0, \quad f(T^2) = 1, \quad f(T^4) = T^2, \quad f(T^6) = T^4.$ This mapping satisfies the requirement that the image is spanned by $\{1, T^2\}$, giving $\dim(\text{Im}(f)) = 2$, and thus $\text{Ker}(f) = \text{Im}(f)$.

Step 2: Choose a Basis $\mathcal{B}$ and Determine the Matrix Representation $[f]_{\mathcal{B}}$

A natural choice for the basis $\mathcal{B}$ of $V$ is $\{1, T^2, T^4, T^6\}$.

Given this basis, we can write the matrix representation of $f$ as follows:

1. $f(1) = 0$ gives the first column of the matrix as $\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$.
2. $f(T^2) = 1$ gives the second column as $\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$.
3. $f(T^4) = T^2$ gives the third column as $\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$.
4. $f(T^6) = T^4$ gives the fourth column as $\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$.

Thus, the matrix representation $[f]_{\mathcal{B}}$ is: $[f]_{\mathcal{B}} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$

Summary

1. We defined $f : V \rightarrow V$ such that $f(1) = 0$, $f(T^2) = 1$, $f(T^4) = T^2$, and $f(T^6) = T^4$, ensuring $\text{Ker}(f) = \text{Im}(f)$.
2. The basis $\mathcal{B} = \{1, T^2, T^4, T^6\}$ was chosen.
3. The matrix representation $[f]_{\mathcal{B}}$ is $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$.

Would you like further details on any specific part of this solution?

Here are some related questions to expand your understanding:

1. How can we determine if a linear transformation has $\text{Ker}(f) = \text{Im}(f)$ in general?
2. What other mappings $f$ might satisfy $\text{Ker}(f) = \text{Im}(f)$ for this space $V$?
3. How would the matrix representation change if a different basis was chosen?
4. Can the mapping $f$ be defined in a way that makes it an isomorphism on a subspace of $V$?
5. What is the significance of the dimensions of the kernel and image in determining the properties of a linear transformation?

Tip: To verify that a linear mapping has $\text{Ker}(f) = \text{Im}(f)$, check that the rank of the mapping is half the dimension of the vector space.