The task is to define a linear mapping $f: \mathbb{R}^2 \to \mathbb{R}^2$ such that the kernel (null space) of $f$, denoted $\text{Kern}(f)$, is equal to the image (range) of $f$, denoted $\text{Bild}(f)$.

This problem implies we are looking for a linear transformation $f$ where the dimension of the kernel is equal to the dimension of the image. Since the mapping is from $\mathbb{R}^2$ to $\mathbb{R}^2$, this suggests that the transformation has rank 1. Here’s the reasoning and construction:

1. Dimensions and Rank-Nullity Theorem: The rank-nullity theorem states that $\text{dim}(\text{Kern}(f)) + \text{dim}(\text{Bild}(f)) = 2$ If we let the image (rank) have dimension 1, then the kernel (nullity) must also have dimension 1.

2. Constructing $f$: We can define a linear map with a 1-dimensional image by using a matrix of rank 1. For example, let $f(x, y) = (x, 0)$ or, equivalently, $f$ can be represented by the matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ This matrix has rank 1, which means the image of $f$ is a 1-dimensional subspace of $\mathbb{R}^2$ (specifically, the span of $(1, 0)$). The kernel of this transformation is the line of vectors that map to zero, i.e., all vectors of the form $(0, y)$ for any $y \in \mathbb{R}$.

3. Verification:

   • The image of $f$ is $\text{Bild}(f) = \{ (x, 0) \mid x \in \mathbb{R} \}$, which is 1-dimensional.
   • The kernel of $f$ is $\text{Kern}(f) = \{ (0, y) \mid y \in \mathbb{R} \}$, which is also 1-dimensional.

Therefore, $\text{Kern}(f) = \text{Bild}(f)$, as required.

Would you like more details on the construction or properties of linear maps? Here are some related questions:

1. How does the rank-nullity theorem apply to this type of problem?
2. What are other examples of rank-1 matrices that could satisfy this requirement?
3. Why does choosing a rank-2 matrix not satisfy $\text{Kern}(f) = \text{Bild}(f)$?
4. Can this solution be generalized to higher-dimensional spaces?
5. What would change if we considered a different target space, such as $\mathbb{R}^3$ instead of $\mathbb{R}^2$?

Tip: To explore properties of linear maps, try computing the kernel and image by solving $A \mathbf{x} = 0$ and finding the span of columns of $A$.