The exercise prompt is in German, and it translates to:

"Define a linear map $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with $\text{Ker}(f) = \text{Im}(f)$."

In this context:

• Ker(f) is the kernel (null space) of the linear map $f$, i.e., the set of vectors in $\mathbb{R}^2$ that are mapped to zero by $f$.
• Im(f) (Bild in German) is the image (range) of the linear map $f$, i.e., the set of all vectors in $\mathbb{R}^2$ that can be reached by applying $f$ to elements of $\mathbb{R}^2$.

To satisfy $\text{Ker}(f) = \text{Im}(f)$, we are looking for a linear map $f$ with specific properties in $\mathbb{R}^2$:

1. The kernel and image must be one-dimensional subspaces (lines through the origin in $\mathbb{R}^2$).
2. This implies that $f$ should map $\mathbb{R}^2$ onto a line, and its kernel should also be a line.

One possible solution is to define a linear transformation matrix $A$ for $f$ in the form: $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ or any scalar multiple of it. This matrix projects vectors onto the $x$-axis:

• Kernel: The $y$-axis (all vectors $(0, y)$).
• Image: The $x$-axis (all vectors $(x, 0)$).

Thus, $\text{Ker}(f) = \text{Im}(f)$ as both are lines, satisfying the condition.

Would you like more details on this solution or any specific part of the reasoning?

Related Questions

1. What other types of linear maps in $\mathbb{R}^2$ satisfy similar properties?
2. How can we determine the kernel and image of a given linear map in general?
3. What does it mean for the kernel and image of a linear map to have equal dimensions?
4. Can we find a matrix $A$ that has a non-trivial kernel in higher dimensions?
5. How does the rank-nullity theorem relate to this problem?

Tip

When defining linear transformations, remember that the matrix representation provides a direct way to analyze the kernel and image by examining row and column spaces.