Can a non-linear transformation, specifically a translation, have eigenvectors?

Question

In trying to grasp the basics of linear algebra, I've encountered the following question:

Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the transformation defined by $(x,y) \mapsto (x+1,y+1)$. Does $f$ have any eigenvectors? Explain. What type of transformation is $f$?

I know that $f$ is a translation, which is a non-linear transformation. Does this mean that it has no eigenvectors? Do eigenvectors only apply to linear transformations?

Or can you actually have eigenvectors for non-linear transformations? If so, would all vectors be eigenvectors for this transformation? The transformation causes points to be translated 1 unit right and 1 unit up, but don't vectors stay the same? For example, the vector $\begin{bmatrix}3\\2\end{bmatrix}$ remains $\begin{bmatrix}3\\2\end{bmatrix}$. Before the transformation, it has initial point $(0,0)$ and terminal point $(3,2)$; after the transformation, it has initial point $(1,1)$ and terminal point $(4,3)$.

Hope you'll answer the quoted question, and also let me know if I've misunderstood anything. Thanks!

Answer 1

Yes, eigenvectors and eigenvalues are defined only for linear maps. And that map $f$ maps elements of $\Bbb R^2$ into elements of $\Bbb R^2$. In particular, it maps $(3,2)$ into $(4,3)$. Thinking about it as acting on vectors, by acting on their initial and final points, is misleading.

Answer 2

The question has already been answered in the straightforward way, but I just wanted to point out another perspective. There is a way to view this as a linear map in some sense. It is an affine map, which can be represented as a matrix by embedding it in $\mathbb{R}^3$ as follows. $$ \begin{pmatrix} 1&0&0 \\ 1&1&0 \\ 1&0&1 \end{pmatrix} \begin{pmatrix} 1 \\ x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ x+1 \\ y+1 \end{pmatrix} $$ Thinking about the eigenvectors and eigenvalues of this matrix may be another instructive perspective on the original problem.