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!
Best Answer
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.