In an attempt to find out how to decide whether a function is a contraction mapping or not, I found several sources that state that a mapping $M : \mathbb{R}^m \to \mathbb{R}^n$ is a contraction mapping if the eigenvalues of the Jacobian are (strictly) less than 1.
When rewriting the condition for a contraction mapping (with $q \in [0, 1[$) \begin{align*} d_{\mathbb{R}^n}(M(x), M(y)) & \leq q \;d_{\mathbb{R}^m}(x, y) \\ \frac{d_{\mathbb{R}^n}(M(x), M(y))}{d_{\mathbb{R}^m}(x, y)} & \leq q\,, \end{align*} I get an intuition for the link with the derivative/Jacobian of $M$, but I do not seem to be able to make the link with the eigenvalues. In this particular case, there would not even be eigenvalues, because the Jacobian is not square (right?).
Therefore my question(s) are: Is there any (easy-to-understand) formal explanation for the usefulness of the Jacobian? Why are we looking at the eigenvalues of the Jacobian? Could we use singular values as well?
Thanks in advance
PS: I'm not quite sure about the tags, so feel free to add/remove tags
At a very basic level, the eigenvectors of a matrix are the directions in which the action of the associated transformation is a simple scaling operation. The associated eigenvalues are the scale factors. By linearity, the overall action of the transformation can be described by a superposition of these “eigen-actions.” If we want the transformation to contract in every direction, then all of its eigenvalues must have absolute value less than unity.
The Jacobian of a map at a point is the best linear approximation to the action of the map near that point and so, per the above explanation, for the map to be a contraction, the eigenvalues of the Jacobian must have absolute values less than unity.