# Definition of a hyperbolic map

In the field of dynamical systems, a hyperbolic map $T: \mathbb{R}^n\rightarrow \mathbb{R}^n$ can be defined in a few equivalent ways. One of them is: T is hyperbolic iff there exists eigenvalues of T with modulus not equal to 1, and there are eigenvalues $\lambda$ such that $|\lambda_1|>1$ and $|\lambda_2|<1$. Furthermore, mathworld wolfram suggests that this means that we can split such a map into a direct sum of generalized eigenspaces.

But the reals are not algebraically closed. This means that any map with a rotational component could possibly be hyperbolic as long as it has a direction in which in expands and another in which it contracts, yet such a map doesn't decompose into generalized eigenspaces.

Should we not be working over the complex field instead for this theory to work out properly?

• say a map on $\mathbb{R}^4$. One expansion direction, one contraction direction, and two involved in rotation – Zhanfeng Lim Apr 16 '18 at 18:08
• The proper way (and common way in good books in dynamics) is to write $\mathbb C^n$ as a direct sum of generalized eigenspaces and then intersecting with $\mathbb R^n$ (think of it as $\mathbb C^n=\mathbb R^n\oplus\mathbb R^n$). Since the map is real, the intersection goes to itself and thus is invariant. – John B Apr 16 '18 at 18:58

Certainly the standard definition is, that a linear map $T\colon \mathbb{R}^n\to \mathbb{R}^n$ is called hyperbolic if $T$ is invertible (sometimes this is omitted) and has no eigenvalue of absolute value $1$. For the decomposition into $A$-invariant subspaces over the real numbers see Theorem 1.1 here. The statement of the Theorem does not refer to (generalised) eigenspaces, and is correct. The remark afterwards with the eigenspaces is not correct.
• Actually it doesn't, the eigenvalues may not be real. It is a quite common mistake by people outside dynamics. The soution is simple: you only need to "intersect" the root spaces with $\mathbb R^n$. By the way, the book that you mention makes the same mistake (although their theorem is correct, as I note above). – John B Apr 16 '18 at 18:56
• @JohnB The Theorem does not speak of generalised eigenspaces, just of $A$-invariant spaces. This should be no problem, right? – Dietrich Burde Apr 16 '18 at 18:58