In this Wikipedia entry on hyperbolic sets of dynamical systems, on the examples section, it was asserted that "more generally, a periodic orbit of $f$ with period $n$ is hyperbolic if and only if $Df^n$ at any point of the orbit has no eigenvalue with absolute value 1, and it is enough to check this condition at a single point of the orbit." I am confused of why it is sufficient to only check at one point.
Asked
Active
Viewed 21 times
0
-
1Because the Jacobians $Df^n$ at the orbit points are conjugate to each other, as they are cyclic permutations of the product of the single Jacobians $Df$ at the orbit points. – Lutz Lehmann Apr 16 '19 at 20:38
-
Thanks. I see it now. – MyCindy2012 Apr 16 '19 at 20:50