Let $M$ be a Riemannian manifold, $U \subset M$ an open subset and $f : U \to M$ a $C^1$ diffeomorphism. I'm trying to show that a periodic orbit with period $n$ is a hyperbolic set iff $Df^n$ at any point of the orbit has no eigenvalue with absolute value $1$.
Let $\Lambda = \{p, f(p), ..., f^{n-1}(p)\}$ be a periodic orbit with period $n$. Suppose that $\Lambda$ is a hyperbolic set. Then there are constants $\lambda \in (0, 1), c > 0$ and for every $x \in \Lambda$, $T_x M = E^s_x \oplus E^u_x$ such that $$ Df_x(E^s_x) = E^s_{f(x)}, Df_x(E^u_x) = E^u_{f(x)} \\ \| Df^k_x(v) \| \leq c \lambda^k \| v \|, \text{for every} \; v \in E^s_x, k \geq 0 \\ \| Df^{-k}_x(v) \| \leq c \lambda^k \| v \|, \text{for every} \; v \in E^u_x, k \geq 0.$$
Let $x = f^i(p) \in \Lambda,\; 0 \le i \le n-1$. Suppose that $Df^n_x$ has an eigenvalue of absolute value $1$, i.e there are $\mu$ with $|\mu| = 1$ and $v \neq 0, v = v^s + v^u$, $v^s \in E^s_x, v^u \in E^u_x$, such that $Df^n_x(v) = \mu v$.
I can't obtain a contradiction. Can someone give me a hint?