Let $A: E \to E$ be a linear isomorphism and $E$ a finite dimension Banach space. Prove that $A$ is hyperbolic if and only if $A$ has no eigenvalue of absolute norm $1$.
This is an exercise of Lan wen's book, Differentiable Dynamical Systems. I did "$A$ hyperbolic so $A$ has no eigenvalue of absolute norm $1$."
For the reciprocal I had an idea, but I don't know how to finish.
Suppose $A$ has no eigenvalue of absolute norm $1$. By the Jordan canonical form Theorem exist a basis $B$ of $E$ such that $A$ can be written in Jordan blocks.
Let $E^1$ be the space generated by the vectors of $B$ that are associated with the eigenvalues with norm $<1$ and $E^2$ the space generated by the vectors of $B$ that are associated with the eigenvalues with norm $> 1$.
It is clear that $E=E^1\oplus E^2$. But I am having a hard time to show that $E^1= E^s$ and $E^2=E^u$.
where $E^s = \{ x \in E \;|\; A^n(x) \to 0, n\to \infty\}$ and $E^u =\{ x \in E \;|\; A^{-n}(x) \to 0, n\to \infty\}$
Thank you very much.