I am reading "Hyperbolic Dynamical Systems" by Vitor Araújo and Marcelo Viana. And when talking about the general solution of a system of linear ode $X' = AX, X(0)=v$, where $A$ is a constant $n\times n$ real matrix, they say:
The linear flow is called hyperbolic if $A$ has no eigenvalues on the imaginary axis. Then the exponential matrix $e^A$ has no eigenvalues with norm $1$.
I didn't figure out this implication. Any help is appreciated.
This is because the exponential map takes the spectrum of the linear map $A$ and sends it onto the spectrum of the linear map $e^A$. That is to say, if $\lambda$ is an eigenvalue of $A$, then $e^\lambda$ is an eigenvalue of $e^A$, and all eigenvalues of $e^A$ can be obtained this way. Thus if $e^\lambda$ were to have modulus $1$, then $\lambda$ would be forced to be a purely complex number.
Sketch of Proof: A proof of this uses Jordan blocks and the fact that for commuting matrices $D,N$ the matrix exponential is multiplicative:
$$e^{D+N}=e^D e^N.$$
Any Jordan block can be written as $J=\lambda I +N$, where $I$ is identity and $N$ is a nilpotent matrix. In particular $e^{\lambda I}= e^\lambda I$ and $e^N$ consists of finitely many terms and it has $1$'s on the diagonal.
For further details see Palis & de Melo's book Geometric Theory of Dynamical Systems, p.45, Prop.2.7.
