What does subspace A-matrix invariance tells me in terms of A Jordan canonical form. I am asked to show that the semi group $(e^{tA})_{t\geq0}$ for $A \in M_n(\mathbb{C})$ is hyperbolic i.e. there exists direct decomposition $\mathbb(C)^n=X_s \oplus  X_u$ in to A-invariant subspaces $X_s$ and $X_u$ and constants $M\geq 1$, $\epsilon>0$ such that
\begin{align}
||e^{tA}x|| &\leq M e^{-\epsilon t}||x|| \qquad for all x\in X_s, t\geq0 \\
||e^{tA}y|| &\geq \frac{1}{M}e^{\epsilon t}||y|| \qquad for all x\in X_s, t\geq0 
\end{align}
only and only if matrix $e^{tA}$ has no eigenvalues of modulus 1.
I know more or less how to approach it. I should prove it by contradiction, I should claim we can change basis such that A is in Jordan canonical form and decompose A into diagonal and the rest then play with it to show inequality does not hold. However I can't see when and how should I use that the $X$-spaces are A invariant.
 A: I would go about proving that if $e^{tA}$ has no eigenvalues of modulus $1$, then we can find such a decomposition (the $\Leftarrow$ direction), then showing that if we have such a decomposition, $e^{tA}$ has now eigenvalues of modulus $1$ (the $\Rightarrow$ direction).
What you generally describe is a reasonable way to handle the $\Leftarrow$ direction of the proof.
For the $\Rightarrow$ direction, we need to use the fact that $X_s$ and $X_u$ are invariant subspaces. In particular, the fact that these are invariant subspaces tells us that, relative to a basis of $\Bbb C^n$ for which $v_1,\dots,v_k$ spans $X_s$ and $v_{k+1} \dots, v_n$ spans $X_u$, we have
$$
e^{At} = \pmatrix{M_1(t) & 0\\0 & M_2(t)},
$$
for some matrix-valued fucntions functions $M_1(t)$, $M_2(t)$.  We note that $M_1(0) = I_k$ and $M_2(0) = I_{n-k}$, and that
$$
\pmatrix{M_1'(t) & 0\\0 &M_2'(t)} = \frac {d}{dt} e^{At} = Ae^{At} = \pmatrix{A_{11} M_1(t) & A_{12} M_{1}(t)\\
A_{21} M_2(t) & A_{22} M_2(t)}.
$$
From the diagonal entries, we see that $M_1'(t) = A_{11} M_1(t)$ and $M_2(t) = A_{22}M_2(t)$. It follows that $M_1(t) = e^{A_{11}t}$ and $M_2(t) = e^{A_{22}t}$.  Because the $M_i(t)$ are invertible for all $t$, we can also deduce that $A_{12},A_{21}$ are zero.  That is, we have
$$
A = \pmatrix{A_{11} & 0\\ 0 & A_{22}}, \quad e^{At} = \pmatrix{e^{A_{11}t} & 0\\ 0 & e^{A_{22}t}}.
$$
Now, from the properties that define $X_s$ and $X_u$, conclude that neither $A_{11}$ nor $A_{22}$ have eigenvalues of magnitude $1$.

Alternatively, you could also deduce that $A$ has the desired block form using the fact that
$$
A = \frac{d}{dt} \left. e^{At}\right|_{t = 0} = \pmatrix{M_1'(0) & 0\\0 & M_2'(0)}.
$$
