Prove that $\mathbb R^n=E^s\oplus E^u$. I need a hint to solve this question:

Let $x'=Ax$ be a hyperbolic system with index of stability $s$,
  $E^s=\{x\in \mathbb R^n;e^{tA}x\to 0$ when $t\to \infty\}$,
  $E^u=\{x\in \mathbb R^n;e^{tA}x\to 0$ when $t\to -\infty\}$.
Show that:
$E^s$ is a vectorial subspace of dimension $s$ and $\mathbb
 R^n=E^s\oplus E^u$.

 A: You can use the Jordan normal form to solve this problem. Let $T$ be a matrix such that
$$ TAT^{-1}=\left(\begin{matrix}S&0\\0&U\end{matrix}\right) $$
where 
$$ S=\left(\begin{matrix}\lambda_1&a_1&0&\cdots&0&0\\0&\lambda_2&a_2&\cdots&0&0\\\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\0&0&0&0&a_{s-2}&0\\0&0&0&0&\lambda_{s-1}&a_{s-1}\\0&0&0&0&0&\lambda_s\end{matrix}\right), U=\left(\begin{matrix}\lambda_{s+1}&a_{s+1}&0&\cdots&0&0\\0&\lambda_{s+2}&a_{s+2}&\cdots&0&0\\\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\0&0&0&0&a_{n-2}&0\\0&0&0&0&\lambda_{n-1}&a_{n-1}\\0&0&0&0&0&\lambda_n\end{matrix}\right) $$
with $\text{Re}\lambda_j<0$ for $1\le j\le s$, $\text{Re}\lambda_j>0$ for $s+1\le j\le n$ and $a_j=0$ or 1 for $1\le j\le n-1$. So
\begin{eqnarray*}
T^{-1}\left(\begin{matrix}S&0\\0&U\end{matrix}\right)T, e^{At}=T^{-1}\left(\begin{matrix}e^{St}&0\\0&e^{Ut}\end{matrix}\right)T.
\end{eqnarray*}
Now for any $x\in\mathbb{R}^n$, let 
$Tx=(X_1,\cdots,X_s,X_{s+1},\cdots,X_n)^t$, $X_S=(X_1,\cdots,X_s,0,\cdots,0)^t$ and $X_S=(0,\cdots,0,X_{s+1},\cdots,X_n)^t$, $x_S=T^{-1}X_S, x_U=T^{-1}X_U$. It is easy to check that $x=x_S+x_U$. Note
\begin{eqnarray*} 
e^{At}x_S=T^{-1}\left(\begin{matrix}e^{St}&0\\0&e^{Ut}\end{matrix}\right)Tx_S=T^{-1}\left(\begin{matrix}e^{St}&0\\0&e^{Ut}\end{matrix}\right)\binom{X_S}{0}=T^{-1}\binom{e^{St}X_s}{0}\to0\text{ as }t\to\infty.\\
e^{At}x_U=T^{-1}\left(\begin{matrix}e^{St}&0\\0&e^{Ut}\end{matrix}\right)Tx_U=T^{-1}\left(\begin{matrix}e^{St}&0\\0&e^{Ut}\end{matrix}\right)\binom{0}{X_U}=T^{-1}\binom{0}{e^{Ut}X_U}\to0\text{ as }t\to-\infty.
\end{eqnarray*}
Thus $x_S\in E^s$ and $X_U\in E^u$. This implies $\mathbb{R}^n=E^s\oplus E^u$. Clearly the dimension of $E^s$ is s.
