Given the system $\bf{x'}=Ax$, where $\bf{A}$$=\begin{bmatrix} -2 &0 &0 \\ 2& 1 & 0\\ 0 &0 &1 \end{bmatrix}$, the solution is $x(t) = \begin{bmatrix} e^{-2t} & 0 &0 \\ \frac{-2}{3} e^{-2t}+\frac{2}{3} e^t & e^t &0 \\ 0& 0 &e^t \end{bmatrix}$. Now, the stable eigenspace $E^s=span\begin{Bmatrix} {\begin{bmatrix} -3\\2 \\ 0 \end{bmatrix}} \end{Bmatrix}$. To find the set of vectors $S:=x\in\mathbb{R}^3:\lim\limits_{t\to\infty} e^{At}x=0$, in my understanding, we need to consider:
$\lim\limits_{t\to\infty}x(t)=\lim\limits_{t\to\infty}e^{At}x = \lim\limits_{t\to\infty}\begin{bmatrix} e^{-2t}x_1\\(-\frac{2}{3}e^{-2t}+\frac{2}{3}e^t)x_1+e^tx_2) \\ e^tx_3 \end{bmatrix}=\vec{0}$. But the only vector satisfying this condition is the vector where $x_1=x_2=x_3= 0$, that is, the zero vector! But how is this possible? If the stable eigenspace $E^s$ has a basis, the set $S$ must contain some vectors besides the zero vector. What am I not doing right?
In the second component you have the term $(\frac23x_1+x_2)e^t$, and so you can take $x_2=-\frac23 x_1$ (the numbers $x_2$ and $x_3$ need not be zero). The third line gives $x_3=0$. So, you get all scalar multiples of the basis for the stable space (as it should).