TITLE: How to find an approximate solution of a three dimensional DE system
QUESTION [6 upvotes]: Suppose A is a $3$ by $3$ matrix. How can I find the approximated solution as $t$ goes to $\infty$ of the system $$X' = AX$$
I think it has something to do with the Jordan decomposition of A, but how? If $A = SJS^{-1}$ can I just take the solution to the system $X' = JX$?

REPLY [2 votes]: If X is diagonalizable with real eigenvalues.
$X' = AX\\
X = e^{At}X_0\\
e^{AT} = \sum \frac {A^n t^n}{n!}\\
A = PDP^{-1}\\
A^n = PD^n P^{-1}\\
e^{At} = P(\sum \frac {D^n t^n}{n!})P^{-1}\\
e^{Dt}= \begin{bmatrix} e^{\lambda_1 t}\\&e^{\lambda_2t}\\&&\ddots \end{bmatrix}\\
X = Pe^{Dt}P^{-1}X_0$
If A can be put into Jordan blocks. 
$J = \begin{bmatrix} \lambda & 1\\&\lambda \end{bmatrix}\\
e^{Jt} = \begin{bmatrix} e^{\lambda t} & te^{\lambda t}\\&e^{\lambda t} \end{bmatrix}$
if A has complex eigenvalues pairs ...
$C = \begin {bmatrix} a & b\\ -b & a\end{bmatrix}\\
e^{Ct} = \begin {bmatrix} e^{at}\cos bt & e^{at} \sin bt\\ -e^{at}\sin bt &e^{ at}\cos bt\end{bmatrix}$