I have:
$\dot{\textbf{x}}=A{\textbf{x}}$
where A is a nxn matrix.
This equation has solution:
$\textbf{x}(t)=e^{\textbf{A}t}\textbf{x}(0)$
A book i'm reading states that:
$\textbf{x}(t_{2})=e^{\textbf{A}(t_{2}-t_{1})}\textbf{x}(t_{1})$
However,i don't know if the last equation is correct since i wasn't able to deduce it by myself. Can you show me all the passages ? Thanks so much
Your equation is autonomous. This means that with $x(t)$ a solution also any function $x_a(t)=x(t+a)$ is a solution with $x_a(0)=x(a)$.
In your instance, $x(t)=e^{tA}x(0)$ and $x_a(t)=e^{tA}x_a(0)=e^{tA}x(a)$. Now replace $t$ with $t-a$ to find $$ x(t)=x((t-a)+a)=e^{(t-a)A}x(a). $$
In general you will find that for the flow $\phi(t;x_0)$ of an autonomous system you have the (semi-)group property $\phi(t+s; x_0)=\phi(t;\phi(s;x_0))$.