Prove that $\|{e^{At}x_o}\| \geq e^{-\lambda t}\|{x_o}\|$

Prove that $\|{e^{At}x_o}\| \geq e^{-\lambda t}\|{x_o}\|$, for some $\lambda \gt 0$, A is $n\times n$ matrix and $x_o$ is a $n \times 1$ vector.

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I'm guessing you want this proved forall $t \geq 0$, $x_0 \in \mathbb{R}^n$ (or maybe $x_0 \in \mathbb{C}^n$?), but you should really state the explicitly.. –  fgp Oct 22 '12 at 17:14

Write $\|x_0\|_2=\|e^{-At}e^{At}x_0\|_2\leq \|e^{-At}\|_2\|e^{At}x_0\|_2$.
Now, $\|e^{-At}\|_2\leq e^{\|At\|_2}=e^{\lambda t}$, where $\lambda:=\|A\|_2$. Notice that if $A$ is nonzero, then $\lambda>0$.
It's not clear what norm you want here, but the same proof should work for any consistant norm, that is: $\|Ax\|\leq \|A\|\|x\|$
thanks Alex and fgp. I ended up doing the same thing but was initially thrown off course since I didn't realize that $||e^{-At}||\leq e^{||At||}$ –  Sanket_Diwale Oct 22 '12 at 18:01