Let $A$ be some matrix and for any $x \in \mathbb{R}^n$ we have for
$$x(t):=e^{At}x$$ that $\frac{d}{dt}||x(t)||<0$ if $x$ was not zero.
Then I was wondering if we can conclude that $||x(t)|| \rightarrow 0$ exponentially fast, that is:
For $||x|| \le 1$:
$$||x(t)||\le \beta e^{-\alpha t}$$ for some $\beta>0,\alpha>0.$
If anything is unclear, please let me know.
Suppose that $A$ has an eigenvalue $a+b\,i$ with $a$, $b$ reals and $a\ge0$, and let $u+i\,v$ be a corresponding eigenvector. Then $$ x(t)=e^{(a+bi)t}(u+i\,v) $$ is a (complex valued) solution whose norm $$ \|x(t)\|=e^{at}\|u+iv\| $$does not decrease. Hence, all eigenvalues of $A$ have negative real part. This implies that the inequality holds with $\alpha$ smaller that the minimum of the absolute value of the real parts of the eigenvalues of $A$. In fact, the following inequality will hold: $$ \|x(t)\|\le C\,e^{-\alpha t}\,\|x(0)\|,\quad t>0, $$ for some constant $C$.