I have a differential equation of a $n\times n$ real matrix $X$: $$\dot{X}=-AX$$ $A$ is also a $n\times n$ real matrix.
Two questions:
What conditions should $A$ satisfy if we want that $X=0$ be stable?
Under conditions of question 1), what conditions should $A$ satisfy if we want $X$ to converge to $0$ faster than $A=I$? and proof?
Update after @skyking 's answer:
I pose an example: $A=\begin{pmatrix}\cos\alpha & -\sin\alpha & x\\ \sin\alpha & \cos\alpha & y\\ 0 &0 & 1\end{pmatrix}$, then we have the eigenvalues: $$\lambda_1=1,\lambda_{2,3}=\cos\alpha\pm\sqrt{\cos^2\alpha-1}$$. The eigenvalues 2 and 3 are complex if $\alpha\ne n\pi$. Now how can we compare its convergence rate and $I$?
Assuming we're talking about complex vector spaces:
The proof is by rewriting it using the Jordan normal form. It would reduce to a system of independent equation in one dimension (unless you have eigenvalues with multiple independent eigenvectors). Anyway you get some independent one dimensional equations on the form
$\dot x = -\lambda x$
These must be stable, and converge faster than $\dot x = -x$.
For the rest of the normal form we have equations on the form
$\dot y = -\lambda y - x$
if $x$ is stable, then $y$ should be stable if $\Re(\lambda)$ is greater than zero and converge faster in case two if $\Re(\lambda)$ is larger than one.