Conditions on the eigenvalues of $A$ for $\lim_\limits{t \to \infty} x(t) = 0$ to hold

Question

Let the $n \times n$ matrix $A$ have real, distinct eigenvalues. Find conditions on the eigenvalues that are necessary and sufficient for $\lim_\limits{t \to \infty} x(t) = 0$ where $x(t)$ is any solution of $\dot x = Ax$.

Answer 1

The necessary and sufficient conditions is $\lambda_i<0$, where the $\lambda_i$'s are the eigenvalues of $A$. Indeed, since $A$ has distinct eigenvalues, it is diagonalisable, i.e. there exists an invertible matrix $P$ such that $A=PDP^{-1}$, where $D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$. Hence, your ODE $x'=Ax$ is equivalent to $$ y'=Dy,\quad\textrm{where}\quad y=Px. $$ Therefore, $y_i'=\lambda_iy_i$ for $1\leq i\leq n$ and thus $y_i(t)=y_i(0)e^{\lambda_i t}$ and since $\lambda_i<0$ we deduce $y_i(t)\to0$ as $t\to\infty$. Finally, using $x= P^{-1}y$ we get $x(t)\to 0$ when $t\to \infty$.

Bonus: Actually we do not need that $A$ has distinct eigenvalues, the minimal assumptions are: $A$ is diagonalisable and its eigenvalues satisfy $Re(\lambda_i)<0$.