Let $x \in \mathbb{R}^{n}$ and $A(t) \in \mathbb{R}^{n\times n}$. If $\dot{x}=A(t)x$ and $\dot{x}=cA(t)x$ with $c>1$ are exponentially stable. Is the convergence rate of $x$ to zero of $\dot{x}=cA(t)x$ faster than that of $\dot{x}=A(t)x$?

This question is initially asked on Mathematics: https://math.stackexchange.com/questions/2881136/is-the-convergence-of-dotx-2atx-faster-than-that-of-dotx-atx

Here is my initial thought:

For linear time-invariant system, the fact $\dot{x}=A(t)x$ is exponentially stable implies $A(t)=A$ is Hurwitz. It is clear that the $i$th eigenvalue $\lambda_{i}(cA)=c\lambda_{i}(A)$, $i,\ldots,n$, thus, the convergence for $\dot{x}=cAx$ with $c>1$ is faster than that of $\dot{x}=Ax$.

For a linear time-varying system. Let us consider the extreme case where $c=0$, $\dot{x}=cA(t)x=0$, which implies it is no longer exponentially stable.

For $c>1$, can we have the same conclusion for exponentially stable linear time-varying systems, i.e., can we conclude that the convergence is faster when $c>1$?

Update 1: For a scalar time-varying system, i.e., $x\in\mathbb{R}$. We can actually prove this conjecture. In fact, from the uniqueness of the equilibrium point ($x=0$ is the only solution that renders $\dot{x}=0$), the solution of $\dot{x}=ca(t)x$ is a monotone function either strictly increasing or strictly decreasing, depending on its initial condition $x(0)$. Thus, for $\dot{x}=ca(t)x$ where $c>1$, the absolute value of the derivative is larger than that of $\dot{x}=a(t)x$, while for both cases the sign of $\dot{x}(t)$ remains unchanged for all $t\ge0$. Thus, $\dot{x}=ca(t)x$ does converge faster to $x=0$ than $\dot{x}=a(t)x$.