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Suppose we have a linear ODE of the form $\dfrac{dy}{dt}= Ay$. We know that $y= e^{tA} y(0)$ is a solution to the ODE, where $e^{tA}$ represents the matrix exponential. Suppose we also knew that this is stable in the sense of Lyapunov (i.e for any $\varepsilon >0$, there exists a $\delta > 0$ such that if $\vert y(0) \vert < \delta$, then $\vert y(t) \vert < \varepsilon$ for all time $t > 0$). How does that affect the norm of the matrix $e^{tA}$.

OK, so the actual problem is as follows. If a linear ODE $y'= Ay$ is Lyapunov stable and if $B= G^{-1}AG$ where $G$ is an invertible matrix, is $B$ also Lyapunov stable?

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"OK, so the actual problem is as follows. If a linear ODE $y'= Ay$ is Lyapunov stable and if $B= G^{-1}AG$ where $G$ is an invertible matrix, is $B$ also Lyapunov stable?"

That is, is the system $z'=Bz$ Lyapunov stable around the origin?

Note that the two systems are related by a change of basis. If $y=Gz$, then the system $y'= Ay$ can be written as $(Gz)' = A(Gz)$, that is, $z'=(G^{-1}AG)z = Bz$. And we know that $|y| = |Gz| \leq ||G||. |z|$, where $||G|| := sup_{|x|=1}|G.x|$ is the norm of the matrix $G$ (prove it!).

Since $G$ is invertible, $z=G^{-1}y$, an analogous argument leads to $|z| \leq ||G^{-1}||.|y|$. I think these inequalities can help you solve the problem.

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