# Matrix exponential and also Stability of Linear ODE

Suppose we have a straight ODE of the kind $\dfrac{dy}{dt}= Ay$. We understand that $y= e^{tA} y(0)$ is a remedy to the ODE, where $e^{tA}$ stands for the matrix rapid. Intend we additionally recognized that this is secure in the feeling of Lyapunov (i.e for any kind of $\varepsilon >0$, there exists a $\delta > 0$ such that if $\vert y(0) \vert < \delta$, after that $\vert y(t) \vert < \varepsilon$ for perpetuity $t > 0$). Just how does that influence the standard of the matrix $e^{tA}$.

OK, so the real trouble is as adheres to. If a straight ODE $y'= Ay$ is Lyapunov secure and also if $B= G^{-1}AG$ where $G$ is an invertible matrix, is $B$ additionally Lyapunov secure?

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2019-05-18 22:54:27
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" OK, so the real trouble is as adheres to. If a straight ODE $y'= Ay$ is Lyapunov secure and also if $B= G^{-1}AG$ where $G$ is an invertible matrix, is $B$ additionally Lyapunov secure?"
That is, is the system $z'=Bz$ Lyapunov secure around the beginning?
Keep in mind that both systems are connected by an adjustment of basis. If $y=Gz$, after that the system $y'= Ay$ can be created as $(Gz)' = A(Gz)$, that is, $z'=(G^{-1}AG)z = Bz$. And also we understand that $|y| = |Gz| \leq ||G||. |z|$, where $||G|| := sup_{|x|=1}|G.x|$ is the standard of the matrix $G$ (confirm it!).
Given that $G$ is invertible, $z=G^{-1}y$, a similar argument brings about $|z| \leq ||G^{-1}||.|y|$. I assume these inequalities can aid you address the trouble.