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Let $A \in M(n)$, let $\lambda \in \mathbb{R}$, let $V_{\lambda} := \ker(\lambda I - A)$ and let $x:\mathbb{R} \to \mathbb{R}^{n}$ be a solution of $\dot x= Ax$ such that $x(t_0) \in V_{\lambda}$ for some $t_0 \in \mathbb{R}$, then $x(t) \in V_{\lambda}$ for every real $t \in \mathbb{R}$.


Since $x(t_0) \in V_{\lambda}$, then $\dot x(t_0)= \lambda x(t_0)$, given that $\dot x(t_0) = Ax(t_0)$. But I don't know how to go on!

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2 Answers 2

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Define $y(t) = \exp(\lambda(t - t_0)) x(t_0)$. Verify that $y' = Ay$ and that $y(t) \in V_\lambda, \forall t \in \mathbb R$. Uniqueness of solutions gives $y = x$ and, therefore, the desired result.

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    $\begingroup$ Thank you for taking the time to contribute. Since the level of this question is inappropriate for MO, it is better not to answer it here, but rather to refer the questioner to MSE and answer there. \\ TeX note: please use $\mathbb R$ \mathbb R, not $I\!R$ I\!R. I have edited accordingly. $\endgroup$
    – LSpice
    Apr 26, 2022 at 22:46
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Derivative of $h(t)=\|\lambda x(t) - A x(t)\|^2$ is $h'(t)=2(\lambda x(t) - A x(t))^T(\lambda A x(t) - A^2 x(t))$ and $h'(t)\le 2 h(t) \|A\|_{op}$. Then $h(t)$ is identically 0 by Gronwall's inequality and $h(0)=0$.

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