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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 $t \in \mathbb{R}$.
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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!



$V_{\lambda}$ is invariant under $A$
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.


