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

Question

From section 1.5 of Doering & Lopes¹:

Exercise 10. 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!

---

1. Claus Ivo Doering, Artur Oscar Lopes, Equações Diferenciais Ordinárias, IMPA, 2016.

Answer 1

once you solve the ODE $x'(t)=Ax(t)$ with initial condition $x(t_0)$

you have $x(t)=\exp((t-t_0)A)x(t_0)$

then you compute $Ax(t)$

$\forall t$ $A$ commute with $\exp((t-t_0)A)$

hence you have $Ax(t)=\exp((t-t_0)A)Ax(t_0)= \lambda \exp((t-t_0)A) x(t_0)= \lambda x(t)$