I'm trying to understand some theory of linear ODE for control theory and I got stuck.
In the book Mathematical Control Theory, Eduardo Sontag shows that there is a unique solution for the initial value problem for a linear ODE:
$$\dot{\xi}(t) = A(t)\xi(t), \quad \xi(\sigma_0) = x_0 \label{1}\tag{I}$$
defined in the interval $[\sigma_0,\infty) \subseteq I$, where $\xi(t) : I \to R^n$ and $A = (\alpha_{ij})$ is a matrix of measurable and locally essentially bounded functions $\alpha_{ij} : I \to R$. My problem comes when he says the following:
Solving the equation \eqref{1} for $t < \sigma_0$ is equivalent to solving the reversed-time equation $\dot{\xi} = −A\xi$, which is also globally Lipschitz. Thus, solutions are defined on $(-\infty, \infty)$. (p.487).
I can see that the equation $\dot{\xi} = −A\xi$ indeed has a unique solution for an initial value, however I don't see clear why is this last equation equivalent to solving $(I)$ for $t < \sigma_0$.
I suppose there should be a IVP of the form $\dot{\xi} = −A\xi$ that is equivalent to solving $(I)$ for $t < \sigma_0$, so I was thinking in the following IVP:
$$\dot{\xi}(t) = -A(t)\xi(t), \quad \xi(\sigma_0) = x_0 \quad \label{2}\tag{II}$$
On one hand, \eqref{2} has a unique solution $\xi_{II}$, and it fulfills that $\dot{\xi}_{II}(\sigma_0) = -A(\sigma_0)x_0$. Also $\dot{\xi}_{II}(t_0) = -A(t_0)\xi_{II}(t_0)$ if $t_0 > \sigma_0$, however this last thing means $-\dot{\xi}_{II}(t_0) = A(t_0)\xi_{II}(t_0)$, therefore $\xi_{II}(t_0)$ is a point solution for \eqref{1} but with a negative in the left, does that mean that a solution for $t_0 \geq \sigma_0$ in \eqref{2} is the same as "solving" the equation \eqref{1} for a $t_0 < \sigma_0$?
As I said, I'm still confused about the equivalence. I would appreciate any clarifications, this must be very simple however I've been stuck thinking of where does that equivalence come from.