How does one prove the following?
Let $A \in \mathbb{R}^{n\times n}$ be a piece-wise continuous matrix, and let $Z(t)\in \mathbb{R}^{n\times n}$ be such that \begin{equation}\tag{1}\dot{Z}(t) = A(t)Z(t).\end{equation} Show that if $Z(t_0)$ is invertible then Z(t) is also invertible for each $t\in \mathbb{R}$.
My intuition so far is that, since (1) is nothing but a $n^2$ scalar ODE system (for the entries of $Z$), it admits a unique global solution by the standard existence theorem. I don't get why, though, this guarantees that $Z(t)$ stays invertible.
Thanks!
It is a well-known in Linear ODE theory equation for fundamental matrix of linear ODE system (please see https://en.wikipedia.org/wiki/Fundamental_matrix_(linear_differential_equation)) The prove of its non-singularity is also can be found in any ODE book.
The schematic proof:
Determinant of the fundamental matrix is a Wronski determinant $W(t) = det(Z(t))$. One can show that it satisfies simple one dimentional linear ODE
$W'(t) = tr(A(t))\cdot W(t)$
which can be solved directly: $W(t) = W(t_0)\cdot \exp\left(\int_{t_0}^t tr(A(s))ds\right)$. Therefore, $W(t)$ is not equals to zero at any $t$ as long it takes place at $t=t_0$.