Given a system of first-order differential equations that is written as $$ \frac{d}{dt} \vec{x}(t) = A(t) \vec{x}(t) $$
One usually looks at the related matrix differential equation $$ \frac{d}{dt} \Psi(t) = A(t) \Psi(t) $$
The difference being that $\vec{x}(t)$ is a column vector and $\Psi(t)$ is a square matrix. According to Wikipedia, the matrix solution is required to be invertible for all $t$. Suppose we start with linearly independent columns for $\Psi(0)$, how is it guaranteed that $\Psi(t)$ will remain linearly independent for all other $t$? Is there a short proof for it?
If the columns of $\Psi(0)$ are linearly independent, then $\det(\Psi(0))\neq 0$. It follows from Liouville's formula $$ \det(\Psi(t))=\det(\Psi(0))\exp\left(\int_0^t\operatorname{tr}A(s)\,ds\right) $$ that $\det(\Psi(t))\neq 0$ too, hence the columns of $\Psi(t)$ are linearly independent.