To test solutions of linear ODEs for lineary independence you can determine the wronski determinant. The theorem says if you have a solution to the linear ODE in the form: $$ \dot{\vec{x}}(t) = A \cdot \vec{x}(t) $$ you can test for linear independence by showing $$\exists t_0 \in I\colon\; \operatorname{det}(W(t_0)) \neq 0 $$ with $W(x)$ being the Wronski-Matrix: $$ W(t) = \begin{pmatrix} \vec{x_1}(t)& \cdots & \vec{x_n}(t) \end{pmatrix} $$ where $x_i$ is the i-th vector of the solution. If you find just one $t_0$ it follows that: $$\forall t \in I\colon\;\operatorname{det}(W(t)) \not\equiv 0$$ (With $\not\equiv 0$ i mean constant not zero). I don't understand why this implication holds true.
Why can I conclude from the fact that the Wronski determinant is at one point equal zero that it must be everywhere? $ det (W (t)) $ is not constant and it depends on $t$.
I'm happy about any answers!
Many Greetings, Sebi2020
This matter may be resolved via Liouville's formula, which affirms that if $X(t)$ an $n \times n$ matrix solution of the differential equation
$\dot X(t) = A(t)X(t) \tag 1$
the determinant of $\det(X(t))$ of $X(t)$ is given by
$\det(X(t)) = \det(X(t_0)) \exp \left ( \displaystyle \int_{t_0}^t \text{trace}(A(s)) \; ds \right ); \tag 2$
since
$\exp \left ( \displaystyle \int_{t_0}^t \text{trace}(A(s)) \; ds \right ) \ne 0, \; \forall t_0, t \in I, \tag 3$
we see that
$\det(X(t)) \ne 0 \Longleftrightarrow \det(X(t_0)) \ne 0. \tag 4$
Now we may conclude with the simple observation that we may take
$W(t) = \det(X(t)) = \det(\vec x_1(t), \vec x_2(t), \ldots, \vec x_n(t)). \tag 5$