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To remain square, the Wronskian picks up an extra derivative whenever a new expression is added. Finding whether $n$ expressions are linearly independent requires taking $n - 1$ derivatives of each. This is vaguely reminiscent of L'Hospital's Rule, where a varying number of derivatives are required depending on the specific problem, but I'm not sure if these are connected.

I assume there is an analogy here between the columns of the Wronskian and geometric vectors; three geometric vectors can't be linearly independent in a plane. What doesn't seem intuitive is the justification for appending the second derivatives to the Wronskian, and by analogy, expanding the geometric space from a plane to a $3$-space. This feels like adding an extra element to a set of vectors, albeit one derived from the vectors themselves, to try to rescue them from linearly dependence. For example, $\begin{bmatrix}1 \\ 2\end{bmatrix}$ and $\begin{bmatrix}2 \\ 4\end{bmatrix}$ aren't linearly independent, but I can append their respective squared lengths to get $\begin{bmatrix}1 \\ 2 \\ 5\end{bmatrix}$ and $\begin{bmatrix}2 \\ 4 \\ 20\end{bmatrix}$, which are linearly dependent, but this alters the nature of the original inquiry, even if I add a third vector $\begin{bmatrix}10 \\ 20 \\ 500\end{bmatrix}$ into the mix. Why then, when a third expression is added to a set, is it correct to append an extra derivative to each to try to preserve the possibility of linear independence? More generally, why is it appropriate to take derivatives into account in an analysis of linear independence in the first place, rather than just the expressions themselves? The question of whether a set of things are linearly independent seems different than the question of whether a set of things and their $n - 1$ derivatives are linearly independent. I am seeking a way to think about this matter intuitively, not a mathematical proof that a nonzero Wronskian corresponds to linear independence.

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