# Dependence implies Wronskian being zero

From Wikipedia:

If the functions $f_i$ are linearly dependent, then so are the columns of the Wronskian as differentiation is a linear operation, so the Wronskian vanishes. Thus, the Wronskian can be used to show that a set of differentiable functions is linearly independent on an interval by showing that it does not vanish identically.

Here's what I don't understand. If $f_i$ are dependent on some $I\subset \mathbb R$, then there is a point $x_0\in I$ at which there is a non-trivial linear dependence relation (because linear independence means that any relation is trivial for all $x\in I$). Then the Wronskian vanishes at the point $x_0$. The negation of this would be: if the Wronskian does not vanish at any point in $I$, then the $f_i$ are independent on $I$. But that's not what Wikipedia says. (It says if there is a point $x_1\in I$ at which the Wronskian is non-zero, then the $f_i$ are independent.) Am I wrong?

Also, at what point does the proof of the converse fail?

## 1 Answer

If $f_i$ are dependent on some $I\subset\mathbb{R}$, then there is a point $x_0\in I$ at which there is a non-trivial linear dependence relation…

No. Linear dependence of a collection of functions $f_i$ on an interval $I\subset\mathbb{R}$ means that there exists a nontrivial linear combination of these functions that is identically equal to zero on $I$ — at all points, not just at some point. This is the only right way to interpret the definition: the result of a linear combination to establish linear dependence must be the zero element of the same vector space. The vector space here is a set of (differentiable) functions on $I$, whose zero element is the zero function — the function $O(x)$ such that $O(x)=0$ for all $x\in I$.

So for linearly dependent functions the Wronskian vanishes at all points of $I$. The negation of this statement is that for linearly independent functions the Wronskian is non-zero at least for one point $x_0\in I$.

• Thanks. Could you explain why does the converse not hold? I don't see why is this proof incorrect: if the Wronskian is identically zero, then the columns of the Wronskian matrix are dependent (because the bottom row of RREF of the matrix is zero), but then there is a non-trivial relation between the first elements of each column. – user419669 Oct 14 '17 at 4:00
• @user419669: I think your mistake is that you're thinking of the elements of the Wronskian as numbers, but they are not numbers -- they are functions. Roughly speaking. the fact that the Wronskian is zero when evaluated at any point $x_1\in I$ gives us a linear dependence between the numerical columns of $W(x_1)$. But the coefficients of such linear dependence may vary from point to point, and as a result there wouldn't be a common linear dependence among the $f_i$'s as functions. – zipirovich Oct 14 '17 at 4:57
• @user419669: The whole idea of RREF works with numbers (or elements of a field), but not necessarily with functions which do not form a field because we can't always divide by a function. For example, imagine $I=(-2,2)$, and you have $x^2$ in the upper-left corner of the Wronskian. It's a non-zero element as a function, even though it is equal to zero at one of the points of the domain. So the whole structure of the RREF of the numerical matrices $W(0)$ and $W(1)$ can be different because one has a zero and one has a nonzero in the upper-left corner. – zipirovich Oct 14 '17 at 5:01