# Finite family of analytic functions linearly dependent if and only if Wronskian is $~0~$

I know that given two analytic functions on some domain $$D$$ of the complex plane, then their Wronskian determinant being $$0$$ is equivalent to them being linearly dependent. I would like to generalize this to a finite family of analytic functions. Naturally, one should use induction. However, the proof of $$n =2$$ is horrifically computational (for me, that is). Is there any clever way to avoid the mess in the induction process? Here are some notes that contain the proof of $$n =2$$ in case one would like to look at it.

Thank you very much in advance. I am much obliged.

Let $M$ be the matrix $\pmatrix{f_1 & \ldots & f_2\cr \ldots & \ldots & \ldots\cr f_1^{(n-1)} & \ldots & f_n^{(n-1)}\cr}$ whose determinant is the Wronskian.
Hint: the existence of a nonzero vector in the null space of the square matrix $M$ is equivalent to $\det M = 0$.
• Indeed. But if $\det M = 0$, what can I say about the equation $Mx = 0$? This might sound trivial, but I the curriculum in my university is quite qeird and we don't have a linear algebra course, meaning that all the linear algebra I know is in the context of multivariable calculus. – user44069 Feb 26 '13 at 2:21
• Here is the basic result. Let $M$ be an $n \times n$ matrix. The following are equivalent: (i) The only solution of $M x = 0$ is $x = 0$. (ii) For all vectors $b$, the equation $M x = b$ has a unique solution. (iii) $M$ has an inverse. (iv) $\det M \ne 0$. – Robert Israel Feb 26 '13 at 21:19