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I want to show linear independence in the wronskian implies linear independence between the functions $f_1(x)$, $f_2(x)$, $f_3(x)$. Let $f_1(x)$, $f_2(x)$, $f_3(x)$ be real-valued functions with first and second order derivatives on the interval $(a, b)$. Consider the following:

$$W(x)=\left( \begin{array}{ccc} f_1(x) & f_1'(x) & f_1''(x) \\ f_2(x) & f_2'(x) & f_2''(x) \\ f_3(x) & f_3'(x) & f_3''(x) \end{array} \right)$$

Formally, if $\exists x$ $\in$ $(a, b)$ s.t. the rows of $W(x)$ are linearly independent then $f_1(x)$, $f_2(x)$, $f_3(x)$ are linearly independent.

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  • $\begingroup$ I think the determinant of that matrix is the Wronskian. $\endgroup$ – James S. Cook Sep 20 '12 at 2:55
  • $\begingroup$ Actually, I would usually take the transpose of the matrix you write since it more naturally appears in the linear combinations $c_1f_1+c_2f_2+c_3f_3=0$, $c_1f_1'+c_2f_2'+c_3f_3'=0$ and $c_1f_1''+c_2f_2''+c_3f_3''=0$ written as a single matrix equation. $\endgroup$ – James S. Cook Sep 20 '12 at 3:38
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Suppose $c_1f_1+c_2f_2+c_3f_3=0$ has only the zero solution. Differentiate twice and write the three equations as a single matrix equation. Since the matrix admits only the zero solution the determinant of the coefficient matrix must be nonzero. This coefficient matrix is precisely the Wronskian's matrix. ( I usually call the determinant of the matrix you write the Wronskian )

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  • $\begingroup$ when I say "zero solution" I mean $c_1=c_2=c_3=0$, this is the standard characterization of linear independence $\endgroup$ – James S. Cook Sep 20 '12 at 2:56

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