# Linear independence and the Wronskian

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.

• I think the determinant of that matrix is the Wronskian. Sep 20, 2012 at 2:55
• 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. Sep 20, 2012 at 3:38

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 )
• when I say "zero solution" I mean $c_1=c_2=c_3=0$, this is the standard characterization of linear independence Sep 20, 2012 at 2:56