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.

  • $\begingroup$ 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

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 )

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.