July 20, 2012

Determinant of the Wronskian

The Wronksian of n functions f_1, \dots, f_n is the matrix determinant \begin{vmatrix}f_1&\dots&f_n\\\dots&\dots&\dots\\f_1^{(n-1)}&\dots&f_n^{(n-1)}}\end{vmatrix}. Its derivative is the matrix determinant \begin{vmatrix}f_1&\dots&f_n\\\dots&\dots&\dots\\f_1^{(n)}&\dots&f_n^{(n)}\end{vmatrix} (that is, the previous matrix with a different bottom row). It’s an interesting exercise to prove this, so let’s do that.

We proceed by our old friend, induction. For n=1 (or 0), the case is obvious. Let it be true through n-1. Expand by the bottom row:
\begin{vmatrix}f_1&\dots&f_n\\\dots&\dots&\dots\\f_1^{(n-1)}&\dots&f_n^{(n-1)}\end{vmatrix} = f_1^{(n-1)}\begin{vmatrix}f_2&\dots&f_n\\\dots&\dots&\dots\\f_2^{(n-2)}&\dots&f_n^{(n-2)}\end{vmatrix} + \dots + f_n^{(n-1)}\begin{vmatrix}f_1&\dots&f_{n-1}\\\dots&\dots&\dots\\f_1^{(n-2)}&\dots&f_{n-1}^{(n-2)}\end{vmatrix}
We take the derivative, applying our induction assumption, obtaining \begin{vmatrix}f_1&\dots&f_n\\\dots&\dots&\dots\\f_1^{(n-2)}&\dots&f_n^{(n-2)}\\f_1^{(n)}&\dots&f_n^{(n)}\end{vmatrix}+\left( f_1^{(n-1)}\begin{vmatrix}f_2&\dots&f_n\\\dots&\dots&\dots\\f_2^{(n-1)}&\dots&f_n^{(n-1)}\end{vmatrix} + \dots + f_n^{(n-1)}\begin{vmatrix}f_1&\dots&f_{n-1}\\\dots&\dots&\dots\\f_1^{(n-1)}&\dots&f_{n-1}^{(n-1)}\end{vmatrix}\right)
But the bracketed part is just \begin{vmatrix}f_1&\dots&f_{n-1}\\\dots&\dots&\dots\\f_1^{(n-1)}&\dots&f_{n}^{(n-1)}\\f_1^{(n-1)}&\dots&f_n^{(n-1)}\end{vmatrix}, which is zero as a matrix with repeated rows is singular. We are done.

Christopher Midgley : 20 Jul 2012 11:39 |  Tags:  |  Comments (1) |  Report a problem

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  1. Nick

    Nice proof! Does this mean you’re going to specialize in analysis and differential equations next year? Or is it the (linear) algebra in this proof that appeals to you most?

    22 Jul 2012, 11:13

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