Why does the Wronskian indicate linear independence when W ≠ 0?

[MathewsMD]

I was just curious and had a question: why does the Wronskian indicate linear independence if $W ≠ 0$ but is linearly dependent if $W = 0$? Is there a proof to help understand the exact operations of the Wronskian and why it conveys these properties based on these results alone? Thank you!

 

[Stephen Tashi]

According to http://en.wikipedia.org/wiki/Wronskian

A common misconception is that W = 0 everywhere implies linear dependence.

 

[Stephen Tashi]

First, think about systems of simultaneous linear equations and how they analyzed with matrices. Traditionally, we write the system in the form $M\ x = b$ with the left hand side being the product of a matrix multiplied on the right by a column vector (as opposed to a matrix multiplied on the left by a row vector). The product $M\ x$ of a matrix times a column vector can be viewed as a linear combination of the columns of $M$ with coefficients taken from the entries of $x$.

Writing the jth column of $M_{*.j}$

$M\ x = x_1 M_{*,1} + x_2 M_{*.2} + ... x_n M_{*,n}$

If we have more or fewer unknowns than equations, the matrix $M$ isn't square, so we can't do an analysis by taking its determinant. When $M$ is a square matrix and its derminant is non-zero, we can find a unique solution for variables $x$. In particular , we can find a unique solution for the system $M\ x = b$ when $b$ is the column vector of zeroes. The unique solution to $M\ x = 0$ would be $x_j = 0$ for all $j$..

If the column vectors of $M$ were dependent then solution for $M\ x = 0$ would not be unique. For example if $M_{*,1} = \sum_{j=2}^n a_j M_{*,j}$ with at least one of the $a_j$ nonzero then the values $x_1 = 1$ and $x_j = -a_j$ for $j > 1$ would be a nonzero solution to $M\ x = 0$.

By analogy, the Wronskian $W$ is the derminant of a square matrix of functions $M$ . A column vector of $M$ gives a function and it's successive derivatives. If $W$ is nonzero then $M\ x = 0$ has the unique solution $x= 0$ , so the column vectors of $M$ are independent.

That's just "by analogy". There would be lots of technicalities to consider if we want to prove anything about solutions to a differential equation. At least the analogy reminds us that a given column has entries all relate to the same function ( and a given row has entries that all relate to the same order of differentiation).