What is the connection between the Wronskian and linear independence?

[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).