Methodology for the Wronskian

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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!
 

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Stephen Tashi
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Stephen Tashi
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First, think about systems of simultaneous linear equations and how they analyzed with matrices. Traditionally, we write the system in the form [itex] M\ x = b [/itex] 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 [itex] M\ x [/itex] of a matrix times a column vector can be viewed as a linear combination of the columns of [itex] M [/itex] with coefficients taken from the entries of [itex] x [/itex].

Writing the jth column of [itex] M_{*.j} [/itex]

[itex] M\ x = x_1 M_{*,1} + x_2 M_{*.2} + ... x_n M_{*,n} [/itex]

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

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

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

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