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Wronskian and linear independence

  1. Sep 4, 2012 #1

    I understand that if we have three functions f, g, and h, they are linearly independent <=> the only c1, c2, and c3 that satisfy (c1)f+(c2)g+(c3)h=0 are c1=c2=c3=0.

    In order to solve for these c1, c2, and c3, we want three equations in the three unknowns. To do this we can differentiate f, g, and h twice and construct the Wronskian. Since this is a square matrix, if the det(W =/= 0, then we know that this system is nonsingular, consistent, and the solution is unique. Furthermore, since its homogeneous we know that unique solution must be c1=c2=c3=0. So if this is the result, we know f,g, and h are linearly independent. But that also means that f', g' and h' are linearly independent, and f'', g'', and h'' are linearly independent, right?

    I guess my confusion is, what if there are functions f,g, and h such that f, g, and h are linearly independent but say, f'', g'' and h'' are linearly dependent? Wouldn't this mean if we construct the Wronskian it will end up inconsistent even though f, h, and h are linearly independent? Is it even possible for that to happen?

    Sorry if the question is confusing.
  2. jcsd
  3. Sep 4, 2012 #2


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    The derivative of ##c_1 f + c_2 g + c_3 h## is just ## c_1 f' + c_2 g' + c_3 h'##, because the c's are constants.

    So the functions are linearly dependent if and only if the derivatives are linearly dependent.
  4. Sep 4, 2012 #3
    ok I want to make sure I understand. Is the reasoning something like this.

    let's assume f' and g' are linearly dependent.

    this means f' = (c)g' for some constant c

    So then we can integrate both sides

    ∫f' = ∫(c)g'

    ∫f' = c∫g'

    f = (c)g

    Which means f and g have to be linearly dependent as well.

    So pretty much if functions are differentiable and linearly dependent, their derivatives are linearly dependent also? And if functions are integrable and linearly dependent, their antiderivatives are linearly dependent also?
  5. Sep 4, 2012 #4


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    Well, this doesn't work for integration, because you forgot about the arbitrary constants and they might mess up the linear dependency.

    But you have got the general idea about what's going on.
  6. Sep 4, 2012 #5
    ok, thank you very much. this clears up the confusion I had!
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