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Proof of derivative of determinant

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Given n2 functions fij, each differentiable on an interval (a,b), define F(x) = det[fij] for each x in (a,b). Prove that the derivative F'(x) is the sum of the n determinants, F'(x) = [tex]\sum_{i=0}^n det(Ai(x))$.[/tex] where Ai(x) is the matrix obtained by differentiating the functions in the ith row of [fij(x)].


    2. Relevant equations
    To be honest I'm not completely sure what equations would be useful in this proof. I cannot get a good intuition on it.

    I suppose Leibniz formula could be handy: http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants

    Expansion by minors might be useful as well: http://en.wikipedia.org/wiki/Laplace_expansion

    However, that is as far as my intuition will take me.
    3. The attempt at a solution
    This has been more of a sit and think process than a pen to paper one. I understand what I am supposed to do, but I have not the slightest idea what my intuition to this proof should be. I can't see how the determinant of the matrix would look (since I cannot diagonalize it or anything to make the determinant easy to calculate). I'm not very strong in my understanding of the Leibniz formula for determinants. I understand the idea behind minor's a bit better (delete i-row,j-column and take determinant of what is left). If I could get a feel for what the determinant of [fij(x)] would look like I might be able to see why the derivative would look like F'(x) = [tex]\sum_{i=0}^n det(Ai(x))$.[/tex]

    I just need some intuition here, basically, and perhaps a good explanation of the weapons in my arsenal that would help me do this proof.

    -Ian
     
  2. jcsd
  3. Nov 5, 2009 #2

    Office_Shredder

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    Think about writing out the determinant as summing over every possible term [tex] \Pi f_{ij}[/tex] where no pair of fij is from the same row or column. Then think about the product rule
     
  4. Nov 5, 2009 #3
    So you are saying use leibniz and match its behavior to that of the product rule? Is there not a way to do this inductively?
     
  5. Nov 5, 2009 #4

    Office_Shredder

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    I guess you can. If you expand by minors along a row, take the derivative of that, and then use your inductive step to find the derivative of the minors
     
  6. Nov 5, 2009 #5
    Alright. Well I am using Leibniz to informally argue it at the moment. I'll see if that pans out.
     
  7. Nov 5, 2009 #6
    I was able to prove it thanks to the general formula for the product rule of n functions. Thanks!
     
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