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Determinant of Matrix Component

  1. Jan 29, 2017 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    Show $$\frac{\partial \det(A)}{\partial A_{pq}} = \frac{1}{2}\epsilon_{pjk}\epsilon_{qmn}A_{jm}A_{kn}$$

    2. Relevant equations
    ##\det(A)=\epsilon_{ijk}A_{1i}A_{2j}A_{3k}##

    3. The attempt at a solution
    $$\frac{\partial \det(A)}{\partial A_{pq}}=\frac{\partial}{\partial A_{pq}}\epsilon_{ijk}A_{1i}A_{2j}A_{3k}\\
    =\epsilon_{ijk}\frac{\partial}{\partial A_{pq}}A_{1i}A_{2j}A_{3k}$$
    but I'm now stuck. I feel like one of the ##A## components on the RHS must go to 1 and the rest would be constant, leaving some sort of ##\epsilon A_{yy} A_{xx}## behind. I'm thinking along the lines of ##\partial A_{ij}/\partial A_{ik} = \delta_{jk}##. Any ideas?
     
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  3. Jan 29, 2017 #2

    PeroK

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    I'd be tempted to move ##A_{pq}## to the top-left of the matrix.
     
  4. Jan 29, 2017 #3

    joshmccraney

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    Are you suggesting take
    $$\frac{\partial \det(A)}{\partial A_{11}}=\frac{\partial}{\partial A_{11}}\epsilon_{ijk}A_{1i}A_{2j}A_{3k}\\
    =\epsilon_{ijk}A_{2j}A_{3k}\frac{\partial}{\partial A_{11}}A_{1i}\\
    =\epsilon_{1jk}A_{2j}A_{3k}$$
     
  5. Jan 29, 2017 #4

    PeroK

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    One way to do it is just to do it for all 9 elements. Not very elegant.

    Otherwise, you need to get a general expression for the terms that have ##A_{pq}## in them. It might be easier to rearrange the matrix, getting ##A_{pq}## to the top left corner.
     
  6. Jan 29, 2017 #5

    joshmccraney

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    Agreed!

    What do you mean "general expression"? Also, you suggest getting ##A_{pq}## to the top left, doesn't this mean making ##A_{pq}=A_{11}##?
     
  7. Jan 29, 2017 #6

    PeroK

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    It means using row and column operations. Alternatively, express the determinant in an equivalent form for expansion via the ##p##th row.
     
  8. Jan 29, 2017 #7

    joshmccraney

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    Since the determinant is a scalar, I'm unsure how to express it for the ##p##th row. Could you elaborate?
     
  9. Jan 29, 2017 #8

    PeroK

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    You can evaluate a determinant using any row as your starting row.

    But, in fact, if you permute the rows to get the ##p##th row at the top, with row ##p+1## second and row ##p+2## third (where these are numbers modulo 3), I think it comes out easily enough. Then permute the columns to get the colums in order ##q, q+1, q+2##.

    Then, if you do the same trick with the Levi-Civita, I think the equation should drop out for general ##p, q##.
     
  10. Jan 29, 2017 #9

    PeroK

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    PS by the same trick, I mean that:

    ##\epsilon_{pjk} = 1## when ##j = p+1## and ##k = p+2##
    ##\epsilon_{pjk} = -1## when ##j = p+2## and ##k = p+1##

    And is ##0## otherwise.
     
  11. Jan 31, 2017 #10

    joshmccraney

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    I agree.

    I don't think I'm understanding what you're saying. Could you illustrate, perhaps with a different problem? I just don't know the technique you suggest.

    I agree.
     
  12. Jan 31, 2017 #11

    PeroK

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    Suppose we take the case ##p, q = 1, 1##:

    ##det(A) = A_{11}(A_{22}A_{33} - A_{23}A_{32}) + \dots##

    Where the rest of the terms do not involve ##A_{11}##, hence you can calculate the required partial derivative.

    You can also check fairly easily that:

    ##\frac{1}{2}\epsilon_{1jk}\epsilon_{1mn}A_{jm}A_{kn} = A_{22}A_{33} - A_{23}A_{32}##

    Now, by either two row operations or no row operations and two column operations or no column operations, you permute ##A_{pq}## to the top left of your determinant. You need to check this for ##p, q = 1, 2, 3##. And, this gives

    ##det(A) = A_{pq}(A_{p+1, q+1}A_{p+2, q+2} - A_{p+1, q+2}A_{p+2, q+1}) + \dots##

    Where we are using modulo 3 here. And that's just about it. All you need to do is manipulate the expression involving the Levi_Civita coefficients using a similar modulo 3 approach.
     
  13. Feb 1, 2017 #12

    joshmccraney

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