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Homework Help: Symbolic Linear algebra conept

  1. Sep 14, 2009 #1
    Symbolic Linear algebra concept

    1. The problem statement, all variables and given/known data
    Evaluate the sum [tex]\Sigma[/tex]k ( [tex]\epsilon[/tex] ijk [tex]\epsilon[/tex] lmk (which contains three terms) by considering the result for all possible combinations of i,j,l,m, that is a) i=j b) i=l c) i=m d) j=l e)j=m f)l=m g) i[tex]\neq[/tex]l or m h) j[tex]\neq[/tex]l or m

    show that

    [tex]\Sigma[/tex]k ( [tex]\epsilon[/tex] ijk [tex]\epsilon[/tex] lmk = [tex]\delta[/tex] il [tex]\delta[/tex] jm - [tex]\delta[/tex] im [tex]\delta[/tex] jl

    and use this result to prove
    A X (B X C)=(A [tex]\bullet[/tex] C)B-(A [tex]\bullet[/tex] B)C

    2. Relevant equations

    3. The attempt at a solution

    i have been working on this problem for over three hours and have also gone to my professor which was of little help, the work I have done would most likely only serve to confuse. I would be EXTREMELY grateful to anyone who would be able to help me understand this problem. I understand for part a) the permittivity will always equal 0 for the first Levi-Civita symbol making the entire sum zero. For question like be im lost as to how you could know which values would be +1 or -1 or 0 out of the 81 possible combinations. As for the proofs im lost on the first and the second I am having a hard time manipulating the concept A X B = [tex]\Sigma[/tex] j, k [tex]\epsilon[/tex] ijk AjBk . Thank you very much to anyone who can shed the slightest of light.
    yours truly
    Last edited: Sep 15, 2009
  2. jcsd
  3. Sep 15, 2009 #2


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    Hi sleventh! :smile:

    (have a sigma: ∑ and a delta: δ and an epsilon: ε and try using the X2 tag just above the Reply box :wink:)
    (what does permittivity have to do with it? :confused:)

    No, there aren't 81 …

    for example, for b) i = l, put i = l = 1, then there are only nine combinations for j and k, and most of them are zero!

    Try again. :smile:
    (you mean (AxB)i = …)

    Just look at the "matrix" definition of AxB (where ex ey and ez are the basis):
    Code (Text):
    ex  ey  ez
    Ax  Ay  Az
    Bx  By  BZ
  4. Sep 15, 2009 #3
    thank you very much tiny tim, It was silly not to set the matrix into its components, i was only oriented to the summation equation. i still am lost on LaTeX Code: \\Sigma k ( LaTeX Code: \\epsilon ijk LaTeX Code: \\epsilon lmk = LaTeX Code: \\delta il LaTeX Code: \\delta jm - LaTeX Code: \\delta im LaTeX Code: \\delta jl
    thank you very much, you helped clear many things up.
  5. Sep 16, 2009 #4


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    hmm … your LaTeX looks a bit weird …

    the PF version is very simple, it doesn't use \\, and for subscripts just type _{ijk}

    (and use \cdot not \bullet)

    For more details, see http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000" [Broken]

    (but it's easier on the server if you use standard Unicode symbols and the X2 and X2 tags instead :wink:)
    Last edited by a moderator: May 4, 2017
  6. Sep 16, 2009 #5


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    If we have n=3 there are 81 as 3^4=81.
    The fancy way to do this would be to observe
    sum epsilon(ijk)epsilon(lmk)
    is isotropic so can be expressed in the form
    is a basis for the isotropic 4th order 3 spance tensors

    The tedious method (though not that tedious) as you mentioned is to group 81 componets into groups based on there possible values -1,0,1.
  7. Sep 16, 2009 #6
    thank you lurflurf, i have been unaware how to put the epsilon notation into delta form, however in retrospect it doesn't seem to have been to difficult. Because of this i did have to sort the 81 possibilities.
    I am though lost on how to actually prove the sum of the two epsilons to equal the Kronecker delta equation. If i figure this out i will post the solution.
    thanks to everyone who helped
  8. Sep 16, 2009 #7


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    If you take the equation as given, it suffices to show the two sides agree for each of the 81 components. Do you mean how to derive the expression?
    sum epsilon(ijk)epsilon(lmk)=A*delta(ij)delta(lm)+B*delta(il)delta(jm)+C*delta( im)delta(jl)
    consider for example
    sum epsilon(11k)epsilon(22k)=A*delta(11)delta(22)+B*delta(12)delta(12)+C*delta( 12)delta(12)
    sum epsilon(12k)epsilon(12k)=A*delta(12)delta(12)+B*delta(11)delta(22)+C*delta( 12)delta(12)
    likewise for C
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