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I Solve the matrix-component eq

  1. Aug 27, 2016 #1

    joshmccraney

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    Solve $$A_{ij} = c B_{ij}+B_{kk} \delta_{ij}$$ for ##B## where ##c## is a known scalar and ##i,j,k## are indices and range either ##1,2,3## and ##\delta_{ij}## which is the Kronecker Delta..

    I've thought to write this into a matrix but I'm unsure what to do with the ##B_{kk}##. Any help or guidance is greatly appreciated.

    (This is from a book, so not exactly homework Thanks!

    Josh
     
  2. jcsd
  3. Aug 27, 2016 #2
    One of the very useful techniques to use when solving such matrix equations is to take the trace of the equation and then substitute the result back in. I am assuming the Einstein summation convention in being applied so that ##B_{kk}## actually means ##B_{11}+B_{22}+B_{33}##. Taking the trace yields ##A_{ii}=(c+3)B_{kk}##, so you just substitute ##B_{kk}## back in and you are done.

    This technique is very useful, so hopefully you will remember it to apply to other problems.
     
  4. Aug 27, 2016 #3

    joshmccraney

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    Thanks for replying Lucas! So are you saying to rewrite the equation as $$A_{ij} = c B_{ij} + \frac{c}{c+3} A_{kk} \delta_{ij} \implies \\ B_{ij} = \frac{1}{c}A_{ij} - \frac{1}{c+3} A_{kk}\delta_{ij}$$ This totally makes sense! So that I understand, you use this trick whenever you have ##X_{ii}##, right? Also, do you have any other fancy tricks? :)
     
  5. Aug 27, 2016 #4
    In the first line it is ##\frac{1}{c+3}## instead of what you wrote. Apart form that it is correct. Also I just realised that ##c+3## must be different than ##0## for this to work, otherwise you find ##A## to be traceless.

    Not necessarely, although that may be a strong indicator to use it. Use it whenever you want a way to reduce the degrees of freedom of the equation. When taking the trace you reduce the freedom from ##n^2## to ##1## for ##n## by ##n## matrices.

    In general a matrix will have ##n^2## components, and it can always be separated into a symmetric and antisymmetric part. The symmetric has ##n(n+1)/2## independent components while an antisymmetric part has ##n(n-1)/2## independent components. The symmetric part can be further separated into traceless part and trace. So taking the trace is analogous to computing a component of a matrix equation. You can also symmetrize or antisymmetrize to reduce the number of equations.

    Another very useful technique is diagonlising one of the matrices. This can help simplify a particular problem.
     
  6. Aug 27, 2016 #5

    joshmccraney

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    Thanks for catching the algebra mistake, and thanks for all your help!!
     
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