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A seemingly simple linear algebra equation that eludes me

  1. Nov 14, 2015 #1
    It is from some famous publications. But I seem can get it from rigorous proof after many hours and different methods of trying and Googling.

    If we have g as the eigenvector of a symmetric matrix and G is the eigenvalue of the symmetric matrix.
    [tex]\left[ {\begin{array}{*{20}{c}} {X11 - G}&{X12}&{X13}\\ {X12}&{X22 - G}&{X23}\\ {X13}&{X23}&{X33 - G} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {g1}\\ {g2}\\ {g3} \end{array}} \right] = 0[/tex]

    and:
    [tex]\begin{array}{l}
    E11 = (X22 - G)(X33 - G) - X_{23}^2\\
    E22 = (X11 - G)(X33 - G) - X_{13}^2\\
    E33 = (X11 - G)(X22 - G) - X_{12}^2
    \end{array}[/tex]

    Then:
    [tex]g1g1 = E11/(E11 + E22 + E33)?[/tex]

    Why?

    Thanks a lot!
     
  2. jcsd
  3. Nov 14, 2015 #2

    fresh_42

    Staff: Mentor

  4. Nov 14, 2015 #3
    Your answer may not be correct. I have figured out the answer:
    g1/g3 = E13/E33 =(g1g3K)(g3g3K)
    g2/g3 = E23/E33 =(g2g3K)(g3g3K)

    Manipulations of similar equations we get: E11=g1g1K, E22=g2g2K, E33=g3g3K ...... (1)

    So E11+E22+E33 =(g1g1+g2g2+g3g3) K = K ......(2)

    From (1) and (2), we obtain E11=g1g1 (E11+E22+E33)
     
  5. Nov 15, 2015 #4

    fresh_42

    Staff: Mentor

    I'm almost 100% certain the inverse of a matrix can be calculated this way although it's been a long time I did it. Have you recognized that the ##A_{ij}## which are obtained by calculating the determinant of ##A## with it's ##i-th## row and ##j-th## column stripped have a) to be transposed and b) have the sign ##(-1)^{i+j}##?

    So your ##E_{ii}## are really the diagonal elements of ##(X - diag(-G,-G,-G))^{-1}##. But I admit I don't know how this might help. Perhaps you can use the facts that invertible symmetric matrices are symmetric again and that they can be diagonalized. Their trace is the sum of the eigenvalues, the determinate the product.

    How exactly is your matrix ##E## defined and are you sure you want to calculate the squared first coordinate of g? If you - just to have a control - think of g as a 3-dimensional vector measured in meter, your equation then has m^2 on its left and no unit on its right side. Looks a little odd.
     
    Last edited: Nov 15, 2015
  6. Nov 15, 2015 #5
    g is dimensionless because it is normalized.
     
  7. Nov 15, 2015 #6

    fresh_42

    Staff: Mentor

    How are ##E## and K defined?
     
  8. Nov 15, 2015 #7
    I have listed E11, E22, and E33 already. E is defined as the cofactor matrix of the left side of the matrix at the first equation. K is an assumed number from g1/g3 = E13/E33 =(g1g3)(g3g3)= (g1g3K)(g3g3K).
     
  9. Nov 15, 2015 #8

    fresh_42

    Staff: Mentor

    Sorry I have absolute no idea where the trace of ##(X - G)^{-1}## comes into play.
     
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