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Find the unique symmetric matrix A such that Y'AY=Y'GY

  1. Feb 6, 2014 #1
    I asked this question here, however the title of the thread (and the thread itself) was sloppy and unclear.I could not find a way to delete or edit.

    This is for a regression analysis course, and I've only taken one introductory course on linear algebra, so when I Google'd "finding a symmetric matrix" a lot of stuff that I didn't understand came up.

    1. The problem statement, all variables and given/known data
    Find the unique symmetric matrix A such that Y'AY=Y'GY


    2. Relevant equations



    3. The attempt at a solution


    Y'AY = Y'GY
    Y'AYY'=Y'GYY'
    Y'AYY'(YY')^-1 = Y'GYY'(YY')^-1
    Y'A=Y'G
    YY'A=YY'G
    (YY')^-1YY'A=YY')^-1YY'G
    A=G

    However, G is not symmetric:
    G=
    [1.0 1 1.0
    -1.0 0 1.0
    0.5 2 -0.5]

    Other than the algebraic manipulation above, I don't know how else to approach this problem.
     
  2. jcsd
  3. Feb 6, 2014 #2

    Ray Vickson

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    Homework Helper

    First: as regards your previous post on this matter, your friend is quite correct: the unique matrix ##G## that gives
    [tex] GY = (Y_1 +Y_2 + Y_3, Y_3-Y_1, (1/2)Y_1 - (1/2)Y_3 +2Y_2)^T [/tex]
    for ALL ##Y_1,Y_2,Y_3## is
    [tex] G = \pmatrix{1&1&1 \\-1&0&1\\1/2&2&-1/2} [/tex]
    You just read this off directly; no work is necessary. The given numerical values of the ##Y_i## are not relevant: they just get in the way. As far as I can see, they were given to you just to confuse you and lead you astray.

    Now you want a symmetric matrix that gives the same quadratic form ##Q(Y) \equiv Y^T G Y##. To clarify, but in a much smaller example with two variables instead of three: you want to know what are the ##a_{ij}## that give
    [tex] g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2
    = a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2[/tex]
    for all ##Y_1,Y_2##.
     
  4. Feb 6, 2014 #3
    Yes, I realized today that YY' is a singular matrix, making it impossible to find the inverse.

    as soon as you said 'quadratic form' I remembered something from the notes (I didn't recognize it as quadratic form because I don't remember learning about them in my linear algebra course. Anyways, in the notes it says "if the matrix A is not symmetric, one may replace [tex]a_{ij}[/tex] with [tex]\frac{a_{ij} + a_{ji}}{2}[/tex]. This gave me the symmetric matrix

    [tex] A = \pmatrix{1&0&\frac{3}{4} \\0&0&\frac{3}{2}\\\frac{3}{4}&\frac{3}{2}&\frac{-1}{2}} [/tex]

    That being said, I'm not really sure what you're doing with the equation

    [tex] g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2
    = a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2[/tex]

    Would it be essentially (for my problem with a 3x3) [tex]g_{11}=a_{11}, g_{12}+g_{21}=2a_{12}, g_{31}+g_{13}=2a_{13}, g_{22}=a_{22}, g_{23}+g_{32}=2a_{23}, and g_{33}=a_{33}[/tex]
     
  5. Feb 7, 2014 #4

    Ray Vickson

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    Yes, exactly: ##A = (G + G^T)/2##, as you have written.
     
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