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Math Help - simetric function problem

  1. #1
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    simetric function problem

    5)there is f:R^{n}xR^{n}->R which is defined by f(x,y)=n\sum_{i=1}^{n}(x_{i}-m_{x})(y_{i}-m_{y}) where
    m_{y}=\frac{y_{1}+..+y_{n}}{n}
    m_{x}=\frac{x_{1}+..+x_{n}}{n}
    x=(x_{1}+..+x_{n}) y=(y_{1}+..+y_{n})
    A)check if this function is a simetric bilenear.
    B)find the representative matrices by the standart basis of R^{n}
    C)find the jordan form of the matrices you found in part B
    D)check if n||x||^{2}>=f(x,x)>=0 for every x\in R^{n}
    ??
    regarding A:
    f is simetric if f(x,y)=f(y,x) so i need to prove that
    f(x,y)=n\sum_{i=1}^{n}(x_{i}-m_{x})(y_{i}-m_{y})=n\sum_{i=1}^{n}(y_{i}-m_{y})(x_{i}-m_{x})=f(y,x)
    but when i put the given m_{y}and m_{x} i get two different expressions
    ?
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  2. #2
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    Re: simetric function problem

    a) since each x_i-m_x,y_i-m_y is a real number, isn't it true that (x_i-m_x)(y_i-m_y) = (y_i-m_y)(x_i-m_x) ? you also need to check linearity in x and y.
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  3. #3
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    Re: simetric function problem

    i made a typing mistake in the questions
    x=(x_{1},..,x_{n}) y=(y_{1},..,y_{n})

    ok i understand part A
    regarding B)

    for standart basis (1,0..,0) .. (0,0..,1)

    so for each vector in this basis

    m_{y}=\frac{0+..+1+..0+0}{n} m_{x}=\frac{0+..+1+..0+0}{n} and

    y_{i}=x_{j}=1

    f(e_i,e_j)=n\sum_{i=1}^{n}(1-\frac{1}{n})(1-\frac{1}{n})=n\sum_{i=1}^{n}(\frac{n-1}{n})^{2}=n^{2}(\frac{n-1}{n})^{2}=(n-2)^{2}
    in transformations each column of the representation matrices would be T(e1)
    what to do here
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  4. #4
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    Re: simetric function problem

    first of all, your sum isn't correct. if you want to calculate f(ei,ej) you should sum over some other index, like k (because k can be anything from 1 to n, but i and j are "constants").

    for k ≠ i,j, we have the term 1/n^2, for k = i or j, we have the term (1 - 1/n)(-1/n) = (1-n)/n^2 (when i ≠ j), so that f(ei,ej) = n[(n-2)/n^2 + (2-2n)/n^2] = -n/n = -1.

    when i = j, we have n[(n-1)/n^2 + (n-1)^2/n^2] = n-1. so for n = 2, for example, you should get the matrix:

    [1 -1]
    [-1 1].
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