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Math Help - Linear Algebra

  1. #1
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    Linear Algebra

    Help.. i can't understand the meaning... May i know how to form the matrix R? Thanks in advance

    The Nth root of 1 is given by w=exp(2*pi*j / N), where j = sqrt -1. It is easily verified as w^N = exp (2*pi*j) = cos 2pi + j sin2pi = 1. Using this N th root of 1, we create an (N*N) square matrix R such that the (a,b) th element of R is given by rab = W^(a-1)(b-1), a = 1,2,....,N; b = 1, 2, ...., N.

    Question--> Write the matrix R in terms of it's element. Clearly, show at least the top 3 X 3 part and all element on the four corner...

    Is this symmetric??
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  2. #2
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    Quote Originally Posted by Chris0724 View Post
    Is this symmetric??
    You need to show a_{ij} = a_{ji}.
    Note, a_{ij} = \omega^{(i-1)(j-1)} = \omega^{(j-1)(i-1)} = a_{ji}.
    Thus it is symmetric.
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  3. #3
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    i had form the matrix.

    | 1 1 1..........1|
    | 1 w^1 w^2 |
    | 1 w^2 w^4 |
    | .
    | .
    |1............W^(n-1)(n-1)|

    Is the matrix above correct if i leave the N-th row & N-th coloum like that?

    Qns 2
    Consider another (N N) square matrix S such that the (c, d)-th element of S is given by Scd = w^(c-1)(d-1), c = 1,2,....,N; d = 1, 2, ...., NShow that determinant of S is equal to the complex conjugate of the determinant of R. Note all you need to show is that S and R are complex conjugate of each other. In other words, show that ω and ω1 are complex conjugate of each other.

    Is this correct?

    w^N = exp (2*pi*j) = cos 2pi + j sin2pi = 1
    w^N = exp (2*pi*j) = cos 2pi - j sin2pi = -1

    Qns 3
    Let T = SR. Write the elements of T.

    May i know wat is w^-2?

    | 3 1+w^-1+w^-2 ........
    |

    Once again many thank you for the help
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  4. #4
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    Help!!

    hi,

    sorry to trouble.. any advice?
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