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Unitary discrete Fourier transform

  1. Dec 11, 2013 #1
    I'm trying to prove that the discrete form of the Fourier transform is a unitary transformation

    So I used the equation for the discrete Fourier transform:
    ##y_k=\frac{1}{\sqrt{N}}\sum^{N-1}_{j=0}{x_je^{i2\pi\frac{jk}{N}}}##

    and I put the fourier transform into a N-1 by N-1 matrix form:
    ##U=\begin{pmatrix}
    e^0 & e^0 & e^0 & ...\\
    e^0 & e^{\frac{i2\pi}{N}} & e^{\frac{i4\pi}{N}} & ...\\
    e^0 & e^{\frac{i4\pi}{N}} & e^{\frac{i8\pi}{N}} & ...\\
    ... & ... & ... & ...\\
    \end{pmatrix}##

    and then found the complex conjugate:
    ##U^*=\begin{pmatrix}
    e^0 & e^0 & e^0 & ...\\
    e^0 & e^{-\frac{i2\pi}{N}} & e^{-\frac{i4\pi}{N}} & ...\\
    e^0 & e^{-\frac{i4\pi}{N}} & e^{-\frac{i8\pi}{N}} & ...\\
    ... & ... & ... & ...\\
    \end{pmatrix}##

    But if I multiply these matrices together I get nothing which even approaches the identity matrix. Anyone have any ideas? Is there something wrong with the matrix?
     
  2. jcsd
  3. Dec 11, 2013 #2

    vela

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    You omitted the factor of ##1/\sqrt{N}##. Without it, the matrix won't be unitary. Also, the inverse of a unitary matrix is its adjoint, not the complex conjugate.
     
  4. Dec 11, 2013 #3
    I understand that I havent put in the ##\frac{1}{\sqrt{N}}##. But why do I need the inverse? I'm trying to calculate whether the matrix is unitary so I need to find the inner product of the matrix and its complex conjugate, wouldn't I?
     
    Last edited: Dec 11, 2013
  5. Dec 11, 2013 #4
    Ok I think I can get the diagonal elements to go to if I add the ##\frac{1}{\sqrt{N}}## factor but I still don't understand how the off diagonal elements go to zero.
     
  6. Dec 11, 2013 #5

    vela

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    No. Unitary means that the inverse of the matrix is its adjoint. In other words, if you multiply a unitary matrix by its adjoint (not conjugate), you get the identity matrix, which is what you're trying to show.
     
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