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Fourier Basis

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Taking the powers of the root of unity

$\displaystyle w_n = \exp(2\pi\mathrm{i}/n)\, , $

we obtain an orthogonal basis of $ \mathbb{C}^n$. The matrix of the basis vectors is

$\displaystyle \left(\begin{array}{ccc} w_n^{0\cdot 0} & \cdots & w_n^{(n-1)\cd... ...w_n^{0\cdot (n-1)} & \cdots & w_n^{(n-1)\cdot (n-1)} \end{array}\right)\, . $

Orthogonality of the columns can easily be verified. The (complex) scalar product of the $ (j+1)$-th and the $ (k+1)$-th basis vector amounts to

$\displaystyle \sum_{\ell=0}^{n-1} w_n^{j\ell} \overline{w_n^{k\ell}} = \sum_\ell w_n^{(j-k)\ell} = \frac{w_n^{(j-k)n} - 1}{w_n^{j-k}-1}\, , $

and the numerator equals zero since $ w_n^n=1$.

For $ n=4$ we have $ w_4 = \exp(2\pi\mathrm{i}/4)=\mathrm{i}$, and we obtain

$\displaystyle \left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & \mathrm{i} & -1... ...& -1 & 1 & -1 \\ 1 & -\mathrm{i} & 1 & \mathrm{i} \end{array}\right)\, . $

(Authors: App/Burkhardt/Höllig)

Example:

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  automatically generated 1/18/2005