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A kernel is said to be symmetric if, $$r(x, y, u, v) = r_1(x, u) \cdot r_1(y, v)$$

My question is, What is the benefit of using symmetric kernels in Discrete Fourier transform?

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    So firstly, it doesn't matter so much how exactly things are paired, just that there is a one-to-one pairing (as long as your integrals are done the right way). The reason for writing it as a product is that it basically drops you down into regular Fourier transform theory on $\Bbb R$. This makes the analysis nearly trivial. – 2017-01-29
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    I don't understand the question. The Fourier transform is an integral transform $Tf(y) = \int_X f(x) k(x,y)dx$ with kernel $k(x,y) = e^{i xy} $. – 2017-01-29
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    @user1952009, I am talking about discrete Fourier transform. – 2017-01-29
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    Then it is almost the same : the discrete Fourier transform is a matrix transform $X(k) = \sum_{n=1}^N x(n) h(n,k)$ with kernel $H(n,k) = e^{-2i \pi nk/N}$. The benefit is that $\frac{H}{\sqrt{N}}$ is [an unitary matrix](https://en.wikipedia.org/wiki/Unitary_matrix), that is $\sum_{n=1}^N \frac{\overline{H(m,n)}}{\sqrt{N}} \frac{H(n,k)}{\sqrt{N}} = \begin{cases} 1 \text{ if } m = k \\ 0 \text{ otherwise} \end{cases}$ i.e. $H H^* = Id$ – 2017-01-29

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