The discrete "Fast Fourier Transform", (FFT) is based on the factorization of the Fourier matrix. If n = 8, we have the 8x8 Fourier matrix, which is this: 1 1 1 1 1 1 1 1 1 w w^2 w^3 w^4 w^5 w^6 w^7 1 w^2 w^4 w^6 w^8 w^10 w^12 w^14 1 w^3 w^6 w^9 w^12 w^15 w^18 w^21 F = 1 w^4 w^8 w^12 w^16 w^20 w^24 w^28 1 w^5 w^10 w^15 w^20 w^25 w^30 w^35 1 w^6 w^12 w^18 w^24 w^30 w^36 w^42 1 w^7 w^14 w^21 w^28 w^35 w^42 w^49 where w = e^(2pi i/8), a primitive complex eighth root of unity We first permute the columns in such a way that all even columns precede all odd columns. This is acocmplished by postmultiplying F by the permutation matrix P, where 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 The 4x4 submatrix in the upper left corner of F is then this 1 1 1 1 1 w^2 w^4 w^6 f = 1 w^4 w^8 w^12 1 w^6 w^12 w^18 Since w^2 = e^(2pi i/4), f is seen to be the 4x4 Fourier matrix. Furthermore, since w^8 =1, the 4x4 submatrix in the lower left corner of F is seen to be another copy of f. The 4x4 submatrix in the upper right corner of F is 1 1 1 1 w w^3 w^5 w^7 w^2 w^6 w^10 w^14 w^3 w^8 w^15 w^21 This is equal to the product of f premultiplied by the appropriate diagonal matrix, 1 0 0 0 0 w 0 0 0 0 w^2 0 0 0 0 w^3 In other words, we have f Df I D f 0 FP = f -Df = I -D 0 f f, in turn, can be similarly factored. In general, if n = 2^k, the nxn Fourier matrix can be factored into a product of three matrices. The first is of the form I D I -D where D is diagonal, with consecutive powers of w on its diagonal. The second is of the form f 0 0 f where f is the n/2 by n/2 Fourier matrix, and the last is a permutation matrix that moves the even columns before the odd columns.