The 
square matrix 
with entries given by
 | (1) |
for 
, 1, 2, ..., 
, where i is the imaginary number 
, and normalized by 
to make it a unitary. The Fourier matrix 
is given by
![F_2=1/(sqrt(2))[1 1; 1 i^2],](https://mathworld.wolfram.com/images/equations/FourierMatrix/NumberedEquation2.svg) | (2) |
and the 
matrix by
 |  | ![1/(sqrt(4))[1 1 1 1; 1 i i^2 i^3; 1 i^2 i^4 i^6; 1 i^3 i^6 i^9]](https://mathworld.wolfram.com/images/equations/FourierMatrix/Inline11.svg) | (3) |
 |  | ![1/2[1 1 ; 1 i; 1 -1 ; 1 -i][1 1 ; 1 i^2 ; 1 1; 1 i^2][1 ; 1 ; 1 ; 1].](https://mathworld.wolfram.com/images/equations/FourierMatrix/Inline14.svg) | (4) |
In general,
![F_(2n)=[I_n D_n; I_n -D_n][F_n ; F_n][ even-odd ; shuffle ],](https://mathworld.wolfram.com/images/equations/FourierMatrix/NumberedEquation3.svg) | (5) |
with
![[F_n ; F_n]=[I_(n/2) D_(n/2) ; I_(n/2) -D_(n/2) ; I_(n/2) D_(n/2); I_(n/2) -D_(n/2)] ×[F_(n/2) ; F_(n/2) ; F_(n/2) ; F_(n/2)][even-odd; 0,2 (mod 4); even-odd; 1,3 (mod 4)],](https://mathworld.wolfram.com/images/equations/FourierMatrix/NumberedEquation4.svg) | (6) |
where 
is the 
identity matrix and 
is the diagonal matrix with entries 1, 
, ..., 
. Note that the factorization (which is the basis of the fast Fourier transform) has two copies of 
in the center factor matrix.
