Page 374 - Matrix Analysis & Applied Linear Algebra
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370              Chapter 5                    Norms, Inner Products, and Orthogonality




                                                  Decomposing the Fourier Matrix

                                               r
                                       If n =2 , then

                                                              F n/2  D n/2 F n/2
                                                       F n =                   P n ,           (5.8.15)
                                                              F n/2  −D n/2 F n/2
                                       where                                   
                                                                1
                                                                  ξ
                                                                     2         
                                                       D n/2 =      ξ          
                                                                       . . .   
                                                                            n −1
                                                                           ξ 2
                                       contains half of the n th  roots of unity and P n is the “even–odd” per-
                                       mutation matrix defined by

                                                   T
                                                 P =[e 0 e 2 e 4 ··· e n−2 | e 1 e 3 e 5 ··· e n−1 ] .
                                                   n

                                        The decomposition (5.8.15) says that a discrete Fourier transform of order
                                         r
                                    n =2 can be accomplished by two Fourier transforms of order n/2=2 r−1 , and
                                    this leads to the FFT algorithm. To get a feel for how the FFT works, consider
                                    the case when n =8, and proceed to “divide and conquer.” If
                                                                               
                                                                            x 0
                                                       x 0
                                                                             x 2 
                                                      x 1 
                                                                               
                                                        
                                                                             x 4       
                                                      x 2                             (0)
                                                                               
                                                                                     x 4
                                                                             x 6 
                                                      x 3 
                                                x 8 =    ,  then   P 8 x 8 =      =      ,
                                                                               
                                                      x 4                             (1)
                                                                             x 1     x
                                                                                      4
                                                                               
                                                      x 5 
                                                                             x 3 
                                                        
                                                       x 6                     
                                                                              x 5
                                                       x 7
                                                                              x 7
                                    so
                                                                                        
                                                                  (0)         (0)        (1)
                                                 F 4   D 4 F 4  x 4        F 4 x 4  + D 4 F 4 x 4
                                        F 8 x 8 =                    =                    .   (5.8.16)
                                                                  (1)         (0)        (1)
                                                 F 4  −D 4 F 4  x 4        F 4 x 4  − D 4 F 4 x 4
                                    But
                                                                                    
                                                   x 0      (0)                    x 1      (2)  
                                                            x                                 x
                                            (0)   x 4      2                (1)   x 5      2
                                         P 4 x 4  =      =       and  P 4 x 4  =      =      ,
                                                           (1)                             (3)
                                                   x 2                               x 3
                                                            x                                 x
                                                             2                                 2
                                                   x 6                               x 7
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