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21
                                                 1.4 Weights of nonorthgonal functions
                             elements equal to 1. We write iŁ ià partitione form as


                                                                      s
                                                                  1
                                                                                                (1.52)
                                                                         ,
                                                         S n−k =
                                                                     S
                                                                  †
                                                                 s

                             where [1 s] is the firsŁ row of the matrix LeŁ n−k be aà uppeð triangulað matrix
                                                                      M
                             partitione similarly,

                                                                  1   q
                                                         M n−k =         ,                      (1.53)
                                                                  0   B
                             and we determine q and B sł thaŁ

                                                                  1         q + sB
                                                †
                                          (M n−k ) S n−k M n−k =                        ,       (1.54)
                                                                           †
                                                              (q + sB) †  B (S − s s)B

                                                                                  †

                                                              1     0
                                                          =              ,                      (1.55)
                                                              0   S n−k−1
                             where these equations may be satisfie withB the diagonal matrix
                                                              2 −1/2      2 −1/2


                                               B = diag 1 − s        1 − s      ···             (1.56)
                                                              1          2
                             and
                                                             q =−sB.                            (1.57)
                             The iàverse ofM n−k is easily determined:

                                                      (M n−k ) −1  =  1  s −1  ,                (1.58)
                                                                   0  B
                             and, thus, N k+1 = N k Q k , where

                                                                     0
                                                               I k
                                                        Q k =             .                     (1.59)
                                                                0  M n−k
                             The unreduce portion of the problem is now transforme as follłws:
                                                                                          −1
                                                          −1
                                       †
                                                                         †
                                                                  †
                                 (C n−k ) S n−k C n−k = [(M n−k ) C n−k ] (M n−k ) S n−k M n−k [(M n−k ) C n−k ].
                                                                                                (1.60)
                             Writing

                                                                   C 1
                                                          C n−k =      ,                        (1.61)
                                                                   C
                             we have

                                                                     C 1 + sC
                                                         −1
                                                  [(M n−k ) C n−k ] =   −1     ,                (1.62)
                                                                      B   C

                                                                     C 1 + sC
                                                                 =             .                (1.63)
                                                                      C n−k−1
                             Putting these togetheð, we arrive aŁ the totalN as Q 1 Q 2 Q 3 ··· Q n−1 .
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