Page 263 - Determinants and Their Applications in Mathematical Physics
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248   6. Applications of Determinants in Mathematical Physics

          Applying double-sum identities (C) and (A),
                       n   n                         n
                              r+s−1
                             [x     +(−1) r+s c]A rs  =  (2r − 1)
                      r=1 s=1                       r=1
                                                  = n 2             (6.3.12)

                            xA            r+s−1
                                   n
                                      n
                                =       x     A rs
                            A
                                  r=1 s=1
                                        n   n

                                   2
                                = n − c       (−1) r+s A .          (6.3.13)
                                                      rs
                                       r=1 s=1
          Differentiating and using (6.3.10),
                            	    
      n  n
                              xA
                                    = c      (−1) r+s
                              A                     α r α s
                                        r  s
                                        2
                                    = cλ .                          (6.3.14)
          It follows from (6.3.5) that
                          xA           1     2          11
                              = 1 −        [n − (c − 1)A ]
                           A         1+ x
                                      (c − 1)xA  + n
                                               11   2
                                 2
                              = n −                   .             (6.3.15)
                                           1+ x
          Hence, eliminating xA /A and using (6.3.14),

                                       11   2
                               (c − 1)xA  + n        2
                                                = −cλ .             (6.3.16)
                                    1+ x
          Differentiating (6.3.7) and using (6.3.10) and the first equation in (6.3.8),
                                      β = λα i .                    (6.3.17)

                                       i
          Differentiating the second equation in (6.3.11) and using (6.3.17),
                                     11
                                 (xA ) =2cλα 1 β 1 .                (6.3.18)
          All preparations for proving the theorem are now complete.
            Put
                                               11
                                  y =4(c − 1)xA .
          Referring to the second equation in (6.3.11),
                                                11
                                 y =4(c − 1)(xA )

                                              2
                                   =4c(c − 1)β .                    (6.3.19)
                                              1
          Referring to the first equation in (6.3.11),
                                  y
                                1 2
                                                 11
                                      =4(c − 1)(A )
                                  x
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