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4.8 Hankelians 1  107

            The second proof illustrates the equivalence of row and column op-
          erations on the one hand and matrix-type products on the other
          (Section 2.3.2).
            Second Proof. Define a triangular matrix P(x) as follows:

                                       i − 1
                             P(x)=           x i−j
                                      j − 1
                                                  n
                                      1
                                                   
                                     x   1         
                                  =  x 2  2x  1     .             (4.8.13)
                                    
                                                    
                                     x   3x   3x 1
                                      3   2        
                                     ................
                                                     n
          Since |P(x)| = |P (x)| = 1 for all values of x.
                         T
                                T
                  A = |P(−h)AP (−h)| n


                                i − 1                     j − 1
                    = (−h) i−j                    (−h) j−i

                                j − 1     |φ i+j−2 | n    i − 1
                                      n                          n
                                                                    (4.8.14)
                    = |α ij | n
          where, applying the formula for the product of three determinants at the
          end of Section 3.3.5,

                       i   j
                                      i − 1                j − 1
                 α ij =     (−h) i−r        φ r+s−2 (−h) j−s
                                     r − 1                 s − 1
                      r=1 s=1
                      i−1 	    
           j−1
                           i − 1                j − 1

                    =            (−h) i−1−r           (−h) j−1−s
                             r                   s             φ r+s
                      r=0                  s=0
                      i−1
                           i − 1            j−1

                    =            (−h) i−1−r ∆
                             r              h  φ r
                      r=0
                           i−1
                        j−1     i − 1
                    =∆                (−h) i−1−r
                        h        r             φ r
                           r=0
                    =∆  j−1 ∆ i−1 φ 0
                        h   h
                    =∆  i+j−2 φ 0 .                                 (4.8.15)
                        n
          The theorem follows. Simple differences are obtained by putting
          h =1.
          Exercise. Prove that
                           n   n
                                  r+s−2
                                 h     A rs (x)= A 11 (x − h).
                          r=1 s=1
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