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4.6 Hessenbergians  91

          If a ij = 0 when j − i> 1, the triangular array of zero elements appears
          in the top right-hand corner. H n can be expressed neatly in column vector
          notation.
            Let


                                                        ,            (4.6.2)
                                                       T
                           C jr = a 1j a 2j a 3j ...a rj O n−r
                                                       n
          where O i represents an unbroken sequence of i zero elements. Then

                         H n = C 12 C 23 C 34 ... C n−1,n C nn .     (4.6.3)


                                                         n
          Hessenbergians satisfy a simple recurrence relation.
          Theorem 4.20.
                                    n−1
                       H n =(−1) n−1    (−1) p r+1,n H r ,  H 0 =1,
                                           r
                                    r=0
          where

                           a ij a j,j−1 a j−1,j−2 ··· a i+2,i+1 a i+1,i ,  j > i
                    p ij =
                           a ii ,                          j = i.
          Proof. Expanding H n by the two nonzero elements in the last row,
                             H n = a nn H n−1 − a n,n−1 K n−1 ,
          where K n−1 is a determinant of order (n − 1) whose last row also contains
          two nonzero elements. Expanding K n−1 in a similar manner,
                         K n−1 = a n−1,n H n−2 − a n−1,n−2 K n−2 ,
          where K n−2 is a determinant of order (n − 2) whose last row also contains
          two nonzero elements. The theorem appears after these expansions are
          repeated a sufficient number of times.
          Illustration.

                H 5 = C 12 C 23 C 34 C 45 C 55 = a 55 H 4 − a 54 C 12 C 23 C 34 C 54 ,






                          C 12 C 23 C 34 C 54 = a 45 H 3 − a 43 C 12 C 23 C 53 ,




                              C 12 C 23 C 53 = a 35 H 2 − a 32 C 12 C 52 ,




                                 C 12 C 52 = a 25 H 1 − a 21 a 15 H 0 .
          Hence,
                     H 5 = a 55 H 4 − (a 45 a 54 )H 3 +(a 35 a 54 a 43 )H 2
                          −(a 25 a 54 a 43 a 32 )H 1 +(a 15 a 54 a 43 a 32 a 21 )H 0
                        = p 55 H 4 − p 45 H 3 + p 35 H 2 − p 25 H 1 + p 15 H 0 .
            Muir and Metzler use the term recurrent without giving a definition of the
          term. A recurrent is any determinant which satisfies a recurrence relation.
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