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Numerical Methods ———  213

                   From which

                                            −
                                             1
                                      =
                                [U k− 1 ] [Q k− 1 ] [ k− 1 ]                                        ...(4.22)
                                               A
                   Substituting Eq. (4.22) in Eq.(4.20), we get
                                             −
                                             1
                                   A =
                                    [ ] [Q k− 1 ] [ k − 1 ][Q k− 1 ]
                                               A
                                    k
                   Hence [A ] is similar to [A ]. This implies that [A ] is similar to [A ] = [A]. Therefore, it has the same
                           k
                                                              k
                                                                            0
                                         k–1
                   eigenvalues as [A]. Also
                                 A   =      − 1  A      ]
                                [ k − 1 ] [Q k − 2 ] [ k − 2 ][Q k − 2
                                           −
                                                       −
                                                                              −
                                                  −
                                                        1
                                           1
                                                                               1
                                                  1
                                                                         =
                                     =
                                                    Q
                                                          A Q
                                                                   ... Q
                                        Q
                                 A
                                                                Q
                                                                           P
                                                                                A P
                   Thus         [ k+ 1 ] [ ] [Q k− 1 ] [ ] [][ ][ ] [ ] [ ] [][ ]
                                                     0
                                                                 1
                                         k
                                                                      k
                                                              0
                                                                            k
                                                                                    k
                                   P =
                                              ... Q
                                            Q
                                        Q
                   where           [ ] [ ][ ] [ ]
                                    k
                                         0
                                             1
                                                  k
                   If k → ∞[P ] exists and we denote it by [P].
                            k
                                                                       [ ])
                                           [ (
                                                −
                                                                 −
                                                                                −
                                                                 1
                                                 1
                                                                                 1
                                                      =
                                                                                   P =
                                                   P
                                                                              P
                   Then         lim Q =  lim P k− 1 ] [ ]) ( lim P  1 ] )( lim P k  = [] [] [ ] I
                                                            [ k−
                                    k
                                                    k
                                k→∞    k→∞               k→∞        k→∞
                   Here, we have two limiting conditions
                                            −
                                             1
                                   A =
                                               A P
                                        P
                   1.             [ ] [ k− 1 ] [][ k− 1 ]
                                    k
                   which means
                                  [ ] (
                                                −
                                                                    −
                                                                    1
                                                 1
                                 lim A =  lim P  1 ] ) [](  lim P  1 ]) = [] [ ][] P
                                                   A
                                                                      A
                                                                  P
                                                         [ k −
                                           [ k −
                                    k
                              k→∞       k→∞           k→∞
                   Therefore, the limit [A ] is similar to [A] and hence has the same eigenvalues as [A].
                                     k
                                   A =
                                            U
                                        Q
                   2.             [ ] [ ][ ]
                                             k
                                    k
                                         k
                                           [ ])(
                                                    [ ])
                                 [ ] (
                   that is         lim A =  lim Q k  lim U k
                                   k
                              k→∞       k→∞     k→∞
                   here we have
                                  [ ] [] I
                                  lim Q =
                                    k
                              k→∞
                   which results in
                                          [ ]
                                 [ ] =
                                  lim A k  lim U k
                              k→∞      k→∞
                   Since every [U ] is an upper triangular matrix, the limit of [A ] is also an upper triangular matrix.
                                                                    k
                               k
                   The accuracy of this method mainly depends on the effectiveness of the algorithm used for decompositions
                   of [A ] into [Q ][U ]. The limit k → ∞ can exist for large size problems.
                                  k
                               k
                       k
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