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Chapter 11

                           THE GENERALISED SYMMETRIC MATRIX
                                        EIGENVALUE PROBLEM




                       Consider now the generalised matrix eigenvalue problem
                                                      Ax = eBx                           (2.63)
                       where A and B are symmetric and B is positive definite. A solution to this
                       problem will be sought by transforming it into a conventional eigenvalue problem.
                       The trivial approach
                                                       - 1
                                                     B Ax = ex                           (11.1)
                                                                   - 1
                       gives an eigenvalue problem of the single matrix B A which is unfortunately not
                                                     - 1
                       symmetric. The approximation to B A generated in a computation may therefore
                       have complex eigenvalues. Furthermore, methods for solving the eigenproblem of
                       non-symmetric matrices require much more work than their symmetric matrix
                       counterparts. Ford and Hall (1974) discuss several transformations that convert
                       (2.63) into a symmetric matrix eigenproblem.
                         Since B is positive definite, its eigenvalues can be written as the squares
                       i = l, 2, . . . , n, so that
                                                           2
                                                      B=ZD Z  T                          (11.2)
                      where Z is the matrix of eigenvectors of B normalised so that
                                                      T  T
                                                   ZZ =Z Z= l .                          (11.3)
                                                               n
                      Then
                                                             - 1 T
                                                    B -1/2 = ZD Z                        (11.4)
                       and
                                             (B -1/2 AB -l/2 )(B 1 / 2 X)=B 1 / 2 XE     (11.5)

                       is equivalent to the complete eigenproblem
                                                      AX = BXE                          (11.5a)
                      which is simply a matrix form which collects together all solutions to (2.63).
                        Equation (11.5) can be solved as a conventional symmetric eigenproblem
                                                      A V = VE                           (11.6)
                                                       1
                      where
                                                        - 1 / 2  - 1 / 2
                                                   A = B   AB                           (11.7a)
                                                    1
                       and
                                                      V = B 1 / 2 X.                    (11.76)
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