Page 203 - Matrices theory and applications
P. 203

10. Approximation of Eigenvalues
                              186
                                                            2
                              The algorithm may be rewritten as
                                                             1
                                                      k
                                                                    k 0
                                                                 M x .
                                                     x =
                                                             k 0
                                                           M x
                                               k
                              We therefore have x  =0.
                                If F  = {0},then ρ(M| F ) <ρ(M) by construction. Hence there exist
                                                                                          k
                                                                                   k
                              (from Theorem 4.2.1) η< ρ(M)and C> 0such that  (M| F )  ≤ Cη for
                                                             k
                                                  k 0
                              every k.Then  (M| F ) z  ≤ C 1 η . On the other hand, ρ((M| E ) −1 )=
                              1/ρ(M), and the same argument as above ensures that  (M| E ) −k   ≤
                                                                     k 0
                                   k
                                                                               k
                              1/C 2µ ,for some µ ∈ (η, ρ(M)), so that  M y  ≥ C 3 µ . Hence,
                                                        k 0
                                                                  k 0
                                                      M z     M y  ,
                              so that
                                                             1
                                                      k             k 0
                                                     x ∼         M y .
                                                             k 0
                                                           M y
                                                                  0
                              We are thus reduced to the case where z =0, that is, where M has no
                              eigenvalue but λ. That will be assumed from now on.
                                Let r be the degree of the minimal polynomial of M. The vector space
                                                    0
                                                         0
                                                                    x contains all the x ’s. Up to
                              spanned by the vectors x ,Mx ,... ,M r−1 0              k
                              the replacement of CC n  by this linear subspace, one may assume that it
                                      n
                              equals CC .Thenwehave r = n. Furthermore, since ker(M − λ) n−1 ,a
                                                                                   0
                              nontrivial linear subspace, is stable under A,wesee that x  ∈ ker(M −
                              λ) n−1 .
                                                 n
                                The vector space CC then admits the basis
                                                             0
                                                                   n
                                                 2
                                         1
                                              0
                                                                                 x }.
                                       {v = x ,v =(M − λ)x ,... ,v =(M − λ)   n−1 0
                              With respect to this basis, M becomes the Jordan matrix
                                                                         
                                                       λ   0   ...  ...
                                                          .    .
                                                          .    .       . 
                                                       1    .   .       . . 
                                                    
                                                                         
                                                ˜        . .  .  .  . .  .  .
                                                                        . 
                                               M =    0    .   .    .  .
                                                     .   .    .    .     
                                                    
                                                     . .  . .  . .  . .  0  
                                                                          
                                                          ...  0    1   λ
                              The matrix λ −k ˜ k
                                            M depends polynomially on k. The coefficient of highest
                              degree, as k → +∞, is at the intersection of the first column and the last
                              row. It equals

                                                           k      1−n
                                                                 λ   ,
                                                         n − 1
                                2                 k
                                 One could normalize x at the end of the computation, but we prefer doing it at
                              each step in order to avoid overflows, and also to ensure (10.6).
   198   199   200   201   202   203   204   205   206   207   208