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                                                                   10.4. The Power Methods
                              The standard method is especially designed for the search for the optimal
                              parameter in the SOR method for a tridiagonal matrix, where we have to
                              compute the spectral radius of the Jacobi iteration matrix (Theorem 9.4.1).
                              10.4.1 The Standard Method
                              Let M ∈ M n (CC) be a matrix. We search for an approximation of its eigen-
                              value of maximum modulus, whenever only one such exists. The standard
                                                                   n               0     n
                              method consists in choosing a norm on CC , a unit vector x ∈ CC ,and
                                                                  k
                              then computing successively the vectors x by the formula
                                                               1      k
                                                      k+1
                                                     x   :=        Mx .
                                                                k
                                                             Mx
                              The justification of this method is given in the following theorem.
                              Theorem 10.4.1 One assumes that Sp M contains only one element λ of
                              maximal modulus (that modulus is thus equal to ρ(M)).
                                                                     k
                                If ρ(M)= 0, the method stops because Mx =0 for some k< n.
                                              n
                                                                               n
                                Otherwise, let CC = E ⊕ F be the decomposition of CC ,where E, F are
                              stable linear subspaces under M,with Sp(M| E )= {λ} and λ  ∈ Sp(M| F ).
                                          0
                                                         k
                              Assume that x  ∈ F.Then Mx  =0 for every k ∈ IN and:
                                1.
                                                                 k
                                                        lim  Mx   = ρ(M).                (10.6)
                                                       k→+∞
                                2.
                                                                    ¯    k
                                                                   λ       k
                                                      V := lim           x
                                                          k→+∞    ρ(M)
                                   is a unit eigenvector of M, associated to the eigenvalue λ.

                                3. If V j  =0,then
                                                                  k
                                                              (Mx ) j
                                                         lim         = λ.
                                                        k→+∞    x k
                                                                 j
                                Proof
                                The case ρ(M) = 0 is obvious because M is then nilpotent. We may thus
                              assume that ρ(M) > 0.
                                                                                         0
                                     0
                                         0
                                                                              0
                                              0
                                                                      0
                                Let x = y + z be the decomposition of x with y ∈ E and z ∈ F.
                                                                          k 0
                                             0
                                                                                         k 0
                              By assumption, y  =0. Since M| E is invertible, M y  =0. Since M x =
                                                                                       k 0
                                        k 0
                                k 0
                                                            k 0
                                              k 0
                              M y + M z , M y ∈ E,and M z ∈ F, we conclude that M x  =0.
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