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208    5. Iterative Methods for Systems of Linear Equations


        which means
                       M ω  :=  ωM +(1 − ω)I,   N ω := ωN ,         (5.31)
        or if the condition of consistency M = I − NA holds,


                  x (k+1)  = ω x (k)  − N Ax (k)  − b 		  +(1 − ω)x (k)

                          = x (k)  − ωN Ax (k)  − b .
          Let us assume for the procedure (5.6) that all eigenvalues of M are real.
        For the smallest one λ min and the largest one λ max we assume
                                 λ min ≤ λ max < 1;

        this is, for example, the case for the Richardson method. Then also the
        eigenvalues of M ω are real, and we conclude that

                    λ i (M ω )= ωλ i (M)+ 1 − ω =1 − ω 1 − λ i (M)
        if the λ i (B) are the eigenvalues of B in an arbitrary ordering. Hence

              (M ω )= max |1 − ω (1 − λ min (M))| , |1 − ω (1 − λ max (M))| ,
        since f(λ):= 1 − ω(1 − λ) is a straight line for a fixed ω (with f(1) = 1
        and f(0) = 1 − ω).

                   f                    ρ(M )
                                           ω
                  1                        1             - 1 + ω ( 1 − λ min )
                                        λ  max
            ω                                                 1 - ω ( 1 − λ max )
             1
                                 λ                     ω     ω
                λ min    λ max
                            1           λ min
                                                     1
              ω 2                                       1 - ω ( 1 − λ min )
               ..
            f (λ) for ω < 1 and ω > 1
                           2
                   1
                             Figure 5.1. Calculation of ¯ω.
          For the optimal ¯ω, i.e., ¯ω with
                                (M ¯ω )= min  (M ω ) ,
                                        ω>0
        we therefore have, as can be seen from Figure 5.1,

                         1 − ¯ω (1 − λ max (M)) = −1+ ¯ω (1 − λ min (M))
                                       2
                 ⇐⇒      ¯ ω =                     .
                             2 − λ max (M) − λ min (M)
        Hence ¯ω> 0and
                          (M ¯ω )= 1 − ¯ω(1 − λ max (M)) < 1;
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