Page 153 - Matrices theory and applications
P. 153

8
                              Matrix Factorizations

















                              The direct solution (by Cramer’s method) of a linear system Mx = b,
                                                      n
                              where M ∈ GL n (k)(b ∈ k ) is computationally expensive, especially if
                              one wishes to solve the system many times with various values of b.In the
                              next chapter we shall study iterative methods for the case k = IR or CC.
                              Here we concentrate on a simple idea: To decompose M as a product PQ
                              in such a way that the resolution of the intermediate systems Py = b and
                              Qx = y is “cheap.” In general, at least one of the matrices is triangular.
                              For example, if P is lower triangular (p ij =0 if i< j), then its diagonal
                              entries p ii are nonzero, and one may solve the system Py = b step by step:

                                                     b 1
                                                 =     ,
                                             y 1
                                                     p 11
                                                 .
                                                 .
                                                 .
                                                     b i − p i1 y 1 − ··· − p i,i−1 y i−1
                                             y i  =                          ,
                                                               p ii
                                                 .
                                                 .
                                                 .
                                                     b n − p n1 y 1 − ··· − p n,n−1y n−1
                                                 =                             .
                                             y n
                                                                p nn
                              The computation of y i needs 2i−1 operations and the final result is obtained
                                 2
                              in n operations. This is not expensive if one notes that computing the
                              product x = M −1 b (assuming that M −1  is computed once and for all, an
                                                   2
                              expensive task) needs 2n − n operations.
   148   149   150   151   152   153   154   155   156   157   158