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3.8 Inverses of Sums and Sensitivity125

                                        The Sherman–Morrison–Woodbury formula (3.8.3) can be verified with di-
                                    rect multiplication, or it can be derived as indicated in Exercise 3.8.6.
                                        To appreciate the utility of the Sherman–Morrison formula, suppose A −1
                                    is known from a previous calculation, but now one entry in A needs to be
                                    changed or updated—say we need to add α to a ij . It’s not necessary to start
                                    from scratch to compute the new inverse because Sherman–Morrison shows how
                                    the previously computed information in A −1  can be updated to produce the
                                    new inverse. Let c = e i and d = αe j , where e i and e j are the i th  and j th
                                    unit columns, respectively. The matrix cd T  has α in the (i, j)-position and
                                    zeros elsewhere so that
                                                                     T
                                                          B = A + cd = A + αe i e T
                                                                                 j
                                    is the updated matrix. According to the Sherman–Morrison formula,
                                                                                T
                                                                          A −1 e i e A −1
                                                            	 −1   −1           j

                                                          T
                                            −1
                                           B   = A + αe i e    = A    − α
                                                          j                    T  −1
                                                                         1+ αe A
                                                                               j    e i
                                                                                                   (3.8.4)
                                                         [A −1 ] ∗i [A −1  ] j∗
                                                   −1
                                               = A    − α               (recall Exercise 3.5.4).
                                                          1+ α[A −1  ] ji
                                    This shows how A −1  changes when a ij is perturbed, and it provides a useful
                                    algorithm for updating A −1 .
                   Example 3.8.1
                                    Problem: Start with A and A −1  given below. Update A by adding 1 to a 21 ,
                                    and then use the Sherman–Morrison formula to update A −1  :

                                                         12              −1       3  −2
                                                   A =            and   A   =            .
                                                         13                     −1    1
                                    Solution: The updated matrix is

                                           12        12        00        12        0                  T
                                     B =         =         +         =         +      (1 0) = A + e 2 e .
                                                                                                      1
                                           23        13        10        13        1
                                    Applying the Sherman–Morrison formula yields the updated inverse
                                                                  T
                                                           A −1 e 2 e A −1  −1   [A −1 ] ∗2 [A −1 ] 1∗
                                                                  1
                                                      −1
                                                −1
                                              B   = A    −              = A   −
                                                                T
                                                           1+ e A  −1             1+[A  −1 ] 12
                                                                1    e 2

                                                                    −2
                                                                        (3 −2)
                                                        3  −2        1               −3    2
                                                  =             −                =            .
                                                      −1    1          1 − 2          2  −1
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