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5.10 Range-Nullspace Decomposition                                                 395


                                        holds, and this forces n< dim N A n+1  , which is impossible. A similar
                                        argument proves equality exists somewhere in the range chain.
                                    Property 2. Once equality is attained, it is maintained throughout the rest of
                                    both chains in (5.10.2). In other words,
                                            0                     k        k+1         k+2
                                      N A     ⊂ N (A) ⊂· · · ⊂ N A  = N A      = N A       = ···
                                                                                                  (5.10.3)
                                           0                      k        k+1        k+2
                                      R A     ⊃ R (A) ⊃ ··· ⊃ R A   = R A      = R A      = ··· .
                                        To prove this for the range chain, observe that if k is the smallest nonneg-

                                        ative integer such that R A k  = R A k+1  , then for all i ≥ 1,
                                              i+k        i  k     i     k     i     k+1       i+k+1
                                         R A      = R A A     = A R A    = A R A       = R A       .
                                        The nullspace chain stops growing at exactly the same place the ranges
                                        stop shrinking because the rank plus nullity theorem (p. 199) insures that
                                                p
                                                                p
                                        dim N (A )= n − dim R (A ).
                                    Property 3. If k is the value at which the ranges stop shrinking and the
                                                                             k        k
                                    nullspaces stop growing in (5.10.3), then R A  ∩ N A  = 0.
                                                          k        k         k                     n
                                        Proof. If x ∈ R A   ∩ N A , then A y = x for some y ∈  , and


                                         k
                                                         2k
                                                                 k
                                        A x = 0. Hence A y = A x = 0 ⇒ y ∈ N A    2k    = N A k     ⇒ x = 0.
                                    Property 4. If k is the value at which the ranges stop shrinking and the
                                                                             k        k     n
                                    nullspaces stop growing in (5.10.3), then R A  + N A  =   .
                                        Proof. Use Property 3 along with (4.4.19), (4.4.15), and (4.4.6), to write
                                              k        k             k           k            k        k
                                     dim R A    + N A     = dim R A    + dim N A   − dim R A    ∩ N A
                                                                     k           k
                                                          = dim R A    + dim N A   = n
                                                                     k        k     n
                                                           =⇒ R A      + N A    =   .
                                    Below is a summary of our observations concerning the index of a square matrix.
                                                                   Index
                                       The index of a square matrix A is the smallest nonnegative integer k
                                       such that any one of the three following statements is true.
                                                   k          k+1
                                       •   rank A    = rank A     .
                                                k        k+1                         k
                                       •   R A    = R A     —i.e., the point where R A  stops shrinking.
                                                k        k+1                          k
                                       •   N A    = N A      —i.e., the point where N A  stops growing.
                                       For nonsingular matrices, index (A)=0. For singular matrices,
                                       index (A)is the smallest positive integer k such that either of the fol-
                                       lowing two statements is true.
                                                k        k
                                       •   R A    ∩ N A    = 0.                                (5.10.4)



                                                              k
                                            n
                                       •     = R A   k     ⊕ N A .
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