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30      1 Linear algebra



                                  A
                    ℜ                       ℜ
                                  A
                       w                      0
                        A 0
                                            Av ≠ 0
                         v
                                   A
                   Figure 1.5 The null space (kernel) of A, K A .

                                    A         ℜ
                      ℜ
                                     A
                          w                      0
                           A  0
                             v                  y  Av ≠ 0
                                                     A
                                      A
                                                  r ∉  A

                                                                     N
                                                                                   N
                   Figure 1.6 Venn diagram of linear transformation by A from domain   into codomain   showing
                   the kernel and the range subspaces.
                   If the null space contains only the null vector, K A is said to be empty.Aswenowshow,if
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                                                                                         N
                   this is the case, then Ax = b must have a unique solution x ∈  for any possible b ∈  .
                   However, if the null space contains any other non zero vectors (i.e., Aw = 0 with w  = 0),
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                   there is no unique solution. Then, depending upon the particular b ∈  , there may be
                   either no solution at all or an infinite number of them.
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                   Theorem Uniqueness of solution for Ax = b Let x ∈  be a solution to the linear
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                   system Ax = b, where b ∈  andAisan N × N real matrix. If the null space (kernel)
                   of A contains only the null vector, K A = 0, this solution is unique.
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                   Proof Let y ∈  be some vector, not necessarily x, that satisfies the system of equations,
                   Ay = b. We then define v = y − x, so that
                                       Ay = A(x + v) = Ax + Av = b + Av              (1.152)

                   If Ay = b, then v ∈ K A ,as Av = 0. If the null space is empty, K A = 0, we must have
                   v = 0 and the solution x is unique.                                QED
                   Now that we have a theorem for uniqueness, let us consider existence. To do so, we define
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                   the range of A, R A , as the subspace of all vectors y ∈   for which there exists some
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                   v ∈  such that Av = y. Formally, we write
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                                        R A ≡ y ∈  ∃v ∈  , Av = y                    (1.153)
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                   Figure 1.6 shows the relationship between the range and the kernel of A.
                   Theorem Existence of solutions for Ax = b LetAbeareal N × N matrix with a null
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                   space (kernel) K A and range R A , and let b ∈  . Then,
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