Page 42 - Numerical Methods for Chemical Engineering
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Existence and uniqueness of solutions                                 31



                   (I) the dimensions of K A and R A satisfy the dimension theorem,

                                           dim(K A ) + dim(R A ) = N                (1.154)
                  (II) If K A contains only the null vector 0, dim(K A ) = 0, and so dim(R A ) = N. The range
                                           N
                                                            N
                      therefore completely fills   so that every b ∈  is in the range, b ∈ R A , and for any
                           N
                      b ∈  , the system Ax = b has a solution.
                                                            [2]
                                                        [1]
                  Proof (I) Let us define an orthonormal basis {u , u ,..., u [P] , u [P+1] ,..., u [N] } for   N
                  such that the first P members of the basis span the kernel of A,
                                                     [1]  [2]  [P]
                                         K A = span u , u ,..., u                   (1.155)
                  Since the kernel is a subspace and thus satisfies all of the properties of a vector space, we
                  can always form such a basis. Therefore, we can write any w ∈ K A as

                                        w = c 1 u [1]  + c 2 u [2]  + ··· + c P u [P]  (1.156)
                  and

                                                dim(K A ) = P                       (1.157)

                                                   N
                  We now write any arbitrary vector v ∈  as an expansion in this orthonormal basis with
                                            [ j]
                  the scalar coefficients v j = v · u ,
                           v = v 1 u [1]  +· · · + v P u [P]  + v P+1 u [P+1]  +· · · + v N u [N]  (1.158)

                  We now operate on this vector by A,
                                  [1]           [P]         [P+1]           [N]
                        Av = v 1 Au  +· · · + v P Au  + v P+1 Au  +· · · + v N Au
                                                                                    (1.159)
                                  [1]         [P]          [P+1]           [N]
                        Av = A v 1 u  +· · · + v P u  + v P+1 Au  +· · · + v N Au
                  As v 1 u [1]  +· · · + v P u [P]  ∈ K A , this equation becomes

                                       Av = v P+1 Au [P+1]  +· · · + v N Au [N]     (1.160)
                                                         N
                  Since the range of A is the set of Av for all v ∈  , we see that any vector in the range can
                  be written as a linear combination of the N − P basis vectors {Au [P+1] ,..., Au [N] }, and
                  so dim(R A ) = N − P. Thus, dim(K A ) + dim(R A ) = N.
                    (II) follows directly from (I).                                   QED

                    What happens to the existence and uniqueness of solutions to Ax = b if K A is not empty?
                  Let us say that dim(K A ) = P > 0, and that we form the orthonormal basis for A as in the
                  proof above, such that we can write any w ∈ K A as
                                               [1]    [2]         [P]
                                        w = c 1 u  + c 2 u  + ··· + c P u           (1.161)
                                                     N
                  We now consider the arbitrary vector v ∈  , written as
                           v = v 1 u [1]  +· · · + v P u [P]  + v P+1 u [P+1]  +· · · + v N u [N]  (1.162)
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