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7.8 Least Squares Vectors and Data Fitting  233


                                        is a matrix of numbers, then
                                                               AX = x 1 C 1 + x 2 C 2 + ··· + x m C m = B

                                        exactly when B is a linear combination of the columns of A, hence is in S.
                                           The following lemma reveals a connection between the least squares vectors for AX=B and
                                        orthogonal projections, as suggested by the inequality (7.2).

                                  LEMMA 7.1

                                        Let B be an n-vector. Then an m-vector X is a least squares vector for AX = B if and only if
                                                                         ∗
                                                                             ∗
                                                                          AX = B S ,
                                        where B S is the orthogonal projection of B onto S.

                                                               ∗
                                        Proof Suppose first that AX = B S . Then
                                                                                      ∗
                                                                      B − B S  =  B − AX
                                                                             ≤  B − C
                                        for all vectors C in S, because B S is the vector in S closest to B. But the vectors C in S are exactly
                                                             m
                                        the vectors AX for X in R ,so
                                                                           ∗
                                                                      B − AX  ≤  B − AX
                                        for every m-vector X, and this proves that X is a least squares vector for AX = B.
                                                                           ∗
                                           Conversely, suppose X is a lease squares vector for AX = B. Then
                                                              ∗
                                                                        ∗
                                                                      AX − B  ≤  AX − B
                                                             ∗
                                        for all X in S. But then AX is the vector in S closest to B. Because B S is the unique vector with
                                        this property, then AX = B S . This completes the proof.
                                                          ∗
                                           We are now able to completely characterize the least squares vectors of AX = B as the
                                        solutions of a system of linear equations obtained using A.


                                  THEOREM 7.17   Least Squares Vectors for AX = B

                                        An m-vector X is a least squares vector of AX = B if and only if X is a solution of the system
                                                                       A AX = A B.
                                                                         t
                                                                                t
                                        Proof  Suppose first that X is a least squares vector of AX = B. By the lemma,
                                                              ∗
                                                                          AX = B S .
                                                                             ∗
                                                                ⊥
                                                                            ∗
                                                                                   ⊥
                                        We know that B − B S is in S ,so B − AX is in S . This means that the columns of A are
                                                                                                          ∗
                                                          ∗
                                        orthogonal to B − AX . Writing the dot product of column j of A with B − AX as a matrix
                                        product, this orthogonality means that
                                                                         t
                                                                                 ∗
                                                                      (C j ) (B − AX ) = 0.
                                                           t
                                                t
                                        Now (C j ) is row j of A ,so
                                                                        t
                                                                                ∗
                                                                       A (B − AX ) = O
                                        in which O is the m × 1 zero matrix. But this equation can be written
                                                                                 t
                                                                         t
                                                                        A AX = A (B)
                                                                             ∗
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                                   October 14, 2010  14:23  THM/NEIL   Page-233        27410_07_ch07_p187-246
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