Page 306 - 35Linear Algebra
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306                                                          Least squares and Singular Values


                            Then he should find a and b to best fit the data.


                                                           11 = a · 1 + b
                                                           19 = a · 2 + b
                                                           31 = a · 3 + b.

                            As a system of linear equations, this becomes:

                                                                        
                                                         1 1           11
                                                                 a   ?
                                                         2 1         =   19  .
                                                                        
                                                                 b
                                                         3 1             31
                                                                                                   T
                                                                                        T
                            There is likely no actual straight line solution, so instead solve M MX = M V .
                                                                                   
                                                         1 1                        11

                                              1 2 3              a       1 2 3
                                                        2 1        =             19   .
                                              1 1 1              b       1 1 1
                                                         3 1                        31
                            This simplifies to

                                                      14 6 142        1 0 10
                                                                  ∼           1  .
                                                       6 3    61      0 1
                                                                              3
                            Thus, the least-squares fit is the line
                                                                       1
                                                            v = 10 t +  .
                                                                       3
                            Notice that this equation implies that Captain Conundrum accelerates towards Italian
                                        2
                            soil at 10 m/s (which is an excellent approximation to reality) and that he started at
                            a downward velocity of  1  m/s (perhaps somebody gave him a shove...)!
                                                  3

                            17.1      Projection Matrices


                                                                                                 T
                                                                                     T
                            We have seen that even if MX = V has no solutions M MX = M V does
                            have solutions. One way to think about this is, since the codomain of M is
                            the direct sum
                                                    codom M = ranM ⊕ ker M      T
                                                                                      T
                            there is a unique way to write V = V r +V k with V k ∈ ker M and V r ∈ ran M,
                            and it is clear that Mx = V only has a solution of V ∈ ran M ⇔ V k = 0.
                            If not, then the closest thing to a solution of MX = V is a solution to


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