Page 307 - 35Linear Algebra
P. 307

17.1 Projection Matrices                                                                      307


                   MX = V r . We learned to find solutions to this in the previous subsection of
                   this book.
                      But here is another question, how can we determine what V r is given M
                   and V ? The answer is simple; suppose X is a solution to MX = V r . Then

                                                                                T
                                                                     T
                                                        T
                                             T
                          MX = V r =⇒ M Mx = M V r =⇒ M Mx = M (V r + 0)
                                                                                      T
                                                         T
                                      T
                                                                   T
                           T
                                                                                           −1
                                                                                               T
                   =⇒ M Mx = M (V r +V k ) =⇒ M Mx = M V =⇒ X = (M M) M V
                               T
                   if indeed M M is invertible. Since, by assumption, X is a solution
                                                             T
                                                        −1
                                                   T
                                             M(M M) M V = V r .
                                                                                          −1
                                                                                               T
                                                                                     T
                   That is, the matrix which projects V onto its ran M part is M(M M) M .
                                                                                
                                             1             1         1           1   1
                                                                 ,
                                                              1
                                             1
                   Example 154 To project     onto span        −1    = ran   1 −1   multi-
                                             1               0       0           0   0
                   ply by the matrix
                                                              −1
                                   1   1                 1   1
                                              1    1 0                  1    1 0
                                   1 −1                    1 −1
                                                            
                                              1 −1 0                    1 −1 0
                                   0   0                   0   0
                                                 
                                             1   1         −1
                                                      2 0       1    1 0
                                        =   1 −1 
                                                      0 2       1 −1 0
                                             0   0
                                                                         
                                          1    1                   2 0 0
                                       1            1    1 0      1
                                    =    1 −1                =    0 2 0    .
                                       2            1 −1 0        2
                                          0    0                     0 0 0
                      This gives
                                                              
                                                2 0 0      1       1
                                            1
                                                                   1
                                                           1
                                               0 2 0       =     .
                                            2
                                                0 0 0      1       0
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