Page 352 - 35Linear Algebra
P. 352

352                                                                          Sample Final Exam


                                   (b) n.

                                   (c) Yes.
                                   (d) n × n.

                                   (e) m × m.
                                                                               T
                                   (f) Yes. This relies on kerM = 0 because if M M had a non-trivial kernel,
                                                                                     T
                                       then there would be a non-zero solution X to M MX = 0. But then
                                                                     T
                                       by multiplying on the left by X we see that ||MX|| = 0. This in turn
                                       implies MX = 0 which contradicts the triviality of the kernel of M.
                                                                               T
                                                      T
                                                                 T
                                                                      T T
                                   (g) Yes because M M     T  = M (M ) = M M.
                                   (h) Yes, all symmetric matrices have a basis of eigenvectors.
                                   (i) No, because otherwise it would not be invertible.
                                   (j) Since the kernel of L is non-trivial, M must have 0 as an eigenvalue.
                                   (k) Since M has a zero eigenvalue in this case, its determinant must vanish.
                                       I.e., det M = 0.

                               4. To begin with the system becomes

                                                                        x
                                                                       
                                                                              
                                                          1 1 1 1               1
                                                                         y
                                                                       
                                                                            
                                                        1 2 2 2   = 1
                                                                       z
                                                          1 2 3 3               1
                                                                        w
                                  Then
                                                                                       
                                                     1 1 1 1          1 0 0      1 1 1 1
                                                                                       
                                              M = 1 2 2 2 = 1 1 0 0 1 1 1
                                                     1 2 3 3          1 0 1      0 1 2 2
                                                                            
                                                         1 0 0      1 1 1 1
                                                     = 1 1 0 0 1 1 1 = LU
                                                                               
                                                                 
                                                       
                                                         1 1 1      0 0 1 1
                                                                                         
                                                                                          a
                                  So now MX = V becomes LW = V where W = UX =               (say). Thus
                                                                                          b
                                                                                          c
                                  we solve LW = V by forward substitution

                                            a = 1, a + b = 1, a + b + c = 1 ⇒ a = 1, b = 0, c = 0 .


                                                      352
   347   348   349   350   351   352   353   354   355   356   357