Page 363 - 35Linear Algebra
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                                  and since ker L = {0 V } we have L = dim ker L = 0, so

                                               dim W = dim V = rank L = dim L(V ).

                                  Since L(V ) is a subspace of W with the same dimension as W, it
                                  must be equal to W. To see why, pick a basis B of L(V ). Each
                                  element of B is a vector in W, so the elements of B form a linearly
                                  independent set in W. Therefore B is a basis of W, since the size
                                  of B is equal to dim W. So L(V ) = span B = W. So L is surjective.

                     14.  (a) F 4 = F 2 + F 3 = 2 + 3 = 5.
                          (b) The number of pairs of doves in any given year equals the number of
                              the previous years plus those that hatch and there are as many of them
                              as pairs of doves in the year before the previous year.

                                     F 1      1             F 2      1
                          (c) X 1 =       =      and X 2 =       =      .
                                     F 0      0             F 1      1

                                                        1 1     1      1
                                              MX 1 =               =      = X 2 .
                                                        1 0     0      1
                          (d) We just need to use the recursion relationship of part (b) in the top
                              slot of X n+1 :

                                          F n+1       F n + F n−1     1 1       F n
                               X n+1 =           =                =                   = MX n .
                                           F n           F n          1 0      F n−1

                          (e) Notice M is symmetric so this is guaranteed to work.

                                            1 − λ   1                          1  2  5
                                       det              = λ(λ − 1) − 1 = λ −      −   ,
                                              1    −λ                          2     4
                                                                                           √ !
                                                      √                                  1± 5
                              so the eigenvalues are  1± 5 . Hence the eigenvectors are    2    ,
                                                      2
                                                                                           1
                                                         √             √    √          √
                              respectively (notice that  1+ 5  + 1 =  1+ 5 1+ 5  and  1− 5  + 1 =
                                                                         .
                                √     √                  2            2     2         2
                              1− 5 1− 5 ). Thus M = PDP    −1  with
                                   .
                                2    2
                                                 √                        √
                                                          !                      √ !
                                               1+ 5    0                1+ 5   1− 5
                                        D =     2       √    and P =      2     2     .
                                                0     1− 5               1      1
                                                       2
                                                                                  n
                                n
                                             ) = PDP
                          (f) M = (PDP    −1 n         −1 PDP −1  . . . PDP −1  = PD P −1 .
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