Page 357 - 35Linear Algebra
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357


                              Finally, for λ =  3
                                             2
                                  3   1                    1    3                        
                                 −         1    0        1       −     0        1 0 −1 0
                                   2   2                      2    2
                                  1         1                 5    5
                                                  
                                                                         
                                                                                              
                                                      
                                     −1        0  ∼  0 −            0  ∼  0 1 −1 0  .
                                                                             
                                 2         2                 4    4
                                  1    1  − 3   0        0    5  − 5   0        0 0    0   0
                                       2    2                 4    4
                                                        
                                                         s
                                                         s
                              So we find the eigenvector     where s 6= 0 is arbitrary.
                                                         s
                                                                                 
                                                                                 1
                         If the mistake X is in the direction of the eigenvector   −2 , then Y = 0.
                                                                                  
                                                                                 1
                         I.e., the satellite returns to the origin O. For all subsequent orbits it will
                         again return to the origin. NASA would be very pleased in this case.
                                                                 
                                                               −1
                         If the mistake X is in the direction    0 , then Y = −X. Hence the
                                                                  
                                                                 1
                         satellite will move to the point opposite to X. After next orbit will move
                         back to X. It will continue this wobbling motion indefinitely. Since this is a
                         stable situation, again, the elite engineers will pat themselves on the back.
                                                                  
                                                                   1
                                                                   1
                         Finally, if the mistake X is in the direction     , the satellite will move to a
                                                                   1
                                   3
                         point Y = X which is further away from the origin. The same will happen
                                   2
                         for all subsequent orbits, with the satellite moving a factor 3/2 further away
                         from O each orbit (in reality, after several orbits, the approximations used
                         by the engineers in their calculations probably fail and a new computation
                         will be needed). In this case, the satellite will be lost in outer space and the
                         engineers will likely lose their jobs!
                                                   
                                               1        0     0 
                                           3
                                                               0
                                                  0
                     10.  (a) A basis for B is       
                                                      ,
                                                            ,
                                                         1
                                                  0      0     1
                                                                 
                          (b) 3.
                               3
                          (c) 2 = 8.
                                   3
                          (d) dimB = 3.
                                                                                         3
                          (e) Because the vectors {v 1 , v 2 , v 3 } are a basis any element v ∈ B can be
                                                                                             1
                                                                                           
                                                                                            b
                                                                 3
                                                     1
                                                           2
                              written uniquely as v = b v 1 +b v 2 +b v 3 for some triplet of bits   2  .
                                                                                            b
                                                                                            b 3
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