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272                                                      Orthonormal Bases and Complements


                            So the set
                                                               1     1  
                                                           1
                                                                  2          
                                                                          3 
                                                                             
                                                        −1      1     1 
                                                              ,
                                                                      ,
                                                                2     3
                                                           0     −1        1 
                                                                  
                                                                             
                                                                          3 
                                                           0       0     −1
                                                                             
                                                     ⊥
                            is an orthogonal basis for L . Dividing each basis vector by its length yields
                                                                         √ 
                                                        1       1        3
                                                        √         √
                                                                               
                                                                  6       √ 6  
                                                         2                     
                                                                            
                                                    − √                     
                                                       1      √ 1       3 
                                                          2      6  ,    √ 6   ,
                                                                    
                                                           ,    2
                                                                            3 
                                                   0  − √                  
                                                                  6       6  
                                                                          √    
                                                                               
                                                       0         0      −   3  
                                                                            2
                                                     ⊥
                            and orthonormal basis for L . Moreover, we have
                                                                                        
                                                  c                 x
                                                                  
                                                             
                                                                                                 
                                                                                            
                                                     c
                                                                      y
                                                                                        
                                             ⊥
                                   4
                                 R = L ⊕ L =         c ∈ R   ⊕       ∈ R x + y + z + w = 0    ,
                                                                              4
                                                                      z
                                                  c                                        
                                                    
                                                                                            
                                                     c               w
                                                                                              
                                               4
                            a decomposition of R into a line and its three dimensional orthogonal complement.
                                                                                 ⊥ ⊥
                               Notice that for any subspace U, the subspace (U ) is just U again. As
                            such, ⊥ is an involution on the set of subspaces of a vector space. (An invo-
                            lution is any mathematical operation which performed twice does nothing.)
                            14.7      Review Problems
                                             Reading Problems        1    , 2   , 3    , 4
                                              Gram–Schmidt                      5
                            Webwork:
                                           Orthogonal eigenbasis               6, 7
                                          Orthogonal complement                 8

                                                  0
                                             λ 1
                               1. Let D =            .
                                             0   λ 2
                                   (a) Write D in terms of the vectors e 1 and e 2 , and their transposes.

                                                       a b
                                  (b) Suppose P =             is invertible. Show that D is similar to
                                                       c d

                                                          1      λ 1 ad − λ 2 bc −(λ 1 − λ 2 )ab
                                                 M =                                           .
                                                       ad − bc   (λ 1 − λ 2 )cd −λ 1 bc + λ 2 ad
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