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452              Chapter 5                    Norms, Inner Products, and Orthogonality

                                            T
                                                                                    T
                                                       )
                                    where U V 1 =(C  −1 T  is nonsingular. To see that U V 2 is also nonsingular,
                                                                                    2
                                            1
                                                                                 T
                                    suppose dim M = r so that dim N = n − r and U V 2 is n − r × n − r. Use
                                                                                 2
                                    the formula for the rank of a product (4.5.1) to write
                                           T              T            T
                                    rank U V 2 = rank U   2  −dim N U 2  ∩R (V 2 )= n−r−dim M∩N = n−r.
                                           2
                                    It now follows from (5.15.7) that P M − P N is nonsingular, and
                                                                        T   −1
                                                                     (U V 1 )        0
                                                 T
                                                             −1
                                               V (P M − P N )  U =      1           T    −1  .
                                                                         0      −(U V 2 )
                                                                                    2
                                                                                             n
                                    (Showing that P M − P N is nonsingular implies M⊕N =        is Exercise
                                    5.15.6.) Formula (5.2.12) on p. 283 for the 2-norm of a block-diagonal matrix
                                    can now be applied to yield
                                                               5                            6
                                         
             
         
   T     
   
   T
                                         
 (P M − P N ) −1 
  = max  
 (U V 1 ) −1 
  , (U V 2 ) −1 
  .  (5.15.8)

                                                        2            1      2      2      2
                                        
   T      
   
   T
                                    But  
 (U V 1 ) −1
 2  = (U V 2 ) −1
 2  because we can again use (5.2.6) to write

                                                           2
                                            1
                                                 1              
  T    
 2         T  T     T
                                                      
 = min   
 U V 1 x 
  = min x V U 1 U V 1 x
                                           
   T       2           1     2             1     1
                                                             2
                                           
 (U V 1 ) −1
   x  =1             x  =1
                                                                                2
                                               1       2
                                                                              T
                                                                  T
                                                                     T
                                                         = min x V (I − U 2 U )V 1 x
                                                                     1
                                                                              2
                                                            x  =1
                                                             2
                                                                     T
                                                                              T
                                                                        T
                                                         = min (1 − x V U 2 U V 1 x)
                                                                              2
                                                                        1
                                                            x  =1
                                                             2
                                                                    
  T    
 2     
  T   
 2
                                                         =1 − max   
 U V 1 x 
  =1 − U V 1 
  .

                                                                       2
                                                                                       2
                                                               x  =1         2              2
                                                                 2
                                                           
         
 2     
      
 2
                                                              T
                                                                                T
                                    By a similar argument, 1/ (U V 2 ) −1
 2  =1− U V 1 
 2  (Exercise 5.15.11(a)).


                                                              2
                                                                                2
                                    Therefore,
                                        
             
   
   T
                                        
 (P M − P N ) −1 
  = (U V 1 ) −1 
  = C T 
  =  C  =  P MN   .


                                                       2      1       2       2       2         2
                                        While the minimal angle works fine for complementary spaces, it may not
                                    convey much information about the separation between noncomplementary sub-
                                    spaces. For example, θ min =0 whenever M and N have a nontrivial inter-
                                    section, but there nevertheless might be a nontrivial “gap” between M and
                                    N —look at Figure 5.15.1. Rather than thinking about angles to measure such a
                                    gap, consider orthogonal distances as discussed in (5.13.13). Define
                                              δ(M, N)= max dist (m, N)= max  (I − P N )m     2
                                                          m∈M               m∈M
                                                          m  2 =1            m  2 =1
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