Page 120 - Advanced Linear Algebra
P. 120

104    Advanced Linear Algebra





                                                                        :
            Proof.  Clearly : q ; ~ ¸ ¹ , since any functional that annihilates both   and
            ;               :  must  annihilate   l  ;  ~  =  : .  Hence,  the sum      b  ;      is direct. The rest
            follows from Theorem 3.14, since

                           i




                         = ~ ¸ ¹ ~ ²: q ;³ ~ :b ; ~ :l ;
            Alternatively,  since       ;  ~    : b   is the identity map, if       =  i , then we can
            write

                                 b    ³ ~ ² k    ³ b ² k    ³  : l ;
                         ~  k ² ;   :        ;         :
                   i

            and so =~ : l ;     .…
            Operator Adjoints
                                                     *

            If  ²= Á > ³ , then we may define a map     d  ¢ > ¦ =  i   by
                 B
                                       d         ² ³ ~   k  ~
                                                          )
                      (
            for   > * .  We will write composition as juxtaposition.  Thus, for any #  =  ,
                                    ´  d ²      ³  µ  ²  #  ³  ~     ²  #    ³
            The map    d  is called the operator adjoint  of   and can be described by the

            phrase “apply   first.”

                        (
            Theorem 3.17  Properties of the Operator Adjoint)
             )
            1  For      B²= Á> ³  and  Á   - ,
                    Á
                                            d
                                   ²  b   ³ ~       d  b      d


             )
                                      B


            2  For  ²= Á > ³  and  ²> Á <³ ,
                       B
                                                 d
                                        ²  ³    d  ~      d
             )

            3   For any invertible  ²= ³ ,
                                  B
                                       ²  c  d  ³   ~  ²  d  ³    c
                                                       )
                             )
            Proof. Proof of part 1  is left for the reader. For part 2 , we have for all   < i ,
                        d
                                                             d
                    ²   ³ ² ³ ~  ²   ³ ~   d  ²  ³ ~   d  ²  d  ² ³³ ~ ²    d  ³² ³

                 )
                                 )
            Part 3  follows from part 2  and
                                  d    ²  c     ³  d    ~  c       ²  ³  d   ~  d   ~
            and in the same way,  ²  c  d d  ~      . Hence  ²  c  d  ~  ²  d c  .…
                                                   ³
                                 ³
                                                           ³
                                                             i
                                                            i
                                          i
                                       i
                                                                 i
                                                                 i
            If  ²= Á > ³ , then     d  B  ( > Á = )  and so     d  d  B ²= Á > ³ . Of course,
                 B

               dd  is not equal to  . However, in the finite-dimensional case, if we use the

            natural maps to identify  =  ii  with   and  >  ii  with  >  , then we can think of   dd
                                       =
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