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Unit 29: Spectral Theory and Properties of Normal Operators




          (iii)  each V  has an orthonormal basis { ,  } with the property that                 Notes
                    j                       j  j
                                    T  = a  + b
                                      j   j    j
                                    T  = – b  + a .
                                      j     j   j
                                  2
          In other words, if r =  a 2  b  and   is chosen so that a = r cos   and b = r sin  , then V is an
          orthogonal direct sum of two-dimensional subspaces  V  on each of which  T acts as  ‘r  times
                                                        j
          rotation through the angle  .
          The proof of Theorem 10 will be based on the following result.

                                                                             2
          Lemma: Let V be a real inner product space and S a normal operator on V such that S  + I = 0. Let
            be any vector in V and   = S . Then
                                       S  *
                                       S  *                                       … (1)

          ( | ) = 0, and       =      .
                                                                              2
                                                                       2
                                     2
                                                             2
          Proof: We have S  =   and S  = S  = –  . Therefore 0 =   S  –     +   S  +     =   S    – 2(S | )
                      2
               2
          +       +   S    + 2(S | ) +   2 .
          Since S is normal, it follows that
                                 2
                                                                  2
                  2
                                                       2
                                        2
          0 =   S*    – 2(S* | ) +       +   S*    + 2(S* | ) +       =   S*  +     +   S*  –    . 2
          This implies (1); hence
                                  ( | ) = (S* | ) = ( |S )
                                       = ( | –  )
                                       = – ( | )
          and ( | ) = 0. Similarly
                                                           2
                                        2  = (S* | ) = ( |S ) =      .
          Proof of Theorem 10:  Let V , …, V  be a maximal collection  of two-dimensional  subspaces
                                 1     s
          satisfying (i) and (ii), and the additional conditions.
                                   T*  = a  – b ,
                                      j   j   j
                                     1  j   s.                                    … (2)

                                   T*  = b  – a
                                      j   j   j
          Let W  = V  + … + V . Then W is the orthogonal direct sum of V , …, V . We shall show that
                   1       s                                   1     s
          W = V. Suppose that this is not the case. Then W    {0}. Moreover, since (iii) and (2) imply that W
                                                                                –1
          is invariant under T and T*, it follows that W  is invariant under T* and T = T**. Let S = b (T – aI).
                                                                                2
                    –1
                                                                             2
          Then S* = b (T* – aI), S*S = SS*, and W  is invariant under S and S*. Since (T – aI)  + b I = 0, it
                     2
          follows that S  + I = 0. Let   be any vector of norm 1 in W  and set   = S . Then   is in W  and
          S  = – . Since T = aI + bS, this implies
                                    T  = a  + b
                                    T  = – b  + a .
          By the lemma, S*  = – , S*  =  , ( | ) = 0, and       = 1. Because T* = aI + bS*, it follows that
                                   T*  = a  – b
                                   T*  = b  + a .




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