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Unit 28: Self Adjoint Operators




                                                                                                Notes
            2 A  A                               0  0, T *  T and  T   T
               n
             0 as A  A
                  n
            A A *   0   A A* 0

                  A A*    A  is self adjoint

                 A S
                  S  is closed.

          Now since S is a closed linear subspace of the Banach space   H , therefore S is a real Banach
          space.                                             ( S is a complete linear space)

          Also  I* I  the identity operator  I S.
          This completes the proof of the theorem.

          Theorem  3:  If  A ,A are  self-adjoint  operators,  then  their  product  A ,A is  self  adjoint
                        1  2                                           1  2
             A ,A  A ,A (i.e. they commute)
              1  2   2  1
          Proof: Let  A ,A be two self adjoint operators in H.
                     1
                       2
          Then  A *  A ,A *  A .
                 1   1  2   2
          Let  A ,A commute, we claim that  A ,A is self-adjoint.
                                           2
                  2
                                         1
               1
                  A ,A * A * A * A A   A A
                    1  2    2  1   2  1  1  2
                  A ,A * A A
                    1  2    1  2
                  A A 2 is self adjoint.
                   1
          Conversely, let  A A is self adjoint, then
                         1
                           2
                  A A * A A  2
                      2
                           1
                    1
                  A * A * A A 2
                    2
                          1
                      1
                  A ,A  commute
                   1  2
          This completes the proof of the theorem.
          Theorem 4: If T is an operator on a Hilbert space H, then T =  T  0  Tx,y  0   x,y H.
          Proof: Let T = 0 (i.e. zero operator). Then for all x and y we have

                  T x,y  Ox,y   O,y   O.

          Conversely,  Tx,y  O  x,y H

                  Tx,Tx  O  x,y H      (taking y = Tx)

                  Tx O  x,y H




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