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Unit 32: Groups Preserving Bilinear Forms




                                                                                                Notes
                 Example 1: Consider a three dimensional co-ordinates (x, y, z). Let us give a rotation
          along z-direction by an angle Q so that the new co-ordinates are x’, y’, z’
          then
                                     x’ = x cos   = y sin
                                     y’ = x sin   + y cos
                                     z’ = z
          We see that the square of the length becomes
                              2
                                  2
                                                                      2
                                                      2
                                      2
                             x’  + y’  + z’ = (x cos   – y sin  )  + (x sin   + y cos  )  + z 2
                                             2
                                         2
                                       = x  + y  + z .
                                                2
          So  the rotation is a transformation that preserves the  bilinear form of the  length. For more
          details see the next section.
          32.2 Groups Preserving Bilinear Forms
          We start this section with a few theorems and examples.
          Theorem 1: Let f be a non-degenerate bilinear form on a finite-dimensional vector space V. The
          set of all linear operators on V which preserve f is a group under the operation of composition.
          Proof: Let G be the set of linear operators preserving f. We observed that the identity operator
          is in G and that whenever S and T are in G the composition ST is also in G. From the fact that f is
          non-degenerate, we shall prove that any operator  T in  G is invertible, and  T   is also  in  G.
                                                                           –1
          Suppose T preserves f. Let   be a vector in the null space of T. Then for any   in V we have
                                 f( ,  ) = f(T , T ) = f(0, T ) = 0.
          Since f is non-degenerate,   = 0. Thus T is invertible. Clearly T  also preserves f; for
                                                             –1
                             f(T –I  , T –1  ) = f(TT –1  , TT –1  ) = f( ,  )
          If f is a non-degenerate bilinear form on the finite-dimensional space V, then each ordered basis
            for V determines a group of matrices 'preserving' f. The set of all matrices [T] , where T is a
          linear operator preserving f, will be a group under matrix multiplication. There is an alternative
          description of this group of matrices, as follows. Let A = [f] , so that if   and   are vectors in V
          with respective coordinate matrices X and Y relative to  , we shall have
                                 f( ,  ) = X’AY.
          Let T be any linear operator on V and M = [T] . Then
                                             t
                               f(T , T ) = (MX)  A (MY)
                                             t
                                          t
                                       = X  (M AM)Y.
                                            t
          Accordingly, T preserves f if and only if M A M = A. In matrix language then, Theorem 1 says the
                                                                              t
          following: If A is an invertible n   n matrix, the set of all n   n matrices M such that M AM = A is
          a group under matrix multiplication. If A = [f] , then M is in this group of matrices if and only if
          M = [T] , where T is a linear operator which preserves f.
          Let f be a bilinear form which is symmetric. A linear operator  T preserves f  If and only if  T
          preserves the quadratic form
                                   g( ) = f( ,  )
          associated with f. If T preserves f, we certainly have
                                  q(T ) = f(T , T ) = f( ,  ) = q( )




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