Page 354 - A Course in Linear Algebra with Applications
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338                 Chapter  Nine:  Advanced  Topics


            Theorem     9.3.4
            Let  f  be a  symmetric  bilinear form  on  an  n-dimensional  real
            vector  space V.  Then  there  is  a  basis BofV  such  that


                 /(u, v)  =  mvi  H    h u kv k  -  u k+iv k+1     UlVi

            where  u\,...,  u n  and  v±,...,  v n  are  the  entries  of  the  coordi-
            nate  vectors  [u]g  and  [v]g  respectively  and  k  and I  are  integers
            satisfying  0 <  k  <  I <  n.

            Proof
            Let  /  be  represented  by  a  matrix  A  with  respect  some  basis
            B'  oiV.  Then  A  is symmetric.  Hence  there  is  an  orthogonal
                                  T
            matrix  S  such that  S AS  =  D  is diagonal,  say  with  diagonal
            entries  di,...  ,d n;  of  course  these  are  the  eigenvalues  of  A.
            Here we can assume that d\,...,   d k  >  0, while d k+i,...  ,di  <  0
            and  di +i  =  •  • • =  d n  =  0,  by  reordering  the  basis  if  necessary.
            Let  E  be the  nxn  diagonal matrix  whose diagonal entries  are
            the  real  numbers


              1/y/di,  ••-,  1/y/dk,  l/y/-d k+1,  ...,1/y/^dl,   1,...,1.

            Then
                            T              T  T
                        (SE) A{SE)     =  E (S AS)E     =  EDE,

            and  the  final  product  is the  matrix

                                  /  h   I     0     |  0  \


                          B          0   I   -Ii-k   I  0
                                    —    I   —       I  —
                                  \  0   I     0      1 0 /


            Now   the  matrix  SE  is  invertible,  so  its  inverse  determines  a
            change  of  basis  from  B'  to  say  B.  Then  /  will  be  represented
                                                                   /
            by the matrix  B  with respect to the basis B.  Finally, (u, v)  =
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