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Linear Algebra




                    Notes          so that the matrix of D in the ordered basis    is


                                                 0 1 0 0
                                                 0 0 2 0
                                           D    0 0 0 3  .
                                                 0 0 0 0

                                   We have seen what happens to representing matrices when transformations are added, namely,
                                   that the matrices add. We should now like to ask what happens when we compose transformations.
                                   More specifically, let V, W and Z be vector spaces over the field F of respective dimensions n, m
                                   and p. Let T be a linear transformation from V into W and U a linear transformation from W into
                                   Z. Suppose we have ordered basis

                                                             
                                                
                                      1 , ..,  n  ,   '  1 , ..,  m  ,    ''  1 , ..,  p
                                   for the respective spaces V, W and Z. Let A be the matrix of T relative to the pair  ', '  and let 
                                   be the matrix of  U relative to the  pair  ', ''.   It is then easy to see that the matrix  C  of  the
                                   transformation UT relative to the pair , ''   is the product of B and A; for, if  is any vector in V.
                                           T   =A
                                              '    
                                           U T    =B T
                                                 ''    ' 
                                   and so  UT      = BA
                                                    ''  
                                   and hence, by the definition and uniqueness of the representing matrix, we must have  C = BA.
                                   One can also see this by carrying out the computation


                                          UT    j =U T  j


                                                      m
                                                 =U     A
                                                         kj  k
                                                     k  1
                                                   m
                                                 =   A U
                                                       kj  k
                                                   k  1

                                                   m    p
                                                 =   A    B
                                                      kj   ik  i
                                                   k  1  i  1
                                                   p   m
                                                 =       B A kj  i
                                                          ik
                                                   i  1  k  1
                                   so that we must have

                                               m
                                          C     B A  .
                                           ij    ik  kj                                                    ...(4)
                                              k  1







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