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Unit 7: Algebra of Linear Transformation




          Theorem 8: Let T be a linear transformation on v  to w  the following statements are equivalent.  Notes
                                                 n   n
          1.   T is non-singular
          2.   For all  ,     v , if  T =  T, then   =  .
                           n
          3.   K  = [ ]
                T
          4.   v(T) = 0
          5.   T is onto, that is, the range of T is w  i.e. p(T) = n.
                                           n
          6.   T maps any basis for v  onto a basis for w .
                                 n              n
          Proof: Let n = dim v = din w. Now

                 rank (T) + nullity (T) = n
          Since T is non-singular if and only if nullity (T) = 0 and rank (T) = n. Therefore T is non-singular
          if and only if T(v ) = w . So, if either condition (1) or (2) holds the other is satisfied as well and T
                        n   n
          is invertible.
          The above equations are also equivalent, there is some basis ( ,  ,  ) for v such that (T , T ,
                                                            1  2  n               1  2
          ....,, T ) is basis for w.
                n
                                                                  2
               Example 10: Let  F be a field and let T be the linear operator on F  defined by
                                T (x , x ) = (x , x , x )
                                   1  2   1  2  1
          Then T is non-singular.
          Proof: If T is singular than T(x , x ) = 0, means we have
                                  1  2
                                 x  + x = 0
                                  1   2
                                     x = 0
                                      1
          so the solution is x  = 0, x  = 0. We also see that T is onto; for let (z , z ) be any vector in F .
                                                                                     2
                          1     2                                 1  2
          To show that (z , z ) is in the range of T we must find scalars z  and z  such that
                      1  2                                  1    2
                                 x  + x  = z
                                  1   2   1
                                     x = z
                                      1  2
                                                                               –1
          and the obvious solution is x , = z , x  = z  – z . This last result gives us an explicit for T , namely
                                 1   2  2  1  2
                                –1
                              T (x , x ) = (z , z  – z )
                                   1  2   2  1  2
          Self Assessment
                                             2
          4.   If T and U be the linear operator on R  defined by
               T(x , x ) = (x , x ) and U(x , x ) = (x , 0)
                  1  2   2  1      1  2   1
               give rules like the ones defining T and U for each of the transformations
               (i)  U + T

               (ii)  UT
               (iii)  TU








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