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96            Chapter  Four:  Introduction  to  Vector  Spaces

                (v)  cd(v)  =  c(dv);
                (vi)  c(u  +  v)  =  cu  +  cv  :  (distributive  law);
                (vii)  (c +  d)v  —  cv + dv;  (distributive  law);
                (viii)  lv  =  v.

           For  economy   of  notation  it  is  customary  to  use  V  to  denote
           the  set  of vectors,  as  well as the  vector  space.  Since the  vector
           space  axioms  just  listed  hold  for  matrices,  they  are  valid  in
             n
           R ;   they  also  hold  in  the  other  examples  of  vector  spaces
           described  in 4.1.
                More  generally,  we  can  define  a  vector  space  over  an  ar-
            bitrary  field of  scalars  F  by  simply  replacing  R  by  F  in  the
           above  axioms.
                Certain  simple  properties  of  vector  spaces  follow  easily
           from  the  axioms.  Since these  are  used  constantly,  it  is as  well
           to  establish  them  at  this  early  stage.
           Lemma     4.2.1
           If  u  and  v  are  vectors  in  a  vector  space,  the  following  state-
            ments  are  true:
                (a)  Ov =  0  and  c  0  =  0  where  c  is  a  scalar;
                (b)  if  u  +  v  =  0,  then  u  =  —v;
                (c)  ( - l ) v = - v .

            Proof
            (a)  In  property  (vii)  above  put  c  =  0  =  d,  to  get  Ov =  Ov  +
            Ov.  Add  — (Ov)   to  both  sides  of  this  equation  and  use  the
            associative  law  (ii)  to  deduce  that


                        0  =  -(Ov)  +  Ov =  (-(Ov)  +  Ov) +  Ov,

            which  leads  to  0  =  Ov.  Proceed  similarly  in the  second  part.
            (b)  Add  —v to both  sides  of  u + v  =  0 and  use the  associative
            law.
            (c)  Using  (vii)  and  (viii),  and  also  (a),  we  obtain


                                     (
                                                    (
                                l
                      (
                 v  + -l)v   = v   + -l)v   =  (1 + -l))v   =  Ov =  0.
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