Show that $(lambda v= vec{0})Rightarrow lambda=0$ or $v=vec{0}$ Vector-space












-1












$begingroup$


If $lambda neq 0$ then $v= 1v=(lambdalambda^{-1})v=lambda v lambda^{-1}v=vec{0}lambda^{-1}v=vec{0}$



Since $vec{0}x=vec{0},forall xin V$



But what if $vneq vec{0}$






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    If a scalar multiple of a nonzero vector is the zero vector, the scalar must be $0$
    $endgroup$
    – pwerth
    Jan 16 at 20:30










  • $begingroup$
    Yes I want to prove that, but I don't know how.
    $endgroup$
    – RM777
    Jan 16 at 20:32
















-1












$begingroup$


If $lambda neq 0$ then $v= 1v=(lambdalambda^{-1})v=lambda v lambda^{-1}v=vec{0}lambda^{-1}v=vec{0}$



Since $vec{0}x=vec{0},forall xin V$



But what if $vneq vec{0}$






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    If a scalar multiple of a nonzero vector is the zero vector, the scalar must be $0$
    $endgroup$
    – pwerth
    Jan 16 at 20:30










  • $begingroup$
    Yes I want to prove that, but I don't know how.
    $endgroup$
    – RM777
    Jan 16 at 20:32














-1












-1








-1





$begingroup$


If $lambda neq 0$ then $v= 1v=(lambdalambda^{-1})v=lambda v lambda^{-1}v=vec{0}lambda^{-1}v=vec{0}$



Since $vec{0}x=vec{0},forall xin V$



But what if $vneq vec{0}$






linear-algebra







share|cite|improve this question









$endgroup$




If $lambda neq 0$ then $v= 1v=(lambdalambda^{-1})v=lambda v lambda^{-1}v=vec{0}lambda^{-1}v=vec{0}$



Since $vec{0}x=vec{0},forall xin V$



But what if $vneq vec{0}$






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 20:26









RM777RM777

38312




38312












  • $begingroup$
    If a scalar multiple of a nonzero vector is the zero vector, the scalar must be $0$
    $endgroup$
    – pwerth
    Jan 16 at 20:30










  • $begingroup$
    Yes I want to prove that, but I don't know how.
    $endgroup$
    – RM777
    Jan 16 at 20:32


















  • $begingroup$
    If a scalar multiple of a nonzero vector is the zero vector, the scalar must be $0$
    $endgroup$
    – pwerth
    Jan 16 at 20:30










  • $begingroup$
    Yes I want to prove that, but I don't know how.
    $endgroup$
    – RM777
    Jan 16 at 20:32
















$begingroup$
If a scalar multiple of a nonzero vector is the zero vector, the scalar must be $0$
$endgroup$
– pwerth
Jan 16 at 20:30




$begingroup$
If a scalar multiple of a nonzero vector is the zero vector, the scalar must be $0$
$endgroup$
– pwerth
Jan 16 at 20:30












$begingroup$
Yes I want to prove that, but I don't know how.
$endgroup$
– RM777
Jan 16 at 20:32




$begingroup$
Yes I want to prove that, but I don't know how.
$endgroup$
– RM777
Jan 16 at 20:32










4 Answers
4






active

oldest

votes


















1












$begingroup$

To show: $$alpha cdot x=0 Rightarrow alpha = 0 or x=0$$




Let $alpha cdot x=0$



If $alpha = 0$ we are done



So let $alpha neq 0$



Then $alpha^{-1}$ exists and



$x=1cdot x=(alpha^{-1}cdotalpha)cdot x=alpha^{-1}cdot(alphacdot x)=alpha^{-1}cdot 0=0$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Suppose that $lambda vec v = vec 0$.



    If $lambda not= 0$ then $vec v = frac{1}lambda lambda vec 0 = vec 0$.



    Otherwise $lambda = 0$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Your proof that $lambdaneq0implies vec v=vec 0$ is enough. It shows that $lambda$ and $vec v$ can't be non-zero simultaneously, and that's all we care about.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        If $v neq vec{0}$ then there exists at least one $v_ineq 0$. We have $lambdacdot v_i=0$, which means $lambda=0$.






        share|cite|improve this answer









        $endgroup$














          Your Answer





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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          To show: $$alpha cdot x=0 Rightarrow alpha = 0 or x=0$$




          Let $alpha cdot x=0$



          If $alpha = 0$ we are done



          So let $alpha neq 0$



          Then $alpha^{-1}$ exists and



          $x=1cdot x=(alpha^{-1}cdotalpha)cdot x=alpha^{-1}cdot(alphacdot x)=alpha^{-1}cdot 0=0$






          share|cite|improve this answer











          $endgroup$


















            1












            $begingroup$

            To show: $$alpha cdot x=0 Rightarrow alpha = 0 or x=0$$




            Let $alpha cdot x=0$



            If $alpha = 0$ we are done



            So let $alpha neq 0$



            Then $alpha^{-1}$ exists and



            $x=1cdot x=(alpha^{-1}cdotalpha)cdot x=alpha^{-1}cdot(alphacdot x)=alpha^{-1}cdot 0=0$






            share|cite|improve this answer











            $endgroup$
















              1












              1








              1





              $begingroup$

              To show: $$alpha cdot x=0 Rightarrow alpha = 0 or x=0$$




              Let $alpha cdot x=0$



              If $alpha = 0$ we are done



              So let $alpha neq 0$



              Then $alpha^{-1}$ exists and



              $x=1cdot x=(alpha^{-1}cdotalpha)cdot x=alpha^{-1}cdot(alphacdot x)=alpha^{-1}cdot 0=0$






              share|cite|improve this answer











              $endgroup$



              To show: $$alpha cdot x=0 Rightarrow alpha = 0 or x=0$$




              Let $alpha cdot x=0$



              If $alpha = 0$ we are done



              So let $alpha neq 0$



              Then $alpha^{-1}$ exists and



              $x=1cdot x=(alpha^{-1}cdotalpha)cdot x=alpha^{-1}cdot(alphacdot x)=alpha^{-1}cdot 0=0$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 16 at 20:46









              RM777

              38312




              38312










              answered Jan 16 at 20:43









              s0ulr3aper07s0ulr3aper07

              683112




              683112























                  1












                  $begingroup$

                  Suppose that $lambda vec v = vec 0$.



                  If $lambda not= 0$ then $vec v = frac{1}lambda lambda vec 0 = vec 0$.



                  Otherwise $lambda = 0$.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Suppose that $lambda vec v = vec 0$.



                    If $lambda not= 0$ then $vec v = frac{1}lambda lambda vec 0 = vec 0$.



                    Otherwise $lambda = 0$.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Suppose that $lambda vec v = vec 0$.



                      If $lambda not= 0$ then $vec v = frac{1}lambda lambda vec 0 = vec 0$.



                      Otherwise $lambda = 0$.






                      share|cite|improve this answer









                      $endgroup$



                      Suppose that $lambda vec v = vec 0$.



                      If $lambda not= 0$ then $vec v = frac{1}lambda lambda vec 0 = vec 0$.



                      Otherwise $lambda = 0$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 16 at 20:34









                      Umberto P.Umberto P.

                      40.4k13370




                      40.4k13370























                          1












                          $begingroup$

                          Your proof that $lambdaneq0implies vec v=vec 0$ is enough. It shows that $lambda$ and $vec v$ can't be non-zero simultaneously, and that's all we care about.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            Your proof that $lambdaneq0implies vec v=vec 0$ is enough. It shows that $lambda$ and $vec v$ can't be non-zero simultaneously, and that's all we care about.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              Your proof that $lambdaneq0implies vec v=vec 0$ is enough. It shows that $lambda$ and $vec v$ can't be non-zero simultaneously, and that's all we care about.






                              share|cite|improve this answer









                              $endgroup$



                              Your proof that $lambdaneq0implies vec v=vec 0$ is enough. It shows that $lambda$ and $vec v$ can't be non-zero simultaneously, and that's all we care about.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 16 at 20:34









                              ArthurArthur

                              122k7122211




                              122k7122211























                                  0












                                  $begingroup$

                                  If $v neq vec{0}$ then there exists at least one $v_ineq 0$. We have $lambdacdot v_i=0$, which means $lambda=0$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    If $v neq vec{0}$ then there exists at least one $v_ineq 0$. We have $lambdacdot v_i=0$, which means $lambda=0$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      If $v neq vec{0}$ then there exists at least one $v_ineq 0$. We have $lambdacdot v_i=0$, which means $lambda=0$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      If $v neq vec{0}$ then there exists at least one $v_ineq 0$. We have $lambdacdot v_i=0$, which means $lambda=0$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 16 at 20:33









                                      EuxhenHEuxhenH

                                      536212




                                      536212






























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