Show that if ${vec{v_1},vec{v_2},vec{v_3}}$ is a linearly independent set, then ${vec{v_1},vec{v_2}}$ is also...












1












$begingroup$


How do you show that if ${vec{v_1},vec{v_2},vec{v_3}}$ is a linearly independent set, then ${vec{v_1},vec{v_2}}$ is also linearly independent.



I need to explain this in complete sentences. I understand how to show linear independence for specific equations but I don't have a high enough level of understand of linear independence to be able to explain this more broadly.






linear-algebra







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$endgroup$








  • 3




    $begingroup$
    Can you state the definitions of "${v_1, v_2, v_3}$ is a linearly independent set" and "${v_1, v_2}$ is a linearly independent set"? If you write out the definitions clearly, you should be able to finish the question in one step.
    $endgroup$
    – angryavian
    Jan 21 at 22:28


















1












$begingroup$


How do you show that if ${vec{v_1},vec{v_2},vec{v_3}}$ is a linearly independent set, then ${vec{v_1},vec{v_2}}$ is also linearly independent.



I need to explain this in complete sentences. I understand how to show linear independence for specific equations but I don't have a high enough level of understand of linear independence to be able to explain this more broadly.






linear-algebra







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Can you state the definitions of "${v_1, v_2, v_3}$ is a linearly independent set" and "${v_1, v_2}$ is a linearly independent set"? If you write out the definitions clearly, you should be able to finish the question in one step.
    $endgroup$
    – angryavian
    Jan 21 at 22:28
















1












1








1





$begingroup$


How do you show that if ${vec{v_1},vec{v_2},vec{v_3}}$ is a linearly independent set, then ${vec{v_1},vec{v_2}}$ is also linearly independent.



I need to explain this in complete sentences. I understand how to show linear independence for specific equations but I don't have a high enough level of understand of linear independence to be able to explain this more broadly.






linear-algebra







share|cite|improve this question











$endgroup$




How do you show that if ${vec{v_1},vec{v_2},vec{v_3}}$ is a linearly independent set, then ${vec{v_1},vec{v_2}}$ is also linearly independent.



I need to explain this in complete sentences. I understand how to show linear independence for specific equations but I don't have a high enough level of understand of linear independence to be able to explain this more broadly.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 22:40









idriskameni

685319




685319










asked Jan 21 at 22:25









hoya2021hoya2021

265




265








  • 3




    $begingroup$
    Can you state the definitions of "${v_1, v_2, v_3}$ is a linearly independent set" and "${v_1, v_2}$ is a linearly independent set"? If you write out the definitions clearly, you should be able to finish the question in one step.
    $endgroup$
    – angryavian
    Jan 21 at 22:28
















  • 3




    $begingroup$
    Can you state the definitions of "${v_1, v_2, v_3}$ is a linearly independent set" and "${v_1, v_2}$ is a linearly independent set"? If you write out the definitions clearly, you should be able to finish the question in one step.
    $endgroup$
    – angryavian
    Jan 21 at 22:28










3




3




$begingroup$
Can you state the definitions of "${v_1, v_2, v_3}$ is a linearly independent set" and "${v_1, v_2}$ is a linearly independent set"? If you write out the definitions clearly, you should be able to finish the question in one step.
$endgroup$
– angryavian
Jan 21 at 22:28






$begingroup$
Can you state the definitions of "${v_1, v_2, v_3}$ is a linearly independent set" and "${v_1, v_2}$ is a linearly independent set"? If you write out the definitions clearly, you should be able to finish the question in one step.
$endgroup$
– angryavian
Jan 21 at 22:28












3 Answers
3






active

oldest

votes


















4












$begingroup$

Suppose that ${v_1,v_2}$ is a linearly dependent set. That means by definition that...




...there is some set of constants $c_1,c_2$ where at least one of $c_1,c_2$ is nonzero such that $c_1v_1 + c_2v_2 = 0$.




Then when considering whether or not ${v_1,v_2,v_3}$ is a linearly independent set or not we see that...




...by using the same $c_1,c_2$ as before and letting $c_3=0$ we have shown that there is an example of $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3v_3=0$ while at least one of $c_1,c_2,c_3$ is nonzero.




So we have shown that ${v_1,v_2}$ linearly dependent implies that ${v_1,v_2,v_3}$ is linearly dependent as well. By contraposition, this is equivalent to have shown that ${v_1,v_2,v_3}$ being linearly independent implies ${v_1,v_2}$ is linearly independent as well.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    IIRC,
    if $v_{1,2,3}$
    are linearly independent then
    $sum a_iv_i = 0
    implies
    a_{1,2,3} = 0
    $    .



    If
    $v_1, v_2$
    are not linearly independent,
    there exist $a, b$
    with at least one of
    $a, b$ nonzero such that
    $av_1+bv_2 =0$.



    Therefore
    $0
    =av_1+bv_2 +0v_3$
    so that
    $v_{1, 2, 3}$
    are not linearly independent.



    Contradiction with assumption.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Here's a short direct proof, without unnecessary negation:
      (equivalent to the other answers)



      Suppose
      $a_1 v_1 + a_2 v_2 =0$.
      We need to show $a_1=a_2=0$.
      Since $0v_3=0$, we also have
      $a_1 v_1 + a_2 v_2 + 0 v_3 =0$.
      Since $v_1,v_2,v_3$ are linearly independent, $a_1=a_2=0$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

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        4












        $begingroup$

        Suppose that ${v_1,v_2}$ is a linearly dependent set. That means by definition that...




        ...there is some set of constants $c_1,c_2$ where at least one of $c_1,c_2$ is nonzero such that $c_1v_1 + c_2v_2 = 0$.




        Then when considering whether or not ${v_1,v_2,v_3}$ is a linearly independent set or not we see that...




        ...by using the same $c_1,c_2$ as before and letting $c_3=0$ we have shown that there is an example of $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3v_3=0$ while at least one of $c_1,c_2,c_3$ is nonzero.




        So we have shown that ${v_1,v_2}$ linearly dependent implies that ${v_1,v_2,v_3}$ is linearly dependent as well. By contraposition, this is equivalent to have shown that ${v_1,v_2,v_3}$ being linearly independent implies ${v_1,v_2}$ is linearly independent as well.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          Suppose that ${v_1,v_2}$ is a linearly dependent set. That means by definition that...




          ...there is some set of constants $c_1,c_2$ where at least one of $c_1,c_2$ is nonzero such that $c_1v_1 + c_2v_2 = 0$.




          Then when considering whether or not ${v_1,v_2,v_3}$ is a linearly independent set or not we see that...




          ...by using the same $c_1,c_2$ as before and letting $c_3=0$ we have shown that there is an example of $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3v_3=0$ while at least one of $c_1,c_2,c_3$ is nonzero.




          So we have shown that ${v_1,v_2}$ linearly dependent implies that ${v_1,v_2,v_3}$ is linearly dependent as well. By contraposition, this is equivalent to have shown that ${v_1,v_2,v_3}$ being linearly independent implies ${v_1,v_2}$ is linearly independent as well.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            Suppose that ${v_1,v_2}$ is a linearly dependent set. That means by definition that...




            ...there is some set of constants $c_1,c_2$ where at least one of $c_1,c_2$ is nonzero such that $c_1v_1 + c_2v_2 = 0$.




            Then when considering whether or not ${v_1,v_2,v_3}$ is a linearly independent set or not we see that...




            ...by using the same $c_1,c_2$ as before and letting $c_3=0$ we have shown that there is an example of $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3v_3=0$ while at least one of $c_1,c_2,c_3$ is nonzero.




            So we have shown that ${v_1,v_2}$ linearly dependent implies that ${v_1,v_2,v_3}$ is linearly dependent as well. By contraposition, this is equivalent to have shown that ${v_1,v_2,v_3}$ being linearly independent implies ${v_1,v_2}$ is linearly independent as well.






            share|cite|improve this answer









            $endgroup$



            Suppose that ${v_1,v_2}$ is a linearly dependent set. That means by definition that...




            ...there is some set of constants $c_1,c_2$ where at least one of $c_1,c_2$ is nonzero such that $c_1v_1 + c_2v_2 = 0$.




            Then when considering whether or not ${v_1,v_2,v_3}$ is a linearly independent set or not we see that...




            ...by using the same $c_1,c_2$ as before and letting $c_3=0$ we have shown that there is an example of $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3v_3=0$ while at least one of $c_1,c_2,c_3$ is nonzero.




            So we have shown that ${v_1,v_2}$ linearly dependent implies that ${v_1,v_2,v_3}$ is linearly dependent as well. By contraposition, this is equivalent to have shown that ${v_1,v_2,v_3}$ being linearly independent implies ${v_1,v_2}$ is linearly independent as well.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 21 at 22:38









            JMoravitzJMoravitz

            47.9k33886




            47.9k33886























                0












                $begingroup$

                IIRC,
                if $v_{1,2,3}$
                are linearly independent then
                $sum a_iv_i = 0
                implies
                a_{1,2,3} = 0
                $                .



                If
                $v_1, v_2$
                are not linearly independent,
                there exist $a, b$
                with at least one of
                $a, b$ nonzero such that
                $av_1+bv_2 =0$.



                Therefore
                $0
                =av_1+bv_2 +0v_3$
                so that
                $v_{1, 2, 3}$
                are not linearly independent.



                Contradiction with assumption.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  IIRC,
                  if $v_{1,2,3}$
                  are linearly independent then
                  $sum a_iv_i = 0
                  implies
                  a_{1,2,3} = 0
                  $                  .



                  If
                  $v_1, v_2$
                  are not linearly independent,
                  there exist $a, b$
                  with at least one of
                  $a, b$ nonzero such that
                  $av_1+bv_2 =0$.



                  Therefore
                  $0
                  =av_1+bv_2 +0v_3$
                  so that
                  $v_{1, 2, 3}$
                  are not linearly independent.



                  Contradiction with assumption.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    IIRC,
                    if $v_{1,2,3}$
                    are linearly independent then
                    $sum a_iv_i = 0
                    implies
                    a_{1,2,3} = 0
                    $                    .



                    If
                    $v_1, v_2$
                    are not linearly independent,
                    there exist $a, b$
                    with at least one of
                    $a, b$ nonzero such that
                    $av_1+bv_2 =0$.



                    Therefore
                    $0
                    =av_1+bv_2 +0v_3$
                    so that
                    $v_{1, 2, 3}$
                    are not linearly independent.



                    Contradiction with assumption.






                    share|cite|improve this answer









                    $endgroup$



                    IIRC,
                    if $v_{1,2,3}$
                    are linearly independent then
                    $sum a_iv_i = 0
                    implies
                    a_{1,2,3} = 0
                    $                    .



                    If
                    $v_1, v_2$
                    are not linearly independent,
                    there exist $a, b$
                    with at least one of
                    $a, b$ nonzero such that
                    $av_1+bv_2 =0$.



                    Therefore
                    $0
                    =av_1+bv_2 +0v_3$
                    so that
                    $v_{1, 2, 3}$
                    are not linearly independent.



                    Contradiction with assumption.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 21 at 22:37









                    marty cohenmarty cohen

                    74k549128




                    74k549128























                        0












                        $begingroup$

                        Here's a short direct proof, without unnecessary negation:
                        (equivalent to the other answers)



                        Suppose
                        $a_1 v_1 + a_2 v_2 =0$.
                        We need to show $a_1=a_2=0$.
                        Since $0v_3=0$, we also have
                        $a_1 v_1 + a_2 v_2 + 0 v_3 =0$.
                        Since $v_1,v_2,v_3$ are linearly independent, $a_1=a_2=0$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Here's a short direct proof, without unnecessary negation:
                          (equivalent to the other answers)



                          Suppose
                          $a_1 v_1 + a_2 v_2 =0$.
                          We need to show $a_1=a_2=0$.
                          Since $0v_3=0$, we also have
                          $a_1 v_1 + a_2 v_2 + 0 v_3 =0$.
                          Since $v_1,v_2,v_3$ are linearly independent, $a_1=a_2=0$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Here's a short direct proof, without unnecessary negation:
                            (equivalent to the other answers)



                            Suppose
                            $a_1 v_1 + a_2 v_2 =0$.
                            We need to show $a_1=a_2=0$.
                            Since $0v_3=0$, we also have
                            $a_1 v_1 + a_2 v_2 + 0 v_3 =0$.
                            Since $v_1,v_2,v_3$ are linearly independent, $a_1=a_2=0$.






                            share|cite|improve this answer









                            $endgroup$



                            Here's a short direct proof, without unnecessary negation:
                            (equivalent to the other answers)



                            Suppose
                            $a_1 v_1 + a_2 v_2 =0$.
                            We need to show $a_1=a_2=0$.
                            Since $0v_3=0$, we also have
                            $a_1 v_1 + a_2 v_2 + 0 v_3 =0$.
                            Since $v_1,v_2,v_3$ are linearly independent, $a_1=a_2=0$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 29 at 20:14









                            ffffforallffffforall

                            36028




                            36028






























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