Showing if a vector in a span is a linear combination of the other vectors, the other vectors still span the...












1












$begingroup$


lets say we have a subspace $W$ and that three vectors, $w1, w2$ and $w3$ span $W$.
I want to show that if $w3$ is a linear combination of $w1$ and $w2$, that $w1$ and $w2$ still span $W$



My attempt so far:
so we can write $w3$ = $(c1*w1)$ + $(c2*w2)$ for $c1,c2$ not all $0$



I'm stuck at this point and can't think of how to prove the first statement, any suggestions?



edit:
question






linear-algebra







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    lets say we have a subspace $W$ and that three vectors, $w1, w2$ and $w3$ span $W$.
    I want to show that if $w3$ is a linear combination of $w1$ and $w2$, that $w1$ and $w2$ still span $W$



    My attempt so far:
    so we can write $w3$ = $(c1*w1)$ + $(c2*w2)$ for $c1,c2$ not all $0$



    I'm stuck at this point and can't think of how to prove the first statement, any suggestions?



    edit:
    question






    linear-algebra







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      lets say we have a subspace $W$ and that three vectors, $w1, w2$ and $w3$ span $W$.
      I want to show that if $w3$ is a linear combination of $w1$ and $w2$, that $w1$ and $w2$ still span $W$



      My attempt so far:
      so we can write $w3$ = $(c1*w1)$ + $(c2*w2)$ for $c1,c2$ not all $0$



      I'm stuck at this point and can't think of how to prove the first statement, any suggestions?



      edit:
      question






      linear-algebra







      share|cite|improve this question











      $endgroup$




      lets say we have a subspace $W$ and that three vectors, $w1, w2$ and $w3$ span $W$.
      I want to show that if $w3$ is a linear combination of $w1$ and $w2$, that $w1$ and $w2$ still span $W$



      My attempt so far:
      so we can write $w3$ = $(c1*w1)$ + $(c2*w2)$ for $c1,c2$ not all $0$



      I'm stuck at this point and can't think of how to prove the first statement, any suggestions?



      edit:
      question






      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 Dec 14 '18 at 5:47







      lohboys

















      asked Dec 14 '18 at 4:27









      lohboyslohboys

      9819




      9819






















          1 Answer
          1






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          1












          $begingroup$

          Take $xin W$.



          Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:46












          • $begingroup$
            im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:48










          • $begingroup$
            They may not be linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 5:14








          • 1




            $begingroup$
            so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
            $endgroup$
            – lohboys
            Dec 14 '18 at 5:26








          • 1




            $begingroup$
            The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 6:16












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          $begingroup$

          Take $xin W$.



          Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:46












          • $begingroup$
            im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:48










          • $begingroup$
            They may not be linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 5:14








          • 1




            $begingroup$
            so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
            $endgroup$
            – lohboys
            Dec 14 '18 at 5:26








          • 1




            $begingroup$
            The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 6:16
















          1












          $begingroup$

          Take $xin W$.



          Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:46












          • $begingroup$
            im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:48










          • $begingroup$
            They may not be linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 5:14








          • 1




            $begingroup$
            so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
            $endgroup$
            – lohboys
            Dec 14 '18 at 5:26








          • 1




            $begingroup$
            The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 6:16














          1












          1








          1





          $begingroup$

          Take $xin W$.



          Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.






          share|cite|improve this answer









          $endgroup$



          Take $xin W$.



          Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 4:42









          Chris CusterChris Custer

          14.3k3827




          14.3k3827












          • $begingroup$
            thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:46












          • $begingroup$
            im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:48










          • $begingroup$
            They may not be linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 5:14








          • 1




            $begingroup$
            so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
            $endgroup$
            – lohboys
            Dec 14 '18 at 5:26








          • 1




            $begingroup$
            The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 6:16


















          • $begingroup$
            thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:46












          • $begingroup$
            im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
            $endgroup$
            – lohboys
            Dec 14 '18 at 4:48










          • $begingroup$
            They may not be linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 5:14








          • 1




            $begingroup$
            so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
            $endgroup$
            – lohboys
            Dec 14 '18 at 5:26








          • 1




            $begingroup$
            The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
            $endgroup$
            – Chris Custer
            Dec 14 '18 at 6:16
















          $begingroup$
          thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
          $endgroup$
          – lohboys
          Dec 14 '18 at 4:46






          $begingroup$
          thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
          $endgroup$
          – lohboys
          Dec 14 '18 at 4:46














          $begingroup$
          im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
          $endgroup$
          – lohboys
          Dec 14 '18 at 4:48




          $begingroup$
          im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
          $endgroup$
          – lohboys
          Dec 14 '18 at 4:48












          $begingroup$
          They may not be linearly independent.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 5:14






          $begingroup$
          They may not be linearly independent.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 5:14






          1




          1




          $begingroup$
          so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
          $endgroup$
          – lohboys
          Dec 14 '18 at 5:26






          $begingroup$
          so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
          $endgroup$
          – lohboys
          Dec 14 '18 at 5:26






          1




          1




          $begingroup$
          The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 6:16




          $begingroup$
          The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 6:16


















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