Proving a set of vectors are a basis












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Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.



So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$



$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$



I'm uncertain how to show the combination of both vectors span $A$.










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    $begingroup$


    Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.



    So I have set $u = (a,b,c) in A$
    such that $a = frac{4c-2b}{2}$, then using this
    $u$ is now $(frac{4c-2b}{2},b,c)$



    $$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$



    I'm uncertain how to show the combination of both vectors span $A$.










    share|cite|improve this question











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      1












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      $begingroup$


      Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.



      So I have set $u = (a,b,c) in A$
      such that $a = frac{4c-2b}{2}$, then using this
      $u$ is now $(frac{4c-2b}{2},b,c)$



      $$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$



      I'm uncertain how to show the combination of both vectors span $A$.










      share|cite|improve this question











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      Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.



      So I have set $u = (a,b,c) in A$
      such that $a = frac{4c-2b}{2}$, then using this
      $u$ is now $(frac{4c-2b}{2},b,c)$



      $$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$



      I'm uncertain how to show the combination of both vectors span $A$.







      linear-algebra






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      edited Dec 21 '18 at 19:04









      hardmath

      29k95299




      29k95299










      asked Dec 15 '18 at 15:52









      Rito LoweRito Lowe

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          3 Answers
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          $begingroup$

          First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.



          To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.






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            $begingroup$

            You mean $a = frac{4c-3b}2$.



            If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.



            If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.



            Since they are not multiple of each other and you know the dimension is $2$, it is a basis.






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              $begingroup$

              Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.






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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.



                To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.



                  To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.



                    To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.






                    share|cite|improve this answer









                    $endgroup$



                    First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.



                    To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 15 '18 at 15:56









                    user3482749user3482749

                    4,266919




                    4,266919























                        1












                        $begingroup$

                        You mean $a = frac{4c-3b}2$.



                        If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.



                        If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.



                        Since they are not multiple of each other and you know the dimension is $2$, it is a basis.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You mean $a = frac{4c-3b}2$.



                          If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.



                          If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.



                          Since they are not multiple of each other and you know the dimension is $2$, it is a basis.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You mean $a = frac{4c-3b}2$.



                            If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.



                            If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.



                            Since they are not multiple of each other and you know the dimension is $2$, it is a basis.






                            share|cite|improve this answer









                            $endgroup$



                            You mean $a = frac{4c-3b}2$.



                            If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.



                            If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.



                            Since they are not multiple of each other and you know the dimension is $2$, it is a basis.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 15 '18 at 15:57









                            Siong Thye GohSiong Thye Goh

                            101k1466118




                            101k1466118























                                1












                                $begingroup$

                                Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 15 '18 at 16:01









                                    Sameer BahetiSameer Baheti

                                    5168




                                    5168






























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