Complete ${(2,3,1),(1,4,3)}$ to a basis of $Bbb R^3$












3












$begingroup$


Having the following set: $C = {(2,3,1),(1,4,3)}$ I want to be able to generate; $V = mathbb{R}^3$



Since $V = mathbb{R}^3$ has dimension $3$, $C$ also needs to have dimension $3$, however, how do I find the vector I need to add to $C$ in order to be able to generate $V$?



Is sufficient to have 3 vectors that are Linearly Independent and then I can generate $V$? And how do I find it?



Thank you!






linear-algebra vector-spaces cross-product







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Having the following set: $C = {(2,3,1),(1,4,3)}$ I want to be able to generate; $V = mathbb{R}^3$



    Since $V = mathbb{R}^3$ has dimension $3$, $C$ also needs to have dimension $3$, however, how do I find the vector I need to add to $C$ in order to be able to generate $V$?



    Is sufficient to have 3 vectors that are Linearly Independent and then I can generate $V$? And how do I find it?



    Thank you!






    linear-algebra vector-spaces cross-product







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Having the following set: $C = {(2,3,1),(1,4,3)}$ I want to be able to generate; $V = mathbb{R}^3$



      Since $V = mathbb{R}^3$ has dimension $3$, $C$ also needs to have dimension $3$, however, how do I find the vector I need to add to $C$ in order to be able to generate $V$?



      Is sufficient to have 3 vectors that are Linearly Independent and then I can generate $V$? And how do I find it?



      Thank you!






      linear-algebra vector-spaces cross-product







      share|cite|improve this question











      $endgroup$




      Having the following set: $C = {(2,3,1),(1,4,3)}$ I want to be able to generate; $V = mathbb{R}^3$



      Since $V = mathbb{R}^3$ has dimension $3$, $C$ also needs to have dimension $3$, however, how do I find the vector I need to add to $C$ in order to be able to generate $V$?



      Is sufficient to have 3 vectors that are Linearly Independent and then I can generate $V$? And how do I find it?



      Thank you!






      linear-algebra vector-spaces cross-product




      linear-algebra vector-spaces cross-product






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 29 '18 at 12:00









      Asaf Karagila♦

      302k32427757




      302k32427757










      asked Dec 28 '18 at 16:23









      Miguel FerreiraMiguel Ferreira

      664




      664






















          4 Answers
          4






          active

          oldest

          votes


















          9












          $begingroup$

          Take any vector $(a,b,c)$ which is not a linear combination of $(2,3,1)$ and $(1,4,3)$. For instance $(1,0,0)$ will do. In fact, any vector $(a,b,c)$ such that $bneq a+c$ will do.






          share|cite|improve this answer











          $endgroup$





















            14












            $begingroup$

            This method works in the general case and does not require using the "cross product":



            Let $C=(v_1,v_2)$ as in the question. Choose any basis of $mathbb{R}^3$. For simplicity we shall take $e_1,e_2,e_3$. Then look at



            $(v_1,v_2,e_1)$, $(v_1,v_2,e_2)$ and $(v_1,v_2,e_3)$.



            One of them is necessarily a linearly independent sequence of vectors (this can be checked by hand, for example you can take $e_1$ as in Jose' answer).



            Indeed, if by contradiction all of them were linearly dependent then since $v_1,v_2$ are independent $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. But this would mean that $mathbb{R}^3 = text{Span} {v_1,v_2}$ which is impossible.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you very much for the help!
              $endgroup$
              – Miguel Ferreira
              Dec 28 '18 at 17:16



















            4












            $begingroup$

            HINT



            If you have 3 linearly independent vectors, they will span all of $mathbb{R}^3$ (why?).



            To get a third one, easiest way is to construct one perpendicular to them both, which is to take a cross product of the two you have






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for the help!:)
              $endgroup$
              – Miguel Ferreira
              Dec 28 '18 at 17:10



















            2












            $begingroup$

            Just take the cross product:
            $$(2,3,1) times (1,4,3) = (5,-5,5)$$



            Then ${(2,3,1),(1,4,3),(5,-5,5)}$ is linearly independent and hence a basis for $mathbb{R}^3$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for the help! Didn't know about this method :)
              $endgroup$
              – Miguel Ferreira
              Dec 28 '18 at 17:17











            Your Answer





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            4 Answers
            4






            active

            oldest

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            4 Answers
            4






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            9












            $begingroup$

            Take any vector $(a,b,c)$ which is not a linear combination of $(2,3,1)$ and $(1,4,3)$. For instance $(1,0,0)$ will do. In fact, any vector $(a,b,c)$ such that $bneq a+c$ will do.






            share|cite|improve this answer











            $endgroup$


















              9












              $begingroup$

              Take any vector $(a,b,c)$ which is not a linear combination of $(2,3,1)$ and $(1,4,3)$. For instance $(1,0,0)$ will do. In fact, any vector $(a,b,c)$ such that $bneq a+c$ will do.






              share|cite|improve this answer











              $endgroup$
















                9












                9








                9





                $begingroup$

                Take any vector $(a,b,c)$ which is not a linear combination of $(2,3,1)$ and $(1,4,3)$. For instance $(1,0,0)$ will do. In fact, any vector $(a,b,c)$ such that $bneq a+c$ will do.






                share|cite|improve this answer











                $endgroup$



                Take any vector $(a,b,c)$ which is not a linear combination of $(2,3,1)$ and $(1,4,3)$. For instance $(1,0,0)$ will do. In fact, any vector $(a,b,c)$ such that $bneq a+c$ will do.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 30 '18 at 10:37

























                answered Dec 28 '18 at 16:27









                José Carlos SantosJosé Carlos Santos

                153k22123225




                153k22123225























                    14












                    $begingroup$

                    This method works in the general case and does not require using the "cross product":



                    Let $C=(v_1,v_2)$ as in the question. Choose any basis of $mathbb{R}^3$. For simplicity we shall take $e_1,e_2,e_3$. Then look at



                    $(v_1,v_2,e_1)$, $(v_1,v_2,e_2)$ and $(v_1,v_2,e_3)$.



                    One of them is necessarily a linearly independent sequence of vectors (this can be checked by hand, for example you can take $e_1$ as in Jose' answer).



                    Indeed, if by contradiction all of them were linearly dependent then since $v_1,v_2$ are independent $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. But this would mean that $mathbb{R}^3 = text{Span} {v_1,v_2}$ which is impossible.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Thank you very much for the help!
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:16
















                    14












                    $begingroup$

                    This method works in the general case and does not require using the "cross product":



                    Let $C=(v_1,v_2)$ as in the question. Choose any basis of $mathbb{R}^3$. For simplicity we shall take $e_1,e_2,e_3$. Then look at



                    $(v_1,v_2,e_1)$, $(v_1,v_2,e_2)$ and $(v_1,v_2,e_3)$.



                    One of them is necessarily a linearly independent sequence of vectors (this can be checked by hand, for example you can take $e_1$ as in Jose' answer).



                    Indeed, if by contradiction all of them were linearly dependent then since $v_1,v_2$ are independent $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. But this would mean that $mathbb{R}^3 = text{Span} {v_1,v_2}$ which is impossible.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Thank you very much for the help!
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:16














                    14












                    14








                    14





                    $begingroup$

                    This method works in the general case and does not require using the "cross product":



                    Let $C=(v_1,v_2)$ as in the question. Choose any basis of $mathbb{R}^3$. For simplicity we shall take $e_1,e_2,e_3$. Then look at



                    $(v_1,v_2,e_1)$, $(v_1,v_2,e_2)$ and $(v_1,v_2,e_3)$.



                    One of them is necessarily a linearly independent sequence of vectors (this can be checked by hand, for example you can take $e_1$ as in Jose' answer).



                    Indeed, if by contradiction all of them were linearly dependent then since $v_1,v_2$ are independent $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. But this would mean that $mathbb{R}^3 = text{Span} {v_1,v_2}$ which is impossible.






                    share|cite|improve this answer











                    $endgroup$



                    This method works in the general case and does not require using the "cross product":



                    Let $C=(v_1,v_2)$ as in the question. Choose any basis of $mathbb{R}^3$. For simplicity we shall take $e_1,e_2,e_3$. Then look at



                    $(v_1,v_2,e_1)$, $(v_1,v_2,e_2)$ and $(v_1,v_2,e_3)$.



                    One of them is necessarily a linearly independent sequence of vectors (this can be checked by hand, for example you can take $e_1$ as in Jose' answer).



                    Indeed, if by contradiction all of them were linearly dependent then since $v_1,v_2$ are independent $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. But this would mean that $mathbb{R}^3 = text{Span} {v_1,v_2}$ which is impossible.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 28 '18 at 19:26

























                    answered Dec 28 '18 at 16:36









                    YankoYanko

                    6,5541728




                    6,5541728












                    • $begingroup$
                      Thank you very much for the help!
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:16


















                    • $begingroup$
                      Thank you very much for the help!
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:16
















                    $begingroup$
                    Thank you very much for the help!
                    $endgroup$
                    – Miguel Ferreira
                    Dec 28 '18 at 17:16




                    $begingroup$
                    Thank you very much for the help!
                    $endgroup$
                    – Miguel Ferreira
                    Dec 28 '18 at 17:16











                    4












                    $begingroup$

                    HINT



                    If you have 3 linearly independent vectors, they will span all of $mathbb{R}^3$ (why?).



                    To get a third one, easiest way is to construct one perpendicular to them both, which is to take a cross product of the two you have






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you for the help!:)
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:10
















                    4












                    $begingroup$

                    HINT



                    If you have 3 linearly independent vectors, they will span all of $mathbb{R}^3$ (why?).



                    To get a third one, easiest way is to construct one perpendicular to them both, which is to take a cross product of the two you have






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you for the help!:)
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:10














                    4












                    4








                    4





                    $begingroup$

                    HINT



                    If you have 3 linearly independent vectors, they will span all of $mathbb{R}^3$ (why?).



                    To get a third one, easiest way is to construct one perpendicular to them both, which is to take a cross product of the two you have






                    share|cite|improve this answer









                    $endgroup$



                    HINT



                    If you have 3 linearly independent vectors, they will span all of $mathbb{R}^3$ (why?).



                    To get a third one, easiest way is to construct one perpendicular to them both, which is to take a cross product of the two you have







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 28 '18 at 16:27









                    gt6989bgt6989b

                    33.3k22452




                    33.3k22452












                    • $begingroup$
                      Thank you for the help!:)
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:10


















                    • $begingroup$
                      Thank you for the help!:)
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:10
















                    $begingroup$
                    Thank you for the help!:)
                    $endgroup$
                    – Miguel Ferreira
                    Dec 28 '18 at 17:10




                    $begingroup$
                    Thank you for the help!:)
                    $endgroup$
                    – Miguel Ferreira
                    Dec 28 '18 at 17:10











                    2












                    $begingroup$

                    Just take the cross product:
                    $$(2,3,1) times (1,4,3) = (5,-5,5)$$



                    Then ${(2,3,1),(1,4,3),(5,-5,5)}$ is linearly independent and hence a basis for $mathbb{R}^3$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you for the help! Didn't know about this method :)
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:17
















                    2












                    $begingroup$

                    Just take the cross product:
                    $$(2,3,1) times (1,4,3) = (5,-5,5)$$



                    Then ${(2,3,1),(1,4,3),(5,-5,5)}$ is linearly independent and hence a basis for $mathbb{R}^3$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you for the help! Didn't know about this method :)
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:17














                    2












                    2








                    2





                    $begingroup$

                    Just take the cross product:
                    $$(2,3,1) times (1,4,3) = (5,-5,5)$$



                    Then ${(2,3,1),(1,4,3),(5,-5,5)}$ is linearly independent and hence a basis for $mathbb{R}^3$.






                    share|cite|improve this answer









                    $endgroup$



                    Just take the cross product:
                    $$(2,3,1) times (1,4,3) = (5,-5,5)$$



                    Then ${(2,3,1),(1,4,3),(5,-5,5)}$ is linearly independent and hence a basis for $mathbb{R}^3$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 28 '18 at 16:52









                    mechanodroidmechanodroid

                    27.1k62446




                    27.1k62446












                    • $begingroup$
                      Thank you for the help! Didn't know about this method :)
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:17


















                    • $begingroup$
                      Thank you for the help! Didn't know about this method :)
                      $endgroup$
                      – Miguel Ferreira
                      Dec 28 '18 at 17:17
















                    $begingroup$
                    Thank you for the help! Didn't know about this method :)
                    $endgroup$
                    – Miguel Ferreira
                    Dec 28 '18 at 17:17




                    $begingroup$
                    Thank you for the help! Didn't know about this method :)
                    $endgroup$
                    – Miguel Ferreira
                    Dec 28 '18 at 17:17


















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