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












3














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





























    3














    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



























      3












      3








      3


      1





      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















      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♦

      301k32425755




      301k32425755










      asked Dec 28 '18 at 16:23









      Miguel Ferreira

      664




      664






















          4 Answers
          4






          active

          oldest

          votes


















          9














          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































            13














            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























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



















            4














            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





















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



















            2














            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





















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











            Your Answer





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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            9














            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




























              9














              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


























                9












                9








                9






                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














                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 2 days ago

























                answered Dec 28 '18 at 16:27









                José Carlos Santos

                150k22121221




                150k22121221























                    13














                    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























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
















                    13














                    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























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














                    13












                    13








                    13






                    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














                    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









                    Yanko

                    6,124724




                    6,124724












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


















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
















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




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











                    4














                    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





















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
















                    4














                    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





















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














                    4












                    4








                    4






                    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












                    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









                    gt6989b

                    33k22452




                    33k22452












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


















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
















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




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











                    2














                    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





















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
















                    2














                    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





















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














                    2












                    2








                    2






                    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












                    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









                    mechanodroid

                    26.2k62245




                    26.2k62245












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


















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
















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




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


















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