Complete the set to get a base












2














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





























    2














    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



























      2












      2








      2







      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 2 hours ago









      gt6989b

      33k22452




      33k22452










      asked 2 hours ago









      Miguel Ferreira

      614




      614






















          4 Answers
          4






          active

          oldest

          votes


















          3














          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. Actually any vector $(a,b,c)$ such that $bneq a+c$ will do.






          share|cite|improve this answer





















          • Thank you for the help!
            – Miguel Ferreira
            1 hour ago










          • I'm glad I could help.
            – José Carlos Santos
            1 hour ago



















          4














          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 the is necessarily an 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 then $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. 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
            1 hour ago



















          3














          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
            1 hour ago



















          1














          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
            1 hour ago











          Your Answer





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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          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. Actually any vector $(a,b,c)$ such that $bneq a+c$ will do.






          share|cite|improve this answer





















          • Thank you for the help!
            – Miguel Ferreira
            1 hour ago










          • I'm glad I could help.
            – José Carlos Santos
            1 hour ago
















          3














          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. Actually any vector $(a,b,c)$ such that $bneq a+c$ will do.






          share|cite|improve this answer





















          • Thank you for the help!
            – Miguel Ferreira
            1 hour ago










          • I'm glad I could help.
            – José Carlos Santos
            1 hour ago














          3












          3








          3






          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. Actually 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. Actually 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










          answered 2 hours ago









          José Carlos Santos

          149k22117219




          149k22117219












          • Thank you for the help!
            – Miguel Ferreira
            1 hour ago










          • I'm glad I could help.
            – José Carlos Santos
            1 hour ago


















          • Thank you for the help!
            – Miguel Ferreira
            1 hour ago










          • I'm glad I could help.
            – José Carlos Santos
            1 hour ago
















          Thank you for the help!
          – Miguel Ferreira
          1 hour ago




          Thank you for the help!
          – Miguel Ferreira
          1 hour ago












          I'm glad I could help.
          – José Carlos Santos
          1 hour ago




          I'm glad I could help.
          – José Carlos Santos
          1 hour ago











          4














          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 the is necessarily an 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 then $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. 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
            1 hour ago
















          4














          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 the is necessarily an 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 then $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. 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
            1 hour ago














          4












          4








          4






          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 the is necessarily an 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 then $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. 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 the is necessarily an 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 then $e_1,e_2,e_3 in text{Span} {v_1,v_2}$. 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 2 hours ago

























          answered 2 hours ago









          Yanko

          5,841723




          5,841723












          • Thank you very much for the help!
            – Miguel Ferreira
            1 hour ago


















          • Thank you very much for the help!
            – Miguel Ferreira
            1 hour ago
















          Thank you very much for the help!
          – Miguel Ferreira
          1 hour ago




          Thank you very much for the help!
          – Miguel Ferreira
          1 hour ago











          3














          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
            1 hour ago
















          3














          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
            1 hour ago














          3












          3








          3






          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 2 hours ago









          gt6989b

          33k22452




          33k22452












          • Thank you for the help!:)
            – Miguel Ferreira
            1 hour ago


















          • Thank you for the help!:)
            – Miguel Ferreira
            1 hour ago
















          Thank you for the help!:)
          – Miguel Ferreira
          1 hour ago




          Thank you for the help!:)
          – Miguel Ferreira
          1 hour ago











          1














          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
            1 hour ago
















          1














          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
            1 hour ago














          1












          1








          1






          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 2 hours ago









          mechanodroid

          25.9k62245




          25.9k62245












          • Thank you for the help! Didn't know about this method :)
            – Miguel Ferreira
            1 hour ago


















          • Thank you for the help! Didn't know about this method :)
            – Miguel Ferreira
            1 hour ago
















          Thank you for the help! Didn't know about this method :)
          – Miguel Ferreira
          1 hour ago




          Thank you for the help! Didn't know about this method :)
          – Miguel Ferreira
          1 hour ago


















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