Find the formula of the linear transformation [closed]












1












$begingroup$


Consider the linear transformation $f:R^3rightarrow R^3$ with $f(0,1,2)=(2,1,1)$ and $ker f={(0,2,1),(1,0,1)}.$ Find:
(i) $f(0,1,0)$ (ii) the general formula of $f$.
Any help, please.






linear-transformations







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$endgroup$



closed as off-topic by Lee David Chung Lin, mrtaurho, José Carlos Santos, Adrian Keister, Theo Bendit Jan 31 at 15:06


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, mrtaurho, José Carlos Santos, Adrian Keister, Theo Bendit

If this question can be reworded to fit the rules in the help center, please edit the question.





















    1












    $begingroup$


    Consider the linear transformation $f:R^3rightarrow R^3$ with $f(0,1,2)=(2,1,1)$ and $ker f={(0,2,1),(1,0,1)}.$ Find:
    (i) $f(0,1,0)$ (ii) the general formula of $f$.
    Any help, please.






    linear-transformations







    share|cite|improve this question











    $endgroup$



    closed as off-topic by Lee David Chung Lin, mrtaurho, José Carlos Santos, Adrian Keister, Theo Bendit Jan 31 at 15:06


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, mrtaurho, José Carlos Santos, Adrian Keister, Theo Bendit

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      1












      1








      1


      2



      $begingroup$


      Consider the linear transformation $f:R^3rightarrow R^3$ with $f(0,1,2)=(2,1,1)$ and $ker f={(0,2,1),(1,0,1)}.$ Find:
      (i) $f(0,1,0)$ (ii) the general formula of $f$.
      Any help, please.






      linear-transformations







      share|cite|improve this question











      $endgroup$




      Consider the linear transformation $f:R^3rightarrow R^3$ with $f(0,1,2)=(2,1,1)$ and $ker f={(0,2,1),(1,0,1)}.$ Find:
      (i) $f(0,1,0)$ (ii) the general formula of $f$.
      Any help, please.






      linear-transformations




      linear-transformations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 28 at 15:15









      Adrian Keister

      5,26771933




      5,26771933










      asked Jan 20 at 18:07









      Jim ArtJim Art

      4719




      4719




      closed as off-topic by Lee David Chung Lin, mrtaurho, José Carlos Santos, Adrian Keister, Theo Bendit Jan 31 at 15:06


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, mrtaurho, José Carlos Santos, Adrian Keister, Theo Bendit

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Lee David Chung Lin, mrtaurho, José Carlos Santos, Adrian Keister, Theo Bendit Jan 31 at 15:06


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, mrtaurho, José Carlos Santos, Adrian Keister, Theo Bendit

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Now,
          $$
          (0,1,0)=frac{1}{3}big[2(0,2,1)-(0,1,2)big],
          $$
          hence
          $$f(0,1,0)
          =frac{1}{3}big[2 cdot f(0,2,1)-f(0,1,2)big]
          =frac{1}{3}big[2 cdot 0 -(2,1,1)big]
          =left(-frac{2}{3},-frac{1}{3},-frac{1}{3}right).
          $$
          Now,
          $$
          (0,0,1)
          =frac{1}{2}big[(0,1,2)-(0,1,0)big],
          $$
          hence
          $$
          f(0,0,1)
          =frac{1}{2}left[(2,1,1)-(-frac{2}{3},-frac{1}{3},-frac{1}{3})right]
          =left(frac{4}{3},frac{2}{3},frac{2}{3}right)
          $$
          Furthermore,
          $$
          (1,0,0)
          =(1,0,1)-(0,0,1),
          $$
          hence
          $$f(1,0,0)
          =f(1,0,1)-f(0,0,1)
          =-f(0,0,1)
          =left(-frac{4}{3},-frac{2}{3},-frac{2}{3}right).
          $$



          We can get the formula of $f$:
          $$
          f(x,y,z)
          = left(-frac{4}{3}x-frac{2}{3}y+frac{4}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}zright).
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $x,y in mathbb{R}^3$, then $f(x+y)=f(x)+f(y)$, and if $alpha in mathbb{R}$ we have $f(alpha x)=alpha f(x)$. If $x in mathrm{ker} f$ then $f(x)=0$. I applied all this properties writing the elements of the standard basis of $mathbb{R}^3$ as linear combinations of elements that we know how $f$ acts on them. Then, once you know the function behaves on the basis element, you know how it acts on any vector.
            $endgroup$
            – user289143
            Jan 22 at 22:49


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Now,
          $$
          (0,1,0)=frac{1}{3}big[2(0,2,1)-(0,1,2)big],
          $$
          hence
          $$f(0,1,0)
          =frac{1}{3}big[2 cdot f(0,2,1)-f(0,1,2)big]
          =frac{1}{3}big[2 cdot 0 -(2,1,1)big]
          =left(-frac{2}{3},-frac{1}{3},-frac{1}{3}right).
          $$
          Now,
          $$
          (0,0,1)
          =frac{1}{2}big[(0,1,2)-(0,1,0)big],
          $$
          hence
          $$
          f(0,0,1)
          =frac{1}{2}left[(2,1,1)-(-frac{2}{3},-frac{1}{3},-frac{1}{3})right]
          =left(frac{4}{3},frac{2}{3},frac{2}{3}right)
          $$
          Furthermore,
          $$
          (1,0,0)
          =(1,0,1)-(0,0,1),
          $$
          hence
          $$f(1,0,0)
          =f(1,0,1)-f(0,0,1)
          =-f(0,0,1)
          =left(-frac{4}{3},-frac{2}{3},-frac{2}{3}right).
          $$



          We can get the formula of $f$:
          $$
          f(x,y,z)
          = left(-frac{4}{3}x-frac{2}{3}y+frac{4}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}zright).
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $x,y in mathbb{R}^3$, then $f(x+y)=f(x)+f(y)$, and if $alpha in mathbb{R}$ we have $f(alpha x)=alpha f(x)$. If $x in mathrm{ker} f$ then $f(x)=0$. I applied all this properties writing the elements of the standard basis of $mathbb{R}^3$ as linear combinations of elements that we know how $f$ acts on them. Then, once you know the function behaves on the basis element, you know how it acts on any vector.
            $endgroup$
            – user289143
            Jan 22 at 22:49
















          4












          $begingroup$

          Now,
          $$
          (0,1,0)=frac{1}{3}big[2(0,2,1)-(0,1,2)big],
          $$
          hence
          $$f(0,1,0)
          =frac{1}{3}big[2 cdot f(0,2,1)-f(0,1,2)big]
          =frac{1}{3}big[2 cdot 0 -(2,1,1)big]
          =left(-frac{2}{3},-frac{1}{3},-frac{1}{3}right).
          $$
          Now,
          $$
          (0,0,1)
          =frac{1}{2}big[(0,1,2)-(0,1,0)big],
          $$
          hence
          $$
          f(0,0,1)
          =frac{1}{2}left[(2,1,1)-(-frac{2}{3},-frac{1}{3},-frac{1}{3})right]
          =left(frac{4}{3},frac{2}{3},frac{2}{3}right)
          $$
          Furthermore,
          $$
          (1,0,0)
          =(1,0,1)-(0,0,1),
          $$
          hence
          $$f(1,0,0)
          =f(1,0,1)-f(0,0,1)
          =-f(0,0,1)
          =left(-frac{4}{3},-frac{2}{3},-frac{2}{3}right).
          $$



          We can get the formula of $f$:
          $$
          f(x,y,z)
          = left(-frac{4}{3}x-frac{2}{3}y+frac{4}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}zright).
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $x,y in mathbb{R}^3$, then $f(x+y)=f(x)+f(y)$, and if $alpha in mathbb{R}$ we have $f(alpha x)=alpha f(x)$. If $x in mathrm{ker} f$ then $f(x)=0$. I applied all this properties writing the elements of the standard basis of $mathbb{R}^3$ as linear combinations of elements that we know how $f$ acts on them. Then, once you know the function behaves on the basis element, you know how it acts on any vector.
            $endgroup$
            – user289143
            Jan 22 at 22:49














          4












          4








          4





          $begingroup$

          Now,
          $$
          (0,1,0)=frac{1}{3}big[2(0,2,1)-(0,1,2)big],
          $$
          hence
          $$f(0,1,0)
          =frac{1}{3}big[2 cdot f(0,2,1)-f(0,1,2)big]
          =frac{1}{3}big[2 cdot 0 -(2,1,1)big]
          =left(-frac{2}{3},-frac{1}{3},-frac{1}{3}right).
          $$
          Now,
          $$
          (0,0,1)
          =frac{1}{2}big[(0,1,2)-(0,1,0)big],
          $$
          hence
          $$
          f(0,0,1)
          =frac{1}{2}left[(2,1,1)-(-frac{2}{3},-frac{1}{3},-frac{1}{3})right]
          =left(frac{4}{3},frac{2}{3},frac{2}{3}right)
          $$
          Furthermore,
          $$
          (1,0,0)
          =(1,0,1)-(0,0,1),
          $$
          hence
          $$f(1,0,0)
          =f(1,0,1)-f(0,0,1)
          =-f(0,0,1)
          =left(-frac{4}{3},-frac{2}{3},-frac{2}{3}right).
          $$



          We can get the formula of $f$:
          $$
          f(x,y,z)
          = left(-frac{4}{3}x-frac{2}{3}y+frac{4}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}zright).
          $$






          share|cite|improve this answer











          $endgroup$



          Now,
          $$
          (0,1,0)=frac{1}{3}big[2(0,2,1)-(0,1,2)big],
          $$
          hence
          $$f(0,1,0)
          =frac{1}{3}big[2 cdot f(0,2,1)-f(0,1,2)big]
          =frac{1}{3}big[2 cdot 0 -(2,1,1)big]
          =left(-frac{2}{3},-frac{1}{3},-frac{1}{3}right).
          $$
          Now,
          $$
          (0,0,1)
          =frac{1}{2}big[(0,1,2)-(0,1,0)big],
          $$
          hence
          $$
          f(0,0,1)
          =frac{1}{2}left[(2,1,1)-(-frac{2}{3},-frac{1}{3},-frac{1}{3})right]
          =left(frac{4}{3},frac{2}{3},frac{2}{3}right)
          $$
          Furthermore,
          $$
          (1,0,0)
          =(1,0,1)-(0,0,1),
          $$
          hence
          $$f(1,0,0)
          =f(1,0,1)-f(0,0,1)
          =-f(0,0,1)
          =left(-frac{4}{3},-frac{2}{3},-frac{2}{3}right).
          $$



          We can get the formula of $f$:
          $$
          f(x,y,z)
          = left(-frac{4}{3}x-frac{2}{3}y+frac{4}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}z,-frac{2}{3}x-frac{1}{3}y+frac{2}{3}zright).
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 28 at 17:30









          Viktor Glombik

          9361527




          9361527










          answered Jan 20 at 18:29









          user289143user289143

          903313




          903313












          • $begingroup$
            If $x,y in mathbb{R}^3$, then $f(x+y)=f(x)+f(y)$, and if $alpha in mathbb{R}$ we have $f(alpha x)=alpha f(x)$. If $x in mathrm{ker} f$ then $f(x)=0$. I applied all this properties writing the elements of the standard basis of $mathbb{R}^3$ as linear combinations of elements that we know how $f$ acts on them. Then, once you know the function behaves on the basis element, you know how it acts on any vector.
            $endgroup$
            – user289143
            Jan 22 at 22:49


















          • $begingroup$
            If $x,y in mathbb{R}^3$, then $f(x+y)=f(x)+f(y)$, and if $alpha in mathbb{R}$ we have $f(alpha x)=alpha f(x)$. If $x in mathrm{ker} f$ then $f(x)=0$. I applied all this properties writing the elements of the standard basis of $mathbb{R}^3$ as linear combinations of elements that we know how $f$ acts on them. Then, once you know the function behaves on the basis element, you know how it acts on any vector.
            $endgroup$
            – user289143
            Jan 22 at 22:49
















          $begingroup$
          If $x,y in mathbb{R}^3$, then $f(x+y)=f(x)+f(y)$, and if $alpha in mathbb{R}$ we have $f(alpha x)=alpha f(x)$. If $x in mathrm{ker} f$ then $f(x)=0$. I applied all this properties writing the elements of the standard basis of $mathbb{R}^3$ as linear combinations of elements that we know how $f$ acts on them. Then, once you know the function behaves on the basis element, you know how it acts on any vector.
          $endgroup$
          – user289143
          Jan 22 at 22:49




          $begingroup$
          If $x,y in mathbb{R}^3$, then $f(x+y)=f(x)+f(y)$, and if $alpha in mathbb{R}$ we have $f(alpha x)=alpha f(x)$. If $x in mathrm{ker} f$ then $f(x)=0$. I applied all this properties writing the elements of the standard basis of $mathbb{R}^3$ as linear combinations of elements that we know how $f$ acts on them. Then, once you know the function behaves on the basis element, you know how it acts on any vector.
          $endgroup$
          – user289143
          Jan 22 at 22:49



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