Linear transformation from $R^3$ to $R^2$.












1












$begingroup$


Find the matrix of the linear transformation $Tcolon {Bbb R}^3 to {Bbb R}^2$ such that



$T(1,1,1) = (1,1)$, $T(1,2,3) = (1,2)$, $T(1,2,4) = (1,4)$.



So far, I have only dealt with transformations in the same R. Any help?






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
    $endgroup$
    – Swapnil Tripathi
    Mar 29 '17 at 21:17


















1












$begingroup$


Find the matrix of the linear transformation $Tcolon {Bbb R}^3 to {Bbb R}^2$ such that



$T(1,1,1) = (1,1)$, $T(1,2,3) = (1,2)$, $T(1,2,4) = (1,4)$.



So far, I have only dealt with transformations in the same R. Any help?






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
    $endgroup$
    – Swapnil Tripathi
    Mar 29 '17 at 21:17
















1












1








1





$begingroup$


Find the matrix of the linear transformation $Tcolon {Bbb R}^3 to {Bbb R}^2$ such that



$T(1,1,1) = (1,1)$, $T(1,2,3) = (1,2)$, $T(1,2,4) = (1,4)$.



So far, I have only dealt with transformations in the same R. Any help?






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$




Find the matrix of the linear transformation $Tcolon {Bbb R}^3 to {Bbb R}^2$ such that



$T(1,1,1) = (1,1)$, $T(1,2,3) = (1,2)$, $T(1,2,4) = (1,4)$.



So far, I have only dealt with transformations in the same R. Any help?






linear-algebra matrices linear-transformations




linear-algebra matrices linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 20:52









A.Γ.

22.9k32656




22.9k32656










asked Mar 29 '17 at 21:08









AmaCAmaC

144110




144110












  • $begingroup$
    The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
    $endgroup$
    – Swapnil Tripathi
    Mar 29 '17 at 21:17




















  • $begingroup$
    The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
    $endgroup$
    – Swapnil Tripathi
    Mar 29 '17 at 21:17


















$begingroup$
The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
$endgroup$
– Swapnil Tripathi
Mar 29 '17 at 21:17






$begingroup$
The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
$endgroup$
– Swapnil Tripathi
Mar 29 '17 at 21:17












3 Answers
3






active

oldest

votes


















1












$begingroup$

Group your 3 constraints into a single one:



$$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$



(where the point means matrix product).



(1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.



Up to you for the last calculations.



You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
    $endgroup$
    – user171326
    Mar 29 '17 at 21:26










  • $begingroup$
    "Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
    $endgroup$
    – Jean Marie
    Mar 29 '17 at 21:32










  • $begingroup$
    I have added the result one must find.
    $endgroup$
    – Jean Marie
    Mar 29 '17 at 22:53



















0












$begingroup$

Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.



Do you see how to find it ? Do you see how to continue ?






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint:
    Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
    $$
    left(
    begin{array}{ccc}
    a & b & c \
    d & e & f \
    end{array}
    right)
    left(
    begin{array}{c}
    1\
    1\
    1
    end{array}
    right)
    =
    left(
    begin{array}{c}
    1\
    1
    end{array}
    right)
    $$
    What other matrix equations can you form?






    share|cite|improve this answer









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Group your 3 constraints into a single one:



      $$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$



      (where the point means matrix product).



      (1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.



      Up to you for the last calculations.



      You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
        $endgroup$
        – user171326
        Mar 29 '17 at 21:26










      • $begingroup$
        "Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
        $endgroup$
        – Jean Marie
        Mar 29 '17 at 21:32










      • $begingroup$
        I have added the result one must find.
        $endgroup$
        – Jean Marie
        Mar 29 '17 at 22:53
















      1












      $begingroup$

      Group your 3 constraints into a single one:



      $$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$



      (where the point means matrix product).



      (1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.



      Up to you for the last calculations.



      You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
        $endgroup$
        – user171326
        Mar 29 '17 at 21:26










      • $begingroup$
        "Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
        $endgroup$
        – Jean Marie
        Mar 29 '17 at 21:32










      • $begingroup$
        I have added the result one must find.
        $endgroup$
        – Jean Marie
        Mar 29 '17 at 22:53














      1












      1








      1





      $begingroup$

      Group your 3 constraints into a single one:



      $$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$



      (where the point means matrix product).



      (1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.



      Up to you for the last calculations.



      You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$






      share|cite|improve this answer











      $endgroup$



      Group your 3 constraints into a single one:



      $$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$



      (where the point means matrix product).



      (1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.



      Up to you for the last calculations.



      You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 29 '17 at 22:52

























      answered Mar 29 '17 at 21:15









      Jean MarieJean Marie

      31k42255




      31k42255












      • $begingroup$
        Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
        $endgroup$
        – user171326
        Mar 29 '17 at 21:26










      • $begingroup$
        "Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
        $endgroup$
        – Jean Marie
        Mar 29 '17 at 21:32










      • $begingroup$
        I have added the result one must find.
        $endgroup$
        – Jean Marie
        Mar 29 '17 at 22:53


















      • $begingroup$
        Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
        $endgroup$
        – user171326
        Mar 29 '17 at 21:26










      • $begingroup$
        "Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
        $endgroup$
        – Jean Marie
        Mar 29 '17 at 21:32










      • $begingroup$
        I have added the result one must find.
        $endgroup$
        – Jean Marie
        Mar 29 '17 at 22:53
















      $begingroup$
      Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
      $endgroup$
      – user171326
      Mar 29 '17 at 21:26




      $begingroup$
      Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
      $endgroup$
      – user171326
      Mar 29 '17 at 21:26












      $begingroup$
      "Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
      $endgroup$
      – Jean Marie
      Mar 29 '17 at 21:32




      $begingroup$
      "Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
      $endgroup$
      – Jean Marie
      Mar 29 '17 at 21:32












      $begingroup$
      I have added the result one must find.
      $endgroup$
      – Jean Marie
      Mar 29 '17 at 22:53




      $begingroup$
      I have added the result one must find.
      $endgroup$
      – Jean Marie
      Mar 29 '17 at 22:53











      0












      $begingroup$

      Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.



      Do you see how to find it ? Do you see how to continue ?






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.



        Do you see how to find it ? Do you see how to continue ?






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.



          Do you see how to find it ? Do you see how to continue ?






          share|cite|improve this answer









          $endgroup$



          Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.



          Do you see how to find it ? Do you see how to continue ?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 29 '17 at 21:12







          user171326






























              0












              $begingroup$

              Hint:
              Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
              $$
              left(
              begin{array}{ccc}
              a & b & c \
              d & e & f \
              end{array}
              right)
              left(
              begin{array}{c}
              1\
              1\
              1
              end{array}
              right)
              =
              left(
              begin{array}{c}
              1\
              1
              end{array}
              right)
              $$
              What other matrix equations can you form?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint:
                Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
                $$
                left(
                begin{array}{ccc}
                a & b & c \
                d & e & f \
                end{array}
                right)
                left(
                begin{array}{c}
                1\
                1\
                1
                end{array}
                right)
                =
                left(
                begin{array}{c}
                1\
                1
                end{array}
                right)
                $$
                What other matrix equations can you form?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint:
                  Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
                  $$
                  left(
                  begin{array}{ccc}
                  a & b & c \
                  d & e & f \
                  end{array}
                  right)
                  left(
                  begin{array}{c}
                  1\
                  1\
                  1
                  end{array}
                  right)
                  =
                  left(
                  begin{array}{c}
                  1\
                  1
                  end{array}
                  right)
                  $$
                  What other matrix equations can you form?






                  share|cite|improve this answer









                  $endgroup$



                  Hint:
                  Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
                  $$
                  left(
                  begin{array}{ccc}
                  a & b & c \
                  d & e & f \
                  end{array}
                  right)
                  left(
                  begin{array}{c}
                  1\
                  1\
                  1
                  end{array}
                  right)
                  =
                  left(
                  begin{array}{c}
                  1\
                  1
                  end{array}
                  right)
                  $$
                  What other matrix equations can you form?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 29 '17 at 21:54









                  D. BritoD. Brito

                  385111




                  385111






























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