Find the transformation matrix of a linear mapping given 2 basis and only one image of one vector of each...












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Linear mapping g: R² ↦ R². They give me one basis:
B = {F1 = (1,1), F2 = (-1,1)} and another C = {e1 = (1,0), e2 = (0,1)}. In addition g(e2) = F1 and g(F2) = e2. I don't know how to get the form of the image, something like g(x, y) = (2x, y-x) (with the correct values). Sorry for my poor format i'm still novice. Thanks in advance!






linear-algebra linear-transformations







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$begingroup$


Linear mapping g: R² ↦ R². They give me one basis:
B = {F1 = (1,1), F2 = (-1,1)} and another C = {e1 = (1,0), e2 = (0,1)}. In addition g(e2) = F1 and g(F2) = e2. I don't know how to get the form of the image, something like g(x, y) = (2x, y-x) (with the correct values). Sorry for my poor format i'm still novice. Thanks in advance!






linear-algebra linear-transformations







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  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 16:59














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$begingroup$


Linear mapping g: R² ↦ R². They give me one basis:
B = {F1 = (1,1), F2 = (-1,1)} and another C = {e1 = (1,0), e2 = (0,1)}. In addition g(e2) = F1 and g(F2) = e2. I don't know how to get the form of the image, something like g(x, y) = (2x, y-x) (with the correct values). Sorry for my poor format i'm still novice. Thanks in advance!






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$




Linear mapping g: R² ↦ R². They give me one basis:
B = {F1 = (1,1), F2 = (-1,1)} and another C = {e1 = (1,0), e2 = (0,1)}. In addition g(e2) = F1 and g(F2) = e2. I don't know how to get the form of the image, something like g(x, y) = (2x, y-x) (with the correct values). Sorry for my poor format i'm still novice. Thanks in advance!






linear-algebra linear-transformations




linear-algebra linear-transformations






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asked Jan 7 at 16:54









S.BuzzoniS.Buzzoni

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  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 16:59


















  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 16:59
















$begingroup$
Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 16:59




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Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
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Jan 7 at 16:59










2 Answers
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Let the matrix of $g:Bbb R^2toBbb R^2$ be $begin{bmatrix}a&b\c&dend{bmatrix}$.$$g(0,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}0\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}=begin{bmatrix}1\1end{bmatrix}\g(-1,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}-1\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}-begin{bmatrix}a\cend{bmatrix}=begin{bmatrix}0\1end{bmatrix}\impliesbegin{bmatrix}a\cend{bmatrix}=begin{bmatrix}1\0end{bmatrix}$$



This gives the matrix of $g$ as $begin{bmatrix}1&1\0&1end{bmatrix}$.$$implies g(x,y)=begin{bmatrix}1&1\0&1end{bmatrix}begin{bmatrix}x\yend{bmatrix}=(x+y,y)$$






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    $begingroup$

    $g(0,1)=(1,1)$
    $ g(-1,1)=-g(1,0)+g(0,1)=(0,1)$
    $-g(1,0) = (-1,0) rightarrow g(e_1)=e_1$
    $g(e_2)=e_1 + e_2$



    So the matrix of the transformation in standard basis looks like this:
    begin{bmatrix}
    1 & 1\
    0 & 1\
    end{bmatrix}






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      2 Answers
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      2 Answers
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      $begingroup$

      Let the matrix of $g:Bbb R^2toBbb R^2$ be $begin{bmatrix}a&b\c&dend{bmatrix}$.$$g(0,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}0\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}=begin{bmatrix}1\1end{bmatrix}\g(-1,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}-1\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}-begin{bmatrix}a\cend{bmatrix}=begin{bmatrix}0\1end{bmatrix}\impliesbegin{bmatrix}a\cend{bmatrix}=begin{bmatrix}1\0end{bmatrix}$$



      This gives the matrix of $g$ as $begin{bmatrix}1&1\0&1end{bmatrix}$.$$implies g(x,y)=begin{bmatrix}1&1\0&1end{bmatrix}begin{bmatrix}x\yend{bmatrix}=(x+y,y)$$






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        $begingroup$

        Let the matrix of $g:Bbb R^2toBbb R^2$ be $begin{bmatrix}a&b\c&dend{bmatrix}$.$$g(0,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}0\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}=begin{bmatrix}1\1end{bmatrix}\g(-1,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}-1\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}-begin{bmatrix}a\cend{bmatrix}=begin{bmatrix}0\1end{bmatrix}\impliesbegin{bmatrix}a\cend{bmatrix}=begin{bmatrix}1\0end{bmatrix}$$



        This gives the matrix of $g$ as $begin{bmatrix}1&1\0&1end{bmatrix}$.$$implies g(x,y)=begin{bmatrix}1&1\0&1end{bmatrix}begin{bmatrix}x\yend{bmatrix}=(x+y,y)$$






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          $begingroup$

          Let the matrix of $g:Bbb R^2toBbb R^2$ be $begin{bmatrix}a&b\c&dend{bmatrix}$.$$g(0,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}0\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}=begin{bmatrix}1\1end{bmatrix}\g(-1,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}-1\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}-begin{bmatrix}a\cend{bmatrix}=begin{bmatrix}0\1end{bmatrix}\impliesbegin{bmatrix}a\cend{bmatrix}=begin{bmatrix}1\0end{bmatrix}$$



          This gives the matrix of $g$ as $begin{bmatrix}1&1\0&1end{bmatrix}$.$$implies g(x,y)=begin{bmatrix}1&1\0&1end{bmatrix}begin{bmatrix}x\yend{bmatrix}=(x+y,y)$$






          share|cite|improve this answer









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          Let the matrix of $g:Bbb R^2toBbb R^2$ be $begin{bmatrix}a&b\c&dend{bmatrix}$.$$g(0,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}0\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}=begin{bmatrix}1\1end{bmatrix}\g(-1,1)=begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}-1\1end{bmatrix}=begin{bmatrix}b\dend{bmatrix}-begin{bmatrix}a\cend{bmatrix}=begin{bmatrix}0\1end{bmatrix}\impliesbegin{bmatrix}a\cend{bmatrix}=begin{bmatrix}1\0end{bmatrix}$$



          This gives the matrix of $g$ as $begin{bmatrix}1&1\0&1end{bmatrix}$.$$implies g(x,y)=begin{bmatrix}1&1\0&1end{bmatrix}begin{bmatrix}x\yend{bmatrix}=(x+y,y)$$







          share|cite|improve this answer












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          answered Jan 7 at 17:07









          Shubham JohriShubham Johri

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              $begingroup$

              $g(0,1)=(1,1)$
              $ g(-1,1)=-g(1,0)+g(0,1)=(0,1)$
              $-g(1,0) = (-1,0) rightarrow g(e_1)=e_1$
              $g(e_2)=e_1 + e_2$



              So the matrix of the transformation in standard basis looks like this:
              begin{bmatrix}
              1 & 1\
              0 & 1\
              end{bmatrix}






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $g(0,1)=(1,1)$
                $ g(-1,1)=-g(1,0)+g(0,1)=(0,1)$
                $-g(1,0) = (-1,0) rightarrow g(e_1)=e_1$
                $g(e_2)=e_1 + e_2$



                So the matrix of the transformation in standard basis looks like this:
                begin{bmatrix}
                1 & 1\
                0 & 1\
                end{bmatrix}






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $g(0,1)=(1,1)$
                  $ g(-1,1)=-g(1,0)+g(0,1)=(0,1)$
                  $-g(1,0) = (-1,0) rightarrow g(e_1)=e_1$
                  $g(e_2)=e_1 + e_2$



                  So the matrix of the transformation in standard basis looks like this:
                  begin{bmatrix}
                  1 & 1\
                  0 & 1\
                  end{bmatrix}






                  share|cite|improve this answer









                  $endgroup$



                  $g(0,1)=(1,1)$
                  $ g(-1,1)=-g(1,0)+g(0,1)=(0,1)$
                  $-g(1,0) = (-1,0) rightarrow g(e_1)=e_1$
                  $g(e_2)=e_1 + e_2$



                  So the matrix of the transformation in standard basis looks like this:
                  begin{bmatrix}
                  1 & 1\
                  0 & 1\
                  end{bmatrix}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 19:01









                  The CatThe Cat

                  18411




                  18411






























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