Determine matrix A with respect to standard basis $f:U->R^2$












0












$begingroup$


I am having trouble understanding my problem and what to calculate



I have been given the subspace
$U={x=$$
begin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
in F^3 | x_1 + x_2 + x_3 = 0} subset F^3
$



and the linear transformation $f: U rightarrow F^2 $
$fbegin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
=begin{vmatrix}
x_1\
x_2+x_3\
end{vmatrix}$



The question is to determine a matrix A that represent $f: U rightarrow F^2 $ with respect to the basis for U and standard basis $(e_1,e_2)$ for $F^2$



My attempt:
I have calculated the basis $B={begin{vmatrix}
1\
0\
-1\
end{vmatrix},begin{vmatrix}
0\
1\
-1\
end{vmatrix}}$



And my matrix A calculated from the linear transformation



$begin{vmatrix}
1&0&0\
0&1&1\
end{vmatrix}$



I'm aware the standard basis are $e_1=begin{vmatrix}
1\
0\
end{vmatrix} ,
e_2=begin{vmatrix}
0\
1\
end{vmatrix}$



I'm not sure about the next step. Do I calculate:
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
-1\
end{vmatrix}$



and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
0\
end{vmatrix}$



and



$fbegin{vmatrix}
0\
1\
-1\
end{vmatrix}
=begin{vmatrix}
0\
1\
end{vmatrix}$



I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $F^2$?






linear-transformations







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am having trouble understanding my problem and what to calculate



    I have been given the subspace
    $U={x=$$
    begin{vmatrix}
    x_1\
    x_2\
    x_3\
    end{vmatrix}
    in F^3 | x_1 + x_2 + x_3 = 0} subset F^3
    $



    and the linear transformation $f: U rightarrow F^2 $
    $fbegin{vmatrix}
    x_1\
    x_2\
    x_3\
    end{vmatrix}
    =begin{vmatrix}
    x_1\
    x_2+x_3\
    end{vmatrix}$



    The question is to determine a matrix A that represent $f: U rightarrow F^2 $ with respect to the basis for U and standard basis $(e_1,e_2)$ for $F^2$



    My attempt:
    I have calculated the basis $B={begin{vmatrix}
    1\
    0\
    -1\
    end{vmatrix},begin{vmatrix}
    0\
    1\
    -1\
    end{vmatrix}}$



    And my matrix A calculated from the linear transformation



    $begin{vmatrix}
    1&0&0\
    0&1&1\
    end{vmatrix}$



    I'm aware the standard basis are $e_1=begin{vmatrix}
    1\
    0\
    end{vmatrix} ,
    e_2=begin{vmatrix}
    0\
    1\
    end{vmatrix}$



    I'm not sure about the next step. Do I calculate:
    $fbegin{vmatrix}
    1\
    0\
    -1\
    end{vmatrix}
    =begin{vmatrix}
    1\
    -1\
    end{vmatrix}$



    and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me
    $fbegin{vmatrix}
    1\
    0\
    -1\
    end{vmatrix}
    =begin{vmatrix}
    1\
    0\
    end{vmatrix}$



    and



    $fbegin{vmatrix}
    0\
    1\
    -1\
    end{vmatrix}
    =begin{vmatrix}
    0\
    1\
    end{vmatrix}$



    I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $F^2$?






    linear-transformations







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am having trouble understanding my problem and what to calculate



      I have been given the subspace
      $U={x=$$
      begin{vmatrix}
      x_1\
      x_2\
      x_3\
      end{vmatrix}
      in F^3 | x_1 + x_2 + x_3 = 0} subset F^3
      $



      and the linear transformation $f: U rightarrow F^2 $
      $fbegin{vmatrix}
      x_1\
      x_2\
      x_3\
      end{vmatrix}
      =begin{vmatrix}
      x_1\
      x_2+x_3\
      end{vmatrix}$



      The question is to determine a matrix A that represent $f: U rightarrow F^2 $ with respect to the basis for U and standard basis $(e_1,e_2)$ for $F^2$



      My attempt:
      I have calculated the basis $B={begin{vmatrix}
      1\
      0\
      -1\
      end{vmatrix},begin{vmatrix}
      0\
      1\
      -1\
      end{vmatrix}}$



      And my matrix A calculated from the linear transformation



      $begin{vmatrix}
      1&0&0\
      0&1&1\
      end{vmatrix}$



      I'm aware the standard basis are $e_1=begin{vmatrix}
      1\
      0\
      end{vmatrix} ,
      e_2=begin{vmatrix}
      0\
      1\
      end{vmatrix}$



      I'm not sure about the next step. Do I calculate:
      $fbegin{vmatrix}
      1\
      0\
      -1\
      end{vmatrix}
      =begin{vmatrix}
      1\
      -1\
      end{vmatrix}$



      and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me
      $fbegin{vmatrix}
      1\
      0\
      -1\
      end{vmatrix}
      =begin{vmatrix}
      1\
      0\
      end{vmatrix}$



      and



      $fbegin{vmatrix}
      0\
      1\
      -1\
      end{vmatrix}
      =begin{vmatrix}
      0\
      1\
      end{vmatrix}$



      I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $F^2$?






      linear-transformations







      share|cite|improve this question











      $endgroup$




      I am having trouble understanding my problem and what to calculate



      I have been given the subspace
      $U={x=$$
      begin{vmatrix}
      x_1\
      x_2\
      x_3\
      end{vmatrix}
      in F^3 | x_1 + x_2 + x_3 = 0} subset F^3
      $



      and the linear transformation $f: U rightarrow F^2 $
      $fbegin{vmatrix}
      x_1\
      x_2\
      x_3\
      end{vmatrix}
      =begin{vmatrix}
      x_1\
      x_2+x_3\
      end{vmatrix}$



      The question is to determine a matrix A that represent $f: U rightarrow F^2 $ with respect to the basis for U and standard basis $(e_1,e_2)$ for $F^2$



      My attempt:
      I have calculated the basis $B={begin{vmatrix}
      1\
      0\
      -1\
      end{vmatrix},begin{vmatrix}
      0\
      1\
      -1\
      end{vmatrix}}$



      And my matrix A calculated from the linear transformation



      $begin{vmatrix}
      1&0&0\
      0&1&1\
      end{vmatrix}$



      I'm aware the standard basis are $e_1=begin{vmatrix}
      1\
      0\
      end{vmatrix} ,
      e_2=begin{vmatrix}
      0\
      1\
      end{vmatrix}$



      I'm not sure about the next step. Do I calculate:
      $fbegin{vmatrix}
      1\
      0\
      -1\
      end{vmatrix}
      =begin{vmatrix}
      1\
      -1\
      end{vmatrix}$



      and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me
      $fbegin{vmatrix}
      1\
      0\
      -1\
      end{vmatrix}
      =begin{vmatrix}
      1\
      0\
      end{vmatrix}$



      and



      $fbegin{vmatrix}
      0\
      1\
      -1\
      end{vmatrix}
      =begin{vmatrix}
      0\
      1\
      end{vmatrix}$



      I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $F^2$?






      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 2 at 23:07







      mahma

















      asked Jan 2 at 22:39









      mahmamahma

      11




      11






















          1 Answer
          1






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          0












          $begingroup$

          The image of the first vector in the basis is
          $$
          begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
          $$
          and the image of the second vector is
          $$
          begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
          $$
          so the matrix is
          $$
          begin{bmatrix}
          1 & 0 \
          -1 & 0
          end{bmatrix}
          $$
          Note that the map $f$ can be more easily described as
          $$
          fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
            $endgroup$
            – mahma
            Jan 3 at 0:08










          • $begingroup$
            @mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
            $endgroup$
            – egreg
            Jan 3 at 0:47











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          The image of the first vector in the basis is
          $$
          begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
          $$
          and the image of the second vector is
          $$
          begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
          $$
          so the matrix is
          $$
          begin{bmatrix}
          1 & 0 \
          -1 & 0
          end{bmatrix}
          $$
          Note that the map $f$ can be more easily described as
          $$
          fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
            $endgroup$
            – mahma
            Jan 3 at 0:08










          • $begingroup$
            @mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
            $endgroup$
            – egreg
            Jan 3 at 0:47
















          0












          $begingroup$

          The image of the first vector in the basis is
          $$
          begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
          $$
          and the image of the second vector is
          $$
          begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
          $$
          so the matrix is
          $$
          begin{bmatrix}
          1 & 0 \
          -1 & 0
          end{bmatrix}
          $$
          Note that the map $f$ can be more easily described as
          $$
          fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
            $endgroup$
            – mahma
            Jan 3 at 0:08










          • $begingroup$
            @mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
            $endgroup$
            – egreg
            Jan 3 at 0:47














          0












          0








          0





          $begingroup$

          The image of the first vector in the basis is
          $$
          begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
          $$
          and the image of the second vector is
          $$
          begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
          $$
          so the matrix is
          $$
          begin{bmatrix}
          1 & 0 \
          -1 & 0
          end{bmatrix}
          $$
          Note that the map $f$ can be more easily described as
          $$
          fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
          $$






          share|cite|improve this answer









          $endgroup$



          The image of the first vector in the basis is
          $$
          begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
          $$
          and the image of the second vector is
          $$
          begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
          $$
          so the matrix is
          $$
          begin{bmatrix}
          1 & 0 \
          -1 & 0
          end{bmatrix}
          $$
          Note that the map $f$ can be more easily described as
          $$
          fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 23:57









          egregegreg

          180k1485202




          180k1485202












          • $begingroup$
            Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
            $endgroup$
            – mahma
            Jan 3 at 0:08










          • $begingroup$
            @mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
            $endgroup$
            – egreg
            Jan 3 at 0:47


















          • $begingroup$
            Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
            $endgroup$
            – mahma
            Jan 3 at 0:08










          • $begingroup$
            @mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
            $endgroup$
            – egreg
            Jan 3 at 0:47
















          $begingroup$
          Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
          $endgroup$
          – mahma
          Jan 3 at 0:08




          $begingroup$
          Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
          $endgroup$
          – mahma
          Jan 3 at 0:08












          $begingroup$
          @mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
          $endgroup$
          – egreg
          Jan 3 at 0:47




          $begingroup$
          @mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
          $endgroup$
          – egreg
          Jan 3 at 0:47


















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