Linear transformation $f in L( mathbb R[t]_{2}, (mathbb R^{2})^{*})$












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Let linear transformation $f in L( mathbb R[t]_{2}, (mathbb R^{2})^{*})$ for which $f(p)( {begin{bmatrix}x_{1}\ x_{2}end{bmatrix}})=p(0)cdot x_{1}+p(1) cdot x_{2}$ for $p in mathbb R[t]_{3},{begin{bmatrix}x_{1}\ x_{2}end{bmatrix}} in mathbb R^{2}$.
(a) Find a basis for subspace $ker f$.
(b) Find transformation matrix $f$ in a basis $1,t,t^{2}$ in space $mathbb R[t]_{2}$ and in the dual base to the base $[1,0]^{T},[0,1]^{T}$ in space $(mathbb R^{2})^{*}$.





I know how to find transformation matrix but I don't understand how to do it when I have functionals and books are unhelpful to me.

Is there anyone who can tell me how to do it?






linear-algebra







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    1












    $begingroup$



    Let linear transformation $f in L( mathbb R[t]_{2}, (mathbb R^{2})^{*})$ for which $f(p)( {begin{bmatrix}x_{1}\ x_{2}end{bmatrix}})=p(0)cdot x_{1}+p(1) cdot x_{2}$ for $p in mathbb R[t]_{3},{begin{bmatrix}x_{1}\ x_{2}end{bmatrix}} in mathbb R^{2}$.
    (a) Find a basis for subspace $ker f$.
    (b) Find transformation matrix $f$ in a basis $1,t,t^{2}$ in space $mathbb R[t]_{2}$ and in the dual base to the base $[1,0]^{T},[0,1]^{T}$ in space $(mathbb R^{2})^{*}$.





    I know how to find transformation matrix but I don't understand how to do it when I have functionals and books are unhelpful to me.

    Is there anyone who can tell me how to do it?






    linear-algebra







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      Let linear transformation $f in L( mathbb R[t]_{2}, (mathbb R^{2})^{*})$ for which $f(p)( {begin{bmatrix}x_{1}\ x_{2}end{bmatrix}})=p(0)cdot x_{1}+p(1) cdot x_{2}$ for $p in mathbb R[t]_{3},{begin{bmatrix}x_{1}\ x_{2}end{bmatrix}} in mathbb R^{2}$.
      (a) Find a basis for subspace $ker f$.
      (b) Find transformation matrix $f$ in a basis $1,t,t^{2}$ in space $mathbb R[t]_{2}$ and in the dual base to the base $[1,0]^{T},[0,1]^{T}$ in space $(mathbb R^{2})^{*}$.





      I know how to find transformation matrix but I don't understand how to do it when I have functionals and books are unhelpful to me.

      Is there anyone who can tell me how to do it?






      linear-algebra







      share|cite|improve this question









      $endgroup$





      Let linear transformation $f in L( mathbb R[t]_{2}, (mathbb R^{2})^{*})$ for which $f(p)( {begin{bmatrix}x_{1}\ x_{2}end{bmatrix}})=p(0)cdot x_{1}+p(1) cdot x_{2}$ for $p in mathbb R[t]_{3},{begin{bmatrix}x_{1}\ x_{2}end{bmatrix}} in mathbb R^{2}$.
      (a) Find a basis for subspace $ker f$.
      (b) Find transformation matrix $f$ in a basis $1,t,t^{2}$ in space $mathbb R[t]_{2}$ and in the dual base to the base $[1,0]^{T},[0,1]^{T}$ in space $(mathbb R^{2})^{*}$.





      I know how to find transformation matrix but I don't understand how to do it when I have functionals and books are unhelpful to me.

      Is there anyone who can tell me how to do it?






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Feb 2 at 13:47









      MP3129MP3129

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          1 Answer
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          $begingroup$

          Do as you usually do. Try to represent the image of basis elements of the domain under $f$ in the linear combination of basis elements of the codomain.
          For example:
          $f(1)(x1, x2) =x1+x2=f1(x1, x2)+f2(x1, x2)=(f1+f2)(x1, x2)$
          This implies $f(1) =f1+f2$, where $f1$ and $f2$ is the dual basis.
          Similarly find $f(t)$ and $f(t^2)$ in the linear combination of $f1$ and $f2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you help me with (a) also? It seemed to me that I could, but probably not
            $endgroup$
            – MP3129
            Feb 2 at 22:24






          • 1




            $begingroup$
            If $p(t)=a+bt+ct^2$ would be in $kerf$ then $ax1+(a+b+c)x2=0$ for all $x1, x2 in {mathbb R}$ and this would imply $a=0, b=-c$. Hence $kerf$ consists of polynomial of the form $bt-bt^2$.
            $endgroup$
            – Afzal Ansari
            Feb 3 at 3:52














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          1 Answer
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          active

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          1












          $begingroup$

          Do as you usually do. Try to represent the image of basis elements of the domain under $f$ in the linear combination of basis elements of the codomain.
          For example:
          $f(1)(x1, x2) =x1+x2=f1(x1, x2)+f2(x1, x2)=(f1+f2)(x1, x2)$
          This implies $f(1) =f1+f2$, where $f1$ and $f2$ is the dual basis.
          Similarly find $f(t)$ and $f(t^2)$ in the linear combination of $f1$ and $f2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you help me with (a) also? It seemed to me that I could, but probably not
            $endgroup$
            – MP3129
            Feb 2 at 22:24






          • 1




            $begingroup$
            If $p(t)=a+bt+ct^2$ would be in $kerf$ then $ax1+(a+b+c)x2=0$ for all $x1, x2 in {mathbb R}$ and this would imply $a=0, b=-c$. Hence $kerf$ consists of polynomial of the form $bt-bt^2$.
            $endgroup$
            – Afzal Ansari
            Feb 3 at 3:52


















          1












          $begingroup$

          Do as you usually do. Try to represent the image of basis elements of the domain under $f$ in the linear combination of basis elements of the codomain.
          For example:
          $f(1)(x1, x2) =x1+x2=f1(x1, x2)+f2(x1, x2)=(f1+f2)(x1, x2)$
          This implies $f(1) =f1+f2$, where $f1$ and $f2$ is the dual basis.
          Similarly find $f(t)$ and $f(t^2)$ in the linear combination of $f1$ and $f2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you help me with (a) also? It seemed to me that I could, but probably not
            $endgroup$
            – MP3129
            Feb 2 at 22:24






          • 1




            $begingroup$
            If $p(t)=a+bt+ct^2$ would be in $kerf$ then $ax1+(a+b+c)x2=0$ for all $x1, x2 in {mathbb R}$ and this would imply $a=0, b=-c$. Hence $kerf$ consists of polynomial of the form $bt-bt^2$.
            $endgroup$
            – Afzal Ansari
            Feb 3 at 3:52
















          1












          1








          1





          $begingroup$

          Do as you usually do. Try to represent the image of basis elements of the domain under $f$ in the linear combination of basis elements of the codomain.
          For example:
          $f(1)(x1, x2) =x1+x2=f1(x1, x2)+f2(x1, x2)=(f1+f2)(x1, x2)$
          This implies $f(1) =f1+f2$, where $f1$ and $f2$ is the dual basis.
          Similarly find $f(t)$ and $f(t^2)$ in the linear combination of $f1$ and $f2$.






          share|cite|improve this answer









          $endgroup$



          Do as you usually do. Try to represent the image of basis elements of the domain under $f$ in the linear combination of basis elements of the codomain.
          For example:
          $f(1)(x1, x2) =x1+x2=f1(x1, x2)+f2(x1, x2)=(f1+f2)(x1, x2)$
          This implies $f(1) =f1+f2$, where $f1$ and $f2$ is the dual basis.
          Similarly find $f(t)$ and $f(t^2)$ in the linear combination of $f1$ and $f2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 2 at 18:30









          Afzal AnsariAfzal Ansari

          856




          856












          • $begingroup$
            Could you help me with (a) also? It seemed to me that I could, but probably not
            $endgroup$
            – MP3129
            Feb 2 at 22:24






          • 1




            $begingroup$
            If $p(t)=a+bt+ct^2$ would be in $kerf$ then $ax1+(a+b+c)x2=0$ for all $x1, x2 in {mathbb R}$ and this would imply $a=0, b=-c$. Hence $kerf$ consists of polynomial of the form $bt-bt^2$.
            $endgroup$
            – Afzal Ansari
            Feb 3 at 3:52




















          • $begingroup$
            Could you help me with (a) also? It seemed to me that I could, but probably not
            $endgroup$
            – MP3129
            Feb 2 at 22:24






          • 1




            $begingroup$
            If $p(t)=a+bt+ct^2$ would be in $kerf$ then $ax1+(a+b+c)x2=0$ for all $x1, x2 in {mathbb R}$ and this would imply $a=0, b=-c$. Hence $kerf$ consists of polynomial of the form $bt-bt^2$.
            $endgroup$
            – Afzal Ansari
            Feb 3 at 3:52


















          $begingroup$
          Could you help me with (a) also? It seemed to me that I could, but probably not
          $endgroup$
          – MP3129
          Feb 2 at 22:24




          $begingroup$
          Could you help me with (a) also? It seemed to me that I could, but probably not
          $endgroup$
          – MP3129
          Feb 2 at 22:24




          1




          1




          $begingroup$
          If $p(t)=a+bt+ct^2$ would be in $kerf$ then $ax1+(a+b+c)x2=0$ for all $x1, x2 in {mathbb R}$ and this would imply $a=0, b=-c$. Hence $kerf$ consists of polynomial of the form $bt-bt^2$.
          $endgroup$
          – Afzal Ansari
          Feb 3 at 3:52






          $begingroup$
          If $p(t)=a+bt+ct^2$ would be in $kerf$ then $ax1+(a+b+c)x2=0$ for all $x1, x2 in {mathbb R}$ and this would imply $a=0, b=-c$. Hence $kerf$ consists of polynomial of the form $bt-bt^2$.
          $endgroup$
          – Afzal Ansari
          Feb 3 at 3:52




















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