Computing the matrix $T$ using the basis $beta$












0












$begingroup$


I have $T(f(x)) = f(1) + f'(0)x + (f'(0)+f''(0))x^2$, and I need to compute the matrix of this linear transformation in base $beta$, which is the standard ordered basis of $P_2(R)$ (namely $(1,x,x^2)$).
I have tried doing it but I get the wrong coordinates and now I don't really know how to properly calculate them.



Thanks in advance for the help :)






linear-algebra







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have $T(f(x)) = f(1) + f'(0)x + (f'(0)+f''(0))x^2$, and I need to compute the matrix of this linear transformation in base $beta$, which is the standard ordered basis of $P_2(R)$ (namely $(1,x,x^2)$).
    I have tried doing it but I get the wrong coordinates and now I don't really know how to properly calculate them.



    Thanks in advance for the help :)






    linear-algebra







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have $T(f(x)) = f(1) + f'(0)x + (f'(0)+f''(0))x^2$, and I need to compute the matrix of this linear transformation in base $beta$, which is the standard ordered basis of $P_2(R)$ (namely $(1,x,x^2)$).
      I have tried doing it but I get the wrong coordinates and now I don't really know how to properly calculate them.



      Thanks in advance for the help :)






      linear-algebra







      share|cite|improve this question











      $endgroup$




      I have $T(f(x)) = f(1) + f'(0)x + (f'(0)+f''(0))x^2$, and I need to compute the matrix of this linear transformation in base $beta$, which is the standard ordered basis of $P_2(R)$ (namely $(1,x,x^2)$).
      I have tried doing it but I get the wrong coordinates and now I don't really know how to properly calculate them.



      Thanks in advance for the help :)






      linear-algebra




      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 15:59









      mike65535

      110116




      110116










      asked Nov 29 '18 at 15:32









      VladVlad

      34




      34






















          1 Answer
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          0












          $begingroup$

          Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,




          $A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you show me how you get to those results? I only got it right for $T(1)$.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:07










          • $begingroup$
            Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
            $endgroup$
            – Chris Custer
            Nov 29 '18 at 16:12










          • $begingroup$
            Oh, I get it now, thank you so much.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:14











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          0












          $begingroup$

          Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,




          $A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you show me how you get to those results? I only got it right for $T(1)$.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:07










          • $begingroup$
            Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
            $endgroup$
            – Chris Custer
            Nov 29 '18 at 16:12










          • $begingroup$
            Oh, I get it now, thank you so much.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:14
















          0












          $begingroup$

          Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,




          $A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you show me how you get to those results? I only got it right for $T(1)$.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:07










          • $begingroup$
            Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
            $endgroup$
            – Chris Custer
            Nov 29 '18 at 16:12










          • $begingroup$
            Oh, I get it now, thank you so much.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:14














          0












          0








          0





          $begingroup$

          Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,




          $A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.







          share|cite|improve this answer









          $endgroup$



          Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,




          $A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 16:05









          Chris CusterChris Custer

          11.3k3824




          11.3k3824












          • $begingroup$
            Could you show me how you get to those results? I only got it right for $T(1)$.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:07










          • $begingroup$
            Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
            $endgroup$
            – Chris Custer
            Nov 29 '18 at 16:12










          • $begingroup$
            Oh, I get it now, thank you so much.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:14


















          • $begingroup$
            Could you show me how you get to those results? I only got it right for $T(1)$.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:07










          • $begingroup$
            Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
            $endgroup$
            – Chris Custer
            Nov 29 '18 at 16:12










          • $begingroup$
            Oh, I get it now, thank you so much.
            $endgroup$
            – Vlad
            Nov 29 '18 at 16:14
















          $begingroup$
          Could you show me how you get to those results? I only got it right for $T(1)$.
          $endgroup$
          – Vlad
          Nov 29 '18 at 16:07




          $begingroup$
          Could you show me how you get to those results? I only got it right for $T(1)$.
          $endgroup$
          – Vlad
          Nov 29 '18 at 16:07












          $begingroup$
          Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
          $endgroup$
          – Chris Custer
          Nov 29 '18 at 16:12




          $begingroup$
          Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
          $endgroup$
          – Chris Custer
          Nov 29 '18 at 16:12












          $begingroup$
          Oh, I get it now, thank you so much.
          $endgroup$
          – Vlad
          Nov 29 '18 at 16:14




          $begingroup$
          Oh, I get it now, thank you so much.
          $endgroup$
          – Vlad
          Nov 29 '18 at 16:14


















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