Find $a_{0}, a_{1}, a_{2}, a_{3} in mathbb C$ for which $f^{*}=a_{0}p_{0}^{*}+ a_{1}p_{1}^{*}+a_{2}p_{2}^{*}+...












1












$begingroup$



In the linear space $mathbb C[x]_{3}$ is a basis: $p_{0}(x)=1$, $p_{1}(x)=x$, $p_{2}(x)=x(x-i)$, $p_{3}(x)=x(x-i)(x-1)$.
$(p_{0}^{*}, p_{1}^{*}, p_{2}^{*}, p_{3}^{*})$ is a basis of space $(mathbb C[x]_{3})^{*}$ which is dual for a basis $(p_{0}, p_{1}, p_{2}, p_{3})$.
Functional $f^{*} in (mathbb C[x]_{3})^{*}$ is such that $f^{*}(p)=p(-i)$.
Find $a_{0}, a_{1}, a_{2}, a_{3} in mathbb C$ for which $f^{*}=a_{0}p_{0}^{*}+ a_{1}p_{1}^{*}+a_{2}p_{2}^{*}+ a_{3}p_{3}^{*}$.






Can you get me some tips to this question? I never had similar task and I do not even know what to look for to start this task. I'm weak from line functionalities and it's too hard for me.






linear-algebra







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    1












    $begingroup$



    In the linear space $mathbb C[x]_{3}$ is a basis: $p_{0}(x)=1$, $p_{1}(x)=x$, $p_{2}(x)=x(x-i)$, $p_{3}(x)=x(x-i)(x-1)$.
    $(p_{0}^{*}, p_{1}^{*}, p_{2}^{*}, p_{3}^{*})$ is a basis of space $(mathbb C[x]_{3})^{*}$ which is dual for a basis $(p_{0}, p_{1}, p_{2}, p_{3})$.
    Functional $f^{*} in (mathbb C[x]_{3})^{*}$ is such that $f^{*}(p)=p(-i)$.
    Find $a_{0}, a_{1}, a_{2}, a_{3} in mathbb C$ for which $f^{*}=a_{0}p_{0}^{*}+ a_{1}p_{1}^{*}+a_{2}p_{2}^{*}+ a_{3}p_{3}^{*}$.






    Can you get me some tips to this question? I never had similar task and I do not even know what to look for to start this task. I'm weak from line functionalities and it's too hard for me.






    linear-algebra







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      In the linear space $mathbb C[x]_{3}$ is a basis: $p_{0}(x)=1$, $p_{1}(x)=x$, $p_{2}(x)=x(x-i)$, $p_{3}(x)=x(x-i)(x-1)$.
      $(p_{0}^{*}, p_{1}^{*}, p_{2}^{*}, p_{3}^{*})$ is a basis of space $(mathbb C[x]_{3})^{*}$ which is dual for a basis $(p_{0}, p_{1}, p_{2}, p_{3})$.
      Functional $f^{*} in (mathbb C[x]_{3})^{*}$ is such that $f^{*}(p)=p(-i)$.
      Find $a_{0}, a_{1}, a_{2}, a_{3} in mathbb C$ for which $f^{*}=a_{0}p_{0}^{*}+ a_{1}p_{1}^{*}+a_{2}p_{2}^{*}+ a_{3}p_{3}^{*}$.






      Can you get me some tips to this question? I never had similar task and I do not even know what to look for to start this task. I'm weak from line functionalities and it's too hard for me.






      linear-algebra







      share|cite|improve this question











      $endgroup$





      In the linear space $mathbb C[x]_{3}$ is a basis: $p_{0}(x)=1$, $p_{1}(x)=x$, $p_{2}(x)=x(x-i)$, $p_{3}(x)=x(x-i)(x-1)$.
      $(p_{0}^{*}, p_{1}^{*}, p_{2}^{*}, p_{3}^{*})$ is a basis of space $(mathbb C[x]_{3})^{*}$ which is dual for a basis $(p_{0}, p_{1}, p_{2}, p_{3})$.
      Functional $f^{*} in (mathbb C[x]_{3})^{*}$ is such that $f^{*}(p)=p(-i)$.
      Find $a_{0}, a_{1}, a_{2}, a_{3} in mathbb C$ for which $f^{*}=a_{0}p_{0}^{*}+ a_{1}p_{1}^{*}+a_{2}p_{2}^{*}+ a_{3}p_{3}^{*}$.






      Can you get me some tips to this question? I never had similar task and I do not even know what to look for to start this task. I'm weak from line functionalities and it's too hard for me.






      linear-algebra




      linear-algebra






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      share|cite|improve this question













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      edited Feb 2 at 23:17







      VirtualUser

















      asked Feb 2 at 19:04









      VirtualUserVirtualUser

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          $begingroup$

          Note that by the definition of the dual basis,
          $$p_j^*(p_i) = begin{cases}0 & i ne j\1 & i = jend{cases}$$



          So if $$f^* = a_0p_0^* + a_1p_1^* + a_2p_2^* + a_3p_3^*$$ then
          $$f^*(p_i) = a_i$$
          for all $i$



          And since $f^*(P) = P(-i)$ for any polynomial $P$, you can calculate $f^*(p_i)$ easily.






          share|cite|improve this answer









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            1












            $begingroup$

            Note that by the definition of the dual basis,
            $$p_j^*(p_i) = begin{cases}0 & i ne j\1 & i = jend{cases}$$



            So if $$f^* = a_0p_0^* + a_1p_1^* + a_2p_2^* + a_3p_3^*$$ then
            $$f^*(p_i) = a_i$$
            for all $i$



            And since $f^*(P) = P(-i)$ for any polynomial $P$, you can calculate $f^*(p_i)$ easily.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Note that by the definition of the dual basis,
              $$p_j^*(p_i) = begin{cases}0 & i ne j\1 & i = jend{cases}$$



              So if $$f^* = a_0p_0^* + a_1p_1^* + a_2p_2^* + a_3p_3^*$$ then
              $$f^*(p_i) = a_i$$
              for all $i$



              And since $f^*(P) = P(-i)$ for any polynomial $P$, you can calculate $f^*(p_i)$ easily.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Note that by the definition of the dual basis,
                $$p_j^*(p_i) = begin{cases}0 & i ne j\1 & i = jend{cases}$$



                So if $$f^* = a_0p_0^* + a_1p_1^* + a_2p_2^* + a_3p_3^*$$ then
                $$f^*(p_i) = a_i$$
                for all $i$



                And since $f^*(P) = P(-i)$ for any polynomial $P$, you can calculate $f^*(p_i)$ easily.






                share|cite|improve this answer









                $endgroup$



                Note that by the definition of the dual basis,
                $$p_j^*(p_i) = begin{cases}0 & i ne j\1 & i = jend{cases}$$



                So if $$f^* = a_0p_0^* + a_1p_1^* + a_2p_2^* + a_3p_3^*$$ then
                $$f^*(p_i) = a_i$$
                for all $i$



                And since $f^*(P) = P(-i)$ for any polynomial $P$, you can calculate $f^*(p_i)$ easily.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 3 at 4:35









                Paul SinclairPaul Sinclair

                20.9k21543




                20.9k21543






























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