Let $a in mathbb C$. In the space $(mathbb C[x]_{2})^{*}$ there are three functionals












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Let $a in mathbb C$. In the space $(mathbb C[x]_{2})^{*}$ there are three functionals: $$f_{1}^{*}(p)=p(0)+ap(1)$$ $$f_{2}^{*}(p)=p(1)-ap(0)$$ $$f_{3}^{*}(p)=p'(1)-p'(0)$$ (a) For which values of parameter $ a$ this system is the basis of space $(mathbb C[x]_{2})^{*}$
(b) Whether there are such polynomials $f_{1},f_{2},f_{3} in mathbb C[x]_2$ for $a = 0$ that $$f_{k}^{*}(f_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases} $$ If exist you find coefficients $ alpha _{1}, alpha _{2}, alpha _{3} in mathbb C$ such that for polynomial $q(x)=x^{2}-3$: $$ q= alpha _{1} f_{1} + alpha _{2} f_{2}+alpha _{3} f_{3}. $$




I understand it but for ordinary polynomials. However this task is really abstract for me - when I must use functionals and I completely don't know what can I do in this case. I know that you can talk me that I am looking for a ready solution but I really don't understand and I find any hint that could lead me to a solution because I do this task by a long time and I know that I cannot do it alone...






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    Let $a in mathbb C$. In the space $(mathbb C[x]_{2})^{*}$ there are three functionals: $$f_{1}^{*}(p)=p(0)+ap(1)$$ $$f_{2}^{*}(p)=p(1)-ap(0)$$ $$f_{3}^{*}(p)=p'(1)-p'(0)$$ (a) For which values of parameter $ a$ this system is the basis of space $(mathbb C[x]_{2})^{*}$
    (b) Whether there are such polynomials $f_{1},f_{2},f_{3} in mathbb C[x]_2$ for $a = 0$ that $$f_{k}^{*}(f_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases} $$ If exist you find coefficients $ alpha _{1}, alpha _{2}, alpha _{3} in mathbb C$ such that for polynomial $q(x)=x^{2}-3$: $$ q= alpha _{1} f_{1} + alpha _{2} f_{2}+alpha _{3} f_{3}. $$




    I understand it but for ordinary polynomials. However this task is really abstract for me - when I must use functionals and I completely don't know what can I do in this case. I know that you can talk me that I am looking for a ready solution but I really don't understand and I find any hint that could lead me to a solution because I do this task by a long time and I know that I cannot do it alone...






    linear-algebra







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      Let $a in mathbb C$. In the space $(mathbb C[x]_{2})^{*}$ there are three functionals: $$f_{1}^{*}(p)=p(0)+ap(1)$$ $$f_{2}^{*}(p)=p(1)-ap(0)$$ $$f_{3}^{*}(p)=p'(1)-p'(0)$$ (a) For which values of parameter $ a$ this system is the basis of space $(mathbb C[x]_{2})^{*}$
      (b) Whether there are such polynomials $f_{1},f_{2},f_{3} in mathbb C[x]_2$ for $a = 0$ that $$f_{k}^{*}(f_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases} $$ If exist you find coefficients $ alpha _{1}, alpha _{2}, alpha _{3} in mathbb C$ such that for polynomial $q(x)=x^{2}-3$: $$ q= alpha _{1} f_{1} + alpha _{2} f_{2}+alpha _{3} f_{3}. $$




      I understand it but for ordinary polynomials. However this task is really abstract for me - when I must use functionals and I completely don't know what can I do in this case. I know that you can talk me that I am looking for a ready solution but I really don't understand and I find any hint that could lead me to a solution because I do this task by a long time and I know that I cannot do it alone...






      linear-algebra







      share|cite|improve this question











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      Let $a in mathbb C$. In the space $(mathbb C[x]_{2})^{*}$ there are three functionals: $$f_{1}^{*}(p)=p(0)+ap(1)$$ $$f_{2}^{*}(p)=p(1)-ap(0)$$ $$f_{3}^{*}(p)=p'(1)-p'(0)$$ (a) For which values of parameter $ a$ this system is the basis of space $(mathbb C[x]_{2})^{*}$
      (b) Whether there are such polynomials $f_{1},f_{2},f_{3} in mathbb C[x]_2$ for $a = 0$ that $$f_{k}^{*}(f_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases} $$ If exist you find coefficients $ alpha _{1}, alpha _{2}, alpha _{3} in mathbb C$ such that for polynomial $q(x)=x^{2}-3$: $$ q= alpha _{1} f_{1} + alpha _{2} f_{2}+alpha _{3} f_{3}. $$




      I understand it but for ordinary polynomials. However this task is really abstract for me - when I must use functionals and I completely don't know what can I do in this case. I know that you can talk me that I am looking for a ready solution but I really don't understand and I find any hint that could lead me to a solution because I do this task by a long time and I know that I cannot do it alone...






      linear-algebra




      linear-algebra






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      edited Feb 2 at 22:02







      MP3129

















      asked Feb 2 at 17:09









      MP3129MP3129

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          I presume that you know that ${1,x,x^2}$ is a basis of $mathbb{C}[x]_2$. Then, the mappings defined as $$g_{k}^{*}(g_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases}$$ are a basis of $(mathbb{C}[x]_2)^*$. Any polynomial $p$ can be written as $$p = bx^2 + cx +d.$$ We then have



          $$begin{pmatrix} f_1^* (p) \ f_2^* (p) \ f_3^* (p) end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} d \ c \ b end{pmatrix},$$
          or, equivalently
          $$ begin{pmatrix} f_1^* \ f_2^* \ f_3^* end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} g_1^* \ g_2^* \ g_3^* end{pmatrix}.$$
          Now if you know that a linear transformation maps a basis to a basis if and only if it is invertible, that should give you a way to check a). For b), think about what the transpose of the matrix above does.






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            $begingroup$

            I presume that you know that ${1,x,x^2}$ is a basis of $mathbb{C}[x]_2$. Then, the mappings defined as $$g_{k}^{*}(g_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases}$$ are a basis of $(mathbb{C}[x]_2)^*$. Any polynomial $p$ can be written as $$p = bx^2 + cx +d.$$ We then have



            $$begin{pmatrix} f_1^* (p) \ f_2^* (p) \ f_3^* (p) end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} d \ c \ b end{pmatrix},$$
            or, equivalently
            $$ begin{pmatrix} f_1^* \ f_2^* \ f_3^* end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} g_1^* \ g_2^* \ g_3^* end{pmatrix}.$$
            Now if you know that a linear transformation maps a basis to a basis if and only if it is invertible, that should give you a way to check a). For b), think about what the transpose of the matrix above does.






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              $begingroup$

              I presume that you know that ${1,x,x^2}$ is a basis of $mathbb{C}[x]_2$. Then, the mappings defined as $$g_{k}^{*}(g_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases}$$ are a basis of $(mathbb{C}[x]_2)^*$. Any polynomial $p$ can be written as $$p = bx^2 + cx +d.$$ We then have



              $$begin{pmatrix} f_1^* (p) \ f_2^* (p) \ f_3^* (p) end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} d \ c \ b end{pmatrix},$$
              or, equivalently
              $$ begin{pmatrix} f_1^* \ f_2^* \ f_3^* end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} g_1^* \ g_2^* \ g_3^* end{pmatrix}.$$
              Now if you know that a linear transformation maps a basis to a basis if and only if it is invertible, that should give you a way to check a). For b), think about what the transpose of the matrix above does.






              share|cite|improve this answer









              $endgroup$
















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                $begingroup$

                I presume that you know that ${1,x,x^2}$ is a basis of $mathbb{C}[x]_2$. Then, the mappings defined as $$g_{k}^{*}(g_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases}$$ are a basis of $(mathbb{C}[x]_2)^*$. Any polynomial $p$ can be written as $$p = bx^2 + cx +d.$$ We then have



                $$begin{pmatrix} f_1^* (p) \ f_2^* (p) \ f_3^* (p) end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} d \ c \ b end{pmatrix},$$
                or, equivalently
                $$ begin{pmatrix} f_1^* \ f_2^* \ f_3^* end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} g_1^* \ g_2^* \ g_3^* end{pmatrix}.$$
                Now if you know that a linear transformation maps a basis to a basis if and only if it is invertible, that should give you a way to check a). For b), think about what the transpose of the matrix above does.






                share|cite|improve this answer









                $endgroup$



                I presume that you know that ${1,x,x^2}$ is a basis of $mathbb{C}[x]_2$. Then, the mappings defined as $$g_{k}^{*}(g_{j})= begin{cases} 1, k=j \ 0,k neq j ?end{cases}$$ are a basis of $(mathbb{C}[x]_2)^*$. Any polynomial $p$ can be written as $$p = bx^2 + cx +d.$$ We then have



                $$begin{pmatrix} f_1^* (p) \ f_2^* (p) \ f_3^* (p) end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} d \ c \ b end{pmatrix},$$
                or, equivalently
                $$ begin{pmatrix} f_1^* \ f_2^* \ f_3^* end{pmatrix} = begin{pmatrix}1+a & a & a \ 1-a & 1 & 1 \ 0 & -1 & 2end{pmatrix} begin{pmatrix} g_1^* \ g_2^* \ g_3^* end{pmatrix}.$$
                Now if you know that a linear transformation maps a basis to a basis if and only if it is invertible, that should give you a way to check a). For b), think about what the transpose of the matrix above does.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Feb 3 at 4:57









                user159517user159517

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