Write a subspace as a Kernel of a linear application












1












$begingroup$


I'd like to discuss the following problem :



Write U = { $f in V | hspace{0.3 cm}x^{2} | f$},where | means "divides", and $ V = mathbb{R}_{k}[x]$



As kernel of the linear application : $$ F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{s}$$



With k that doens't depend on s,



And write down the matrix of change basis in the canonical basis of $mathbb{R}_{k}[x]$ and $mathbb{R}^{s}$.



The second part should be quite easier once found what $F$ does.



My attempt was to pass in coordinates because the kernel of $ F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{s}$ should be equal to the kernel of $ phi : mathbb{R}^{k+1} longmapsto mathbb{R}^{s}$,



Trying to find $U$ in $mathbb{R}^{k+1}$ writing $$mathbb{R}^{k+1} = Span begin{pmatrix}a \ b \ c+1 \ cdots \ k end{pmatrix} bigoplus Span {e_{1}, cdots e_{s} }$$ with $e_{3}$ missing, and ${a,b,cdots,k} in mathbb{R}$,



And defining $phi$ to be zero on the vectors of $Span begin{pmatrix}a \ b \ c+1 \ cdots \ k end{pmatrix}$,



This will work ?



Any help would be appreciated, thank you all!






linear-algebra vector-spaces linear-transformations direct-sum







share|cite|improve this question









$endgroup$












  • $begingroup$
    Looks rather weird. What is ${Bbb R}_k[x]$? First study the structure of $U$.
    $endgroup$
    – Wuestenfux
    Dec 21 '18 at 10:08












  • $begingroup$
    @Wuestenfux U should be the ideal generated by $x^{2}$ ?
    $endgroup$
    – jacopoburelli
    Dec 21 '18 at 10:09










  • $begingroup$
    What about $Fcolon p(x)=p_0+p_1x+ldotsmapsto (p_0,p_1)$?
    $endgroup$
    – A.Γ.
    Dec 21 '18 at 10:15
















1












$begingroup$


I'd like to discuss the following problem :



Write U = { $f in V | hspace{0.3 cm}x^{2} | f$},where | means "divides", and $ V = mathbb{R}_{k}[x]$



As kernel of the linear application : $$ F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{s}$$



With k that doens't depend on s,



And write down the matrix of change basis in the canonical basis of $mathbb{R}_{k}[x]$ and $mathbb{R}^{s}$.



The second part should be quite easier once found what $F$ does.



My attempt was to pass in coordinates because the kernel of $ F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{s}$ should be equal to the kernel of $ phi : mathbb{R}^{k+1} longmapsto mathbb{R}^{s}$,



Trying to find $U$ in $mathbb{R}^{k+1}$ writing $$mathbb{R}^{k+1} = Span begin{pmatrix}a \ b \ c+1 \ cdots \ k end{pmatrix} bigoplus Span {e_{1}, cdots e_{s} }$$ with $e_{3}$ missing, and ${a,b,cdots,k} in mathbb{R}$,



And defining $phi$ to be zero on the vectors of $Span begin{pmatrix}a \ b \ c+1 \ cdots \ k end{pmatrix}$,



This will work ?



Any help would be appreciated, thank you all!






linear-algebra vector-spaces linear-transformations direct-sum







share|cite|improve this question









$endgroup$












  • $begingroup$
    Looks rather weird. What is ${Bbb R}_k[x]$? First study the structure of $U$.
    $endgroup$
    – Wuestenfux
    Dec 21 '18 at 10:08












  • $begingroup$
    @Wuestenfux U should be the ideal generated by $x^{2}$ ?
    $endgroup$
    – jacopoburelli
    Dec 21 '18 at 10:09










  • $begingroup$
    What about $Fcolon p(x)=p_0+p_1x+ldotsmapsto (p_0,p_1)$?
    $endgroup$
    – A.Γ.
    Dec 21 '18 at 10:15














1












1








1





$begingroup$


I'd like to discuss the following problem :



Write U = { $f in V | hspace{0.3 cm}x^{2} | f$},where | means "divides", and $ V = mathbb{R}_{k}[x]$



As kernel of the linear application : $$ F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{s}$$



With k that doens't depend on s,



And write down the matrix of change basis in the canonical basis of $mathbb{R}_{k}[x]$ and $mathbb{R}^{s}$.



The second part should be quite easier once found what $F$ does.



My attempt was to pass in coordinates because the kernel of $ F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{s}$ should be equal to the kernel of $ phi : mathbb{R}^{k+1} longmapsto mathbb{R}^{s}$,



Trying to find $U$ in $mathbb{R}^{k+1}$ writing $$mathbb{R}^{k+1} = Span begin{pmatrix}a \ b \ c+1 \ cdots \ k end{pmatrix} bigoplus Span {e_{1}, cdots e_{s} }$$ with $e_{3}$ missing, and ${a,b,cdots,k} in mathbb{R}$,



And defining $phi$ to be zero on the vectors of $Span begin{pmatrix}a \ b \ c+1 \ cdots \ k end{pmatrix}$,



This will work ?



Any help would be appreciated, thank you all!






linear-algebra vector-spaces linear-transformations direct-sum







share|cite|improve this question









$endgroup$




I'd like to discuss the following problem :



Write U = { $f in V | hspace{0.3 cm}x^{2} | f$},where | means "divides", and $ V = mathbb{R}_{k}[x]$



As kernel of the linear application : $$ F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{s}$$



With k that doens't depend on s,



And write down the matrix of change basis in the canonical basis of $mathbb{R}_{k}[x]$ and $mathbb{R}^{s}$.



The second part should be quite easier once found what $F$ does.



My attempt was to pass in coordinates because the kernel of $ F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{s}$ should be equal to the kernel of $ phi : mathbb{R}^{k+1} longmapsto mathbb{R}^{s}$,



Trying to find $U$ in $mathbb{R}^{k+1}$ writing $$mathbb{R}^{k+1} = Span begin{pmatrix}a \ b \ c+1 \ cdots \ k end{pmatrix} bigoplus Span {e_{1}, cdots e_{s} }$$ with $e_{3}$ missing, and ${a,b,cdots,k} in mathbb{R}$,



And defining $phi$ to be zero on the vectors of $Span begin{pmatrix}a \ b \ c+1 \ cdots \ k end{pmatrix}$,



This will work ?



Any help would be appreciated, thank you all!






linear-algebra vector-spaces linear-transformations direct-sum




linear-algebra vector-spaces linear-transformations direct-sum






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 10:04









jacopoburellijacopoburelli

1687




1687












  • $begingroup$
    Looks rather weird. What is ${Bbb R}_k[x]$? First study the structure of $U$.
    $endgroup$
    – Wuestenfux
    Dec 21 '18 at 10:08












  • $begingroup$
    @Wuestenfux U should be the ideal generated by $x^{2}$ ?
    $endgroup$
    – jacopoburelli
    Dec 21 '18 at 10:09










  • $begingroup$
    What about $Fcolon p(x)=p_0+p_1x+ldotsmapsto (p_0,p_1)$?
    $endgroup$
    – A.Γ.
    Dec 21 '18 at 10:15


















  • $begingroup$
    Looks rather weird. What is ${Bbb R}_k[x]$? First study the structure of $U$.
    $endgroup$
    – Wuestenfux
    Dec 21 '18 at 10:08












  • $begingroup$
    @Wuestenfux U should be the ideal generated by $x^{2}$ ?
    $endgroup$
    – jacopoburelli
    Dec 21 '18 at 10:09










  • $begingroup$
    What about $Fcolon p(x)=p_0+p_1x+ldotsmapsto (p_0,p_1)$?
    $endgroup$
    – A.Γ.
    Dec 21 '18 at 10:15
















$begingroup$
Looks rather weird. What is ${Bbb R}_k[x]$? First study the structure of $U$.
$endgroup$
– Wuestenfux
Dec 21 '18 at 10:08






$begingroup$
Looks rather weird. What is ${Bbb R}_k[x]$? First study the structure of $U$.
$endgroup$
– Wuestenfux
Dec 21 '18 at 10:08














$begingroup$
@Wuestenfux U should be the ideal generated by $x^{2}$ ?
$endgroup$
– jacopoburelli
Dec 21 '18 at 10:09




$begingroup$
@Wuestenfux U should be the ideal generated by $x^{2}$ ?
$endgroup$
– jacopoburelli
Dec 21 '18 at 10:09












$begingroup$
What about $Fcolon p(x)=p_0+p_1x+ldotsmapsto (p_0,p_1)$?
$endgroup$
– A.Γ.
Dec 21 '18 at 10:15




$begingroup$
What about $Fcolon p(x)=p_0+p_1x+ldotsmapsto (p_0,p_1)$?
$endgroup$
– A.Γ.
Dec 21 '18 at 10:15










1 Answer
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$begingroup$

Observe that $U$ consists of all polynomials of the form $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2$.



Now let $s=2$ and define $F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{2}$ as follows:



If $f in V$ and $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2+a_1x+a_0$,



then $F(f):=(a_1,a_0)$.



Show that $F$ has the desired properties.






share|cite|improve this answer









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    $begingroup$

    Observe that $U$ consists of all polynomials of the form $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2$.



    Now let $s=2$ and define $F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{2}$ as follows:



    If $f in V$ and $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2+a_1x+a_0$,



    then $F(f):=(a_1,a_0)$.



    Show that $F$ has the desired properties.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Observe that $U$ consists of all polynomials of the form $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2$.



      Now let $s=2$ and define $F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{2}$ as follows:



      If $f in V$ and $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2+a_1x+a_0$,



      then $F(f):=(a_1,a_0)$.



      Show that $F$ has the desired properties.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Observe that $U$ consists of all polynomials of the form $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2$.



        Now let $s=2$ and define $F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{2}$ as follows:



        If $f in V$ and $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2+a_1x+a_0$,



        then $F(f):=(a_1,a_0)$.



        Show that $F$ has the desired properties.






        share|cite|improve this answer









        $endgroup$



        Observe that $U$ consists of all polynomials of the form $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2$.



        Now let $s=2$ and define $F : mathbb{R}_{k}[x] longmapsto mathbb{R}^{2}$ as follows:



        If $f in V$ and $f(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_2x^2+a_1x+a_0$,



        then $F(f):=(a_1,a_0)$.



        Show that $F$ has the desired properties.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 10:18









        FredFred

        45.4k1848




        45.4k1848






























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