Linear Algebra: Dimension of kernel












0












$begingroup$


Suppose that we have the vector space



$V={f/f:Rrightarrow R, text{every class derivative is defined in R}}$



and $φ: Vrightarrow W$ with $φ(f)=f+f'$ is linear.
I want to find the dimension of the $Ker φ$ and if $Βleq V$ with $Βcap A={0}$ to show that the restriction of $φ$ in $B$ is one to one.
Any ideas please?






linear-algebra vector-fields







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  • $begingroup$
    what's a class derivative
    $endgroup$
    – seanv507
    Jan 8 at 10:14
















0












$begingroup$


Suppose that we have the vector space



$V={f/f:Rrightarrow R, text{every class derivative is defined in R}}$



and $φ: Vrightarrow W$ with $φ(f)=f+f'$ is linear.
I want to find the dimension of the $Ker φ$ and if $Βleq V$ with $Βcap A={0}$ to show that the restriction of $φ$ in $B$ is one to one.
Any ideas please?






linear-algebra vector-fields







share|cite|improve this question









$endgroup$



migrated from stats.stackexchange.com Jan 8 at 11:22


This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.


















  • $begingroup$
    what's a class derivative
    $endgroup$
    – seanv507
    Jan 8 at 10:14














0












0








0





$begingroup$


Suppose that we have the vector space



$V={f/f:Rrightarrow R, text{every class derivative is defined in R}}$



and $φ: Vrightarrow W$ with $φ(f)=f+f'$ is linear.
I want to find the dimension of the $Ker φ$ and if $Βleq V$ with $Βcap A={0}$ to show that the restriction of $φ$ in $B$ is one to one.
Any ideas please?






linear-algebra vector-fields







share|cite|improve this question









$endgroup$




Suppose that we have the vector space



$V={f/f:Rrightarrow R, text{every class derivative is defined in R}}$



and $φ: Vrightarrow W$ with $φ(f)=f+f'$ is linear.
I want to find the dimension of the $Ker φ$ and if $Βleq V$ with $Βcap A={0}$ to show that the restriction of $φ$ in $B$ is one to one.
Any ideas please?






linear-algebra vector-fields




linear-algebra vector-fields






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asked Jan 8 at 8:20









GeorgeGeorge

32




32




migrated from stats.stackexchange.com Jan 8 at 11:22


This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.









migrated from stats.stackexchange.com Jan 8 at 11:22


This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.














  • $begingroup$
    what's a class derivative
    $endgroup$
    – seanv507
    Jan 8 at 10:14


















  • $begingroup$
    what's a class derivative
    $endgroup$
    – seanv507
    Jan 8 at 10:14
















$begingroup$
what's a class derivative
$endgroup$
– seanv507
Jan 8 at 10:14




$begingroup$
what's a class derivative
$endgroup$
– seanv507
Jan 8 at 10:14










2 Answers
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Let's $f in V$ and $g in text{Ker}{varphi} subset V$, thus we have $varphi(f+g) = varphi(f)$



So we can write:
$varphi(f+g) = varphi(f) + varphi(g) = f +f' + g +g' = f +f' Rightarrow g +g' = 0$



Which is an ODE with solution $g = ke^{-t}$. This spans the kernel which has dimension 1.






share|cite|improve this answer









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    $begingroup$

    I agree with your solution. How can I prove now that the restriction of $phi$ in $B$ is one to one?






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      0












      $begingroup$

      Let's $f in V$ and $g in text{Ker}{varphi} subset V$, thus we have $varphi(f+g) = varphi(f)$



      So we can write:
      $varphi(f+g) = varphi(f) + varphi(g) = f +f' + g +g' = f +f' Rightarrow g +g' = 0$



      Which is an ODE with solution $g = ke^{-t}$. This spans the kernel which has dimension 1.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let's $f in V$ and $g in text{Ker}{varphi} subset V$, thus we have $varphi(f+g) = varphi(f)$



        So we can write:
        $varphi(f+g) = varphi(f) + varphi(g) = f +f' + g +g' = f +f' Rightarrow g +g' = 0$



        Which is an ODE with solution $g = ke^{-t}$. This spans the kernel which has dimension 1.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let's $f in V$ and $g in text{Ker}{varphi} subset V$, thus we have $varphi(f+g) = varphi(f)$



          So we can write:
          $varphi(f+g) = varphi(f) + varphi(g) = f +f' + g +g' = f +f' Rightarrow g +g' = 0$



          Which is an ODE with solution $g = ke^{-t}$. This spans the kernel which has dimension 1.






          share|cite|improve this answer









          $endgroup$



          Let's $f in V$ and $g in text{Ker}{varphi} subset V$, thus we have $varphi(f+g) = varphi(f)$



          So we can write:
          $varphi(f+g) = varphi(f) + varphi(g) = f +f' + g +g' = f +f' Rightarrow g +g' = 0$



          Which is an ODE with solution $g = ke^{-t}$. This spans the kernel which has dimension 1.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 11:20









          Carlos CamposCarlos Campos

          58439




          58439























              0












              $begingroup$

              I agree with your solution. How can I prove now that the restriction of $phi$ in $B$ is one to one?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I agree with your solution. How can I prove now that the restriction of $phi$ in $B$ is one to one?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I agree with your solution. How can I prove now that the restriction of $phi$ in $B$ is one to one?






                  share|cite|improve this answer









                  $endgroup$



                  I agree with your solution. How can I prove now that the restriction of $phi$ in $B$ is one to one?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 7:40









                  GeorgeGeorge

                  32




                  32






























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