Investigate whether the given transformation is a monomorphism / epimorphism. Find image and kernel












2












$begingroup$


I have serious doubts - I will be very grateful if someone will help me here



Investigate whether the given transformation is a monomorphism / epimorphism. Find its image and kernel.
$$ F in L(mathbb R[x]_3,mathbb R[x]_3), F(p)(t) = p(t+1) - p(t) $$



My try



Okay, let
$$ p(t) = ax^3 + bx^2 + cx + d $$
then
$$F(p)(t) = ... = 3ax^2 + 3ax + 2bx + 3 = x^2(3a)+x(3a+2b) +3 = a(3x^2+3x) + b(2x) + 3 $$
Is it monomorphism?
let $m_0 = 0 wedge m_1 = 3a wedge m_2 = 3a+b wedge m_3 = 3$


so $a = frac{m_1}{3} wedge b = frac{m_2-m_1}{2} $

so $a$ and $b$ are determined unambiguously so it is monomorphism


It is not epimorphism because $ x^3 inmathbb R[x]_3 $ but I can't get $x^3$ in use of $F$

$ker F = left{ 0 right} $ because of part $ 3 $ it will be never polynomial zero



If it comes to image, we checked that:
$$F(p)(t) = ... = a(3x^2+3x) + b(2x) + 3 $$
so $ im(F) = span(3x^2+3,2x,3) $
Moreover this system is lineary independent so $rank(F) = 3$



Have I done this correctly? Or there is something to fix?






linear-algebra monomorphisms







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I have serious doubts - I will be very grateful if someone will help me here



    Investigate whether the given transformation is a monomorphism / epimorphism. Find its image and kernel.
    $$ F in L(mathbb R[x]_3,mathbb R[x]_3), F(p)(t) = p(t+1) - p(t) $$



    My try



    Okay, let
    $$ p(t) = ax^3 + bx^2 + cx + d $$
    then
    $$F(p)(t) = ... = 3ax^2 + 3ax + 2bx + 3 = x^2(3a)+x(3a+2b) +3 = a(3x^2+3x) + b(2x) + 3 $$
    Is it monomorphism?
    let $m_0 = 0 wedge m_1 = 3a wedge m_2 = 3a+b wedge m_3 = 3$


    so $a = frac{m_1}{3} wedge b = frac{m_2-m_1}{2} $

    so $a$ and $b$ are determined unambiguously so it is monomorphism


    It is not epimorphism because $ x^3 inmathbb R[x]_3 $ but I can't get $x^3$ in use of $F$

    $ker F = left{ 0 right} $ because of part $ 3 $ it will be never polynomial zero



    If it comes to image, we checked that:
    $$F(p)(t) = ... = a(3x^2+3x) + b(2x) + 3 $$
    so $ im(F) = span(3x^2+3,2x,3) $
    Moreover this system is lineary independent so $rank(F) = 3$



    Have I done this correctly? Or there is something to fix?






    linear-algebra monomorphisms







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I have serious doubts - I will be very grateful if someone will help me here



      Investigate whether the given transformation is a monomorphism / epimorphism. Find its image and kernel.
      $$ F in L(mathbb R[x]_3,mathbb R[x]_3), F(p)(t) = p(t+1) - p(t) $$



      My try



      Okay, let
      $$ p(t) = ax^3 + bx^2 + cx + d $$
      then
      $$F(p)(t) = ... = 3ax^2 + 3ax + 2bx + 3 = x^2(3a)+x(3a+2b) +3 = a(3x^2+3x) + b(2x) + 3 $$
      Is it monomorphism?
      let $m_0 = 0 wedge m_1 = 3a wedge m_2 = 3a+b wedge m_3 = 3$


      so $a = frac{m_1}{3} wedge b = frac{m_2-m_1}{2} $

      so $a$ and $b$ are determined unambiguously so it is monomorphism


      It is not epimorphism because $ x^3 inmathbb R[x]_3 $ but I can't get $x^3$ in use of $F$

      $ker F = left{ 0 right} $ because of part $ 3 $ it will be never polynomial zero



      If it comes to image, we checked that:
      $$F(p)(t) = ... = a(3x^2+3x) + b(2x) + 3 $$
      so $ im(F) = span(3x^2+3,2x,3) $
      Moreover this system is lineary independent so $rank(F) = 3$



      Have I done this correctly? Or there is something to fix?






      linear-algebra monomorphisms







      share|cite|improve this question









      $endgroup$




      I have serious doubts - I will be very grateful if someone will help me here



      Investigate whether the given transformation is a monomorphism / epimorphism. Find its image and kernel.
      $$ F in L(mathbb R[x]_3,mathbb R[x]_3), F(p)(t) = p(t+1) - p(t) $$



      My try



      Okay, let
      $$ p(t) = ax^3 + bx^2 + cx + d $$
      then
      $$F(p)(t) = ... = 3ax^2 + 3ax + 2bx + 3 = x^2(3a)+x(3a+2b) +3 = a(3x^2+3x) + b(2x) + 3 $$
      Is it monomorphism?
      let $m_0 = 0 wedge m_1 = 3a wedge m_2 = 3a+b wedge m_3 = 3$


      so $a = frac{m_1}{3} wedge b = frac{m_2-m_1}{2} $

      so $a$ and $b$ are determined unambiguously so it is monomorphism


      It is not epimorphism because $ x^3 inmathbb R[x]_3 $ but I can't get $x^3$ in use of $F$

      $ker F = left{ 0 right} $ because of part $ 3 $ it will be never polynomial zero



      If it comes to image, we checked that:
      $$F(p)(t) = ... = a(3x^2+3x) + b(2x) + 3 $$
      so $ im(F) = span(3x^2+3,2x,3) $
      Moreover this system is lineary independent so $rank(F) = 3$



      Have I done this correctly? Or there is something to fix?






      linear-algebra monomorphisms




      linear-algebra monomorphisms






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 12:42









      VirtualUserVirtualUser

      1,096116




      1,096116






















          1 Answer
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          $begingroup$

          You have a contradiction in your solution: if the map is injective it must also be surjective, because it is a linear map of a finite dimensional vector space to itself.



          It's much easier with matrices. If you consider the standard basis ${p_0=1,p_1=t,p_2=t^2,p_3=t^3}$, then you see that
          $$
          F(p_0)=0,quad
          F(p_1)=1,quad
          F(p_2)=2t+1quad
          F(p_3)=3t^2+3t+1
          $$
          and therefore the matrix of $F$ is
          begin{bmatrix}
          0 & 1 & 1 & 1 \
          0 & 0 & 2 & 3 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}



          What's the rank of this matrix?




          Where did you go wrong? The image of $p(t)=at^3+bt^2+ct+d$ is
          begin{align}
          p(t+1)-p(t)
          &=at^3+3at^2+3at+a+bt^2+2bt+b+ct+d-at^3-bt^2-ct-d\
          &=3at^2+(3a+2b)t+(a+b+c)
          end{align}
          which is zero when
          begin{cases}
          3a=0\
          3a+2b=0\
          a+b+c=0
          end{cases}
          so when $a=b=c=0$. But $d$ can be anything. Thus the nullity is $1$ and the rank is $3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            rank of this matrix is 3. Can you point me which poin twhere exactly do I have the mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:30












          • $begingroup$
            @VirtualUser The rank is $3$.
            $endgroup$
            – egreg
            Jan 4 at 16:31










          • $begingroup$
            Yes, stupid mistake, I had on mind $3$. But what with the moment of mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:34












          • $begingroup$
            @VirtualUser I added the analysis.
            $endgroup$
            – egreg
            Jan 4 at 16:36










          • $begingroup$
            Ahh, there is mistake but in calculating. It is not good but still better than mistake in understanding the problem. Anyway, thanks for tip and analysis
            $endgroup$
            – VirtualUser
            Jan 4 at 16:45











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          1 Answer
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          1 Answer
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          active

          oldest

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          1












          $begingroup$

          You have a contradiction in your solution: if the map is injective it must also be surjective, because it is a linear map of a finite dimensional vector space to itself.



          It's much easier with matrices. If you consider the standard basis ${p_0=1,p_1=t,p_2=t^2,p_3=t^3}$, then you see that
          $$
          F(p_0)=0,quad
          F(p_1)=1,quad
          F(p_2)=2t+1quad
          F(p_3)=3t^2+3t+1
          $$
          and therefore the matrix of $F$ is
          begin{bmatrix}
          0 & 1 & 1 & 1 \
          0 & 0 & 2 & 3 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}



          What's the rank of this matrix?




          Where did you go wrong? The image of $p(t)=at^3+bt^2+ct+d$ is
          begin{align}
          p(t+1)-p(t)
          &=at^3+3at^2+3at+a+bt^2+2bt+b+ct+d-at^3-bt^2-ct-d\
          &=3at^2+(3a+2b)t+(a+b+c)
          end{align}
          which is zero when
          begin{cases}
          3a=0\
          3a+2b=0\
          a+b+c=0
          end{cases}
          so when $a=b=c=0$. But $d$ can be anything. Thus the nullity is $1$ and the rank is $3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            rank of this matrix is 3. Can you point me which poin twhere exactly do I have the mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:30












          • $begingroup$
            @VirtualUser The rank is $3$.
            $endgroup$
            – egreg
            Jan 4 at 16:31










          • $begingroup$
            Yes, stupid mistake, I had on mind $3$. But what with the moment of mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:34












          • $begingroup$
            @VirtualUser I added the analysis.
            $endgroup$
            – egreg
            Jan 4 at 16:36










          • $begingroup$
            Ahh, there is mistake but in calculating. It is not good but still better than mistake in understanding the problem. Anyway, thanks for tip and analysis
            $endgroup$
            – VirtualUser
            Jan 4 at 16:45
















          1












          $begingroup$

          You have a contradiction in your solution: if the map is injective it must also be surjective, because it is a linear map of a finite dimensional vector space to itself.



          It's much easier with matrices. If you consider the standard basis ${p_0=1,p_1=t,p_2=t^2,p_3=t^3}$, then you see that
          $$
          F(p_0)=0,quad
          F(p_1)=1,quad
          F(p_2)=2t+1quad
          F(p_3)=3t^2+3t+1
          $$
          and therefore the matrix of $F$ is
          begin{bmatrix}
          0 & 1 & 1 & 1 \
          0 & 0 & 2 & 3 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}



          What's the rank of this matrix?




          Where did you go wrong? The image of $p(t)=at^3+bt^2+ct+d$ is
          begin{align}
          p(t+1)-p(t)
          &=at^3+3at^2+3at+a+bt^2+2bt+b+ct+d-at^3-bt^2-ct-d\
          &=3at^2+(3a+2b)t+(a+b+c)
          end{align}
          which is zero when
          begin{cases}
          3a=0\
          3a+2b=0\
          a+b+c=0
          end{cases}
          so when $a=b=c=0$. But $d$ can be anything. Thus the nullity is $1$ and the rank is $3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            rank of this matrix is 3. Can you point me which poin twhere exactly do I have the mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:30












          • $begingroup$
            @VirtualUser The rank is $3$.
            $endgroup$
            – egreg
            Jan 4 at 16:31










          • $begingroup$
            Yes, stupid mistake, I had on mind $3$. But what with the moment of mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:34












          • $begingroup$
            @VirtualUser I added the analysis.
            $endgroup$
            – egreg
            Jan 4 at 16:36










          • $begingroup$
            Ahh, there is mistake but in calculating. It is not good but still better than mistake in understanding the problem. Anyway, thanks for tip and analysis
            $endgroup$
            – VirtualUser
            Jan 4 at 16:45














          1












          1








          1





          $begingroup$

          You have a contradiction in your solution: if the map is injective it must also be surjective, because it is a linear map of a finite dimensional vector space to itself.



          It's much easier with matrices. If you consider the standard basis ${p_0=1,p_1=t,p_2=t^2,p_3=t^3}$, then you see that
          $$
          F(p_0)=0,quad
          F(p_1)=1,quad
          F(p_2)=2t+1quad
          F(p_3)=3t^2+3t+1
          $$
          and therefore the matrix of $F$ is
          begin{bmatrix}
          0 & 1 & 1 & 1 \
          0 & 0 & 2 & 3 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}



          What's the rank of this matrix?




          Where did you go wrong? The image of $p(t)=at^3+bt^2+ct+d$ is
          begin{align}
          p(t+1)-p(t)
          &=at^3+3at^2+3at+a+bt^2+2bt+b+ct+d-at^3-bt^2-ct-d\
          &=3at^2+(3a+2b)t+(a+b+c)
          end{align}
          which is zero when
          begin{cases}
          3a=0\
          3a+2b=0\
          a+b+c=0
          end{cases}
          so when $a=b=c=0$. But $d$ can be anything. Thus the nullity is $1$ and the rank is $3$.






          share|cite|improve this answer











          $endgroup$



          You have a contradiction in your solution: if the map is injective it must also be surjective, because it is a linear map of a finite dimensional vector space to itself.



          It's much easier with matrices. If you consider the standard basis ${p_0=1,p_1=t,p_2=t^2,p_3=t^3}$, then you see that
          $$
          F(p_0)=0,quad
          F(p_1)=1,quad
          F(p_2)=2t+1quad
          F(p_3)=3t^2+3t+1
          $$
          and therefore the matrix of $F$ is
          begin{bmatrix}
          0 & 1 & 1 & 1 \
          0 & 0 & 2 & 3 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}



          What's the rank of this matrix?




          Where did you go wrong? The image of $p(t)=at^3+bt^2+ct+d$ is
          begin{align}
          p(t+1)-p(t)
          &=at^3+3at^2+3at+a+bt^2+2bt+b+ct+d-at^3-bt^2-ct-d\
          &=3at^2+(3a+2b)t+(a+b+c)
          end{align}
          which is zero when
          begin{cases}
          3a=0\
          3a+2b=0\
          a+b+c=0
          end{cases}
          so when $a=b=c=0$. But $d$ can be anything. Thus the nullity is $1$ and the rank is $3$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 16:35

























          answered Jan 4 at 15:50









          egregegreg

          184k1486205




          184k1486205












          • $begingroup$
            rank of this matrix is 3. Can you point me which poin twhere exactly do I have the mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:30












          • $begingroup$
            @VirtualUser The rank is $3$.
            $endgroup$
            – egreg
            Jan 4 at 16:31










          • $begingroup$
            Yes, stupid mistake, I had on mind $3$. But what with the moment of mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:34












          • $begingroup$
            @VirtualUser I added the analysis.
            $endgroup$
            – egreg
            Jan 4 at 16:36










          • $begingroup$
            Ahh, there is mistake but in calculating. It is not good but still better than mistake in understanding the problem. Anyway, thanks for tip and analysis
            $endgroup$
            – VirtualUser
            Jan 4 at 16:45


















          • $begingroup$
            rank of this matrix is 3. Can you point me which poin twhere exactly do I have the mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:30












          • $begingroup$
            @VirtualUser The rank is $3$.
            $endgroup$
            – egreg
            Jan 4 at 16:31










          • $begingroup$
            Yes, stupid mistake, I had on mind $3$. But what with the moment of mistake?
            $endgroup$
            – VirtualUser
            Jan 4 at 16:34












          • $begingroup$
            @VirtualUser I added the analysis.
            $endgroup$
            – egreg
            Jan 4 at 16:36










          • $begingroup$
            Ahh, there is mistake but in calculating. It is not good but still better than mistake in understanding the problem. Anyway, thanks for tip and analysis
            $endgroup$
            – VirtualUser
            Jan 4 at 16:45
















          $begingroup$
          rank of this matrix is 3. Can you point me which poin twhere exactly do I have the mistake?
          $endgroup$
          – VirtualUser
          Jan 4 at 16:30






          $begingroup$
          rank of this matrix is 3. Can you point me which poin twhere exactly do I have the mistake?
          $endgroup$
          – VirtualUser
          Jan 4 at 16:30














          $begingroup$
          @VirtualUser The rank is $3$.
          $endgroup$
          – egreg
          Jan 4 at 16:31




          $begingroup$
          @VirtualUser The rank is $3$.
          $endgroup$
          – egreg
          Jan 4 at 16:31












          $begingroup$
          Yes, stupid mistake, I had on mind $3$. But what with the moment of mistake?
          $endgroup$
          – VirtualUser
          Jan 4 at 16:34






          $begingroup$
          Yes, stupid mistake, I had on mind $3$. But what with the moment of mistake?
          $endgroup$
          – VirtualUser
          Jan 4 at 16:34














          $begingroup$
          @VirtualUser I added the analysis.
          $endgroup$
          – egreg
          Jan 4 at 16:36




          $begingroup$
          @VirtualUser I added the analysis.
          $endgroup$
          – egreg
          Jan 4 at 16:36












          $begingroup$
          Ahh, there is mistake but in calculating. It is not good but still better than mistake in understanding the problem. Anyway, thanks for tip and analysis
          $endgroup$
          – VirtualUser
          Jan 4 at 16:45




          $begingroup$
          Ahh, there is mistake but in calculating. It is not good but still better than mistake in understanding the problem. Anyway, thanks for tip and analysis
          $endgroup$
          – VirtualUser
          Jan 4 at 16:45


















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          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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