The differential Linear operator of a space of polynomials of degree at most $n$












0












$begingroup$


I've come across this statement made about the differentiation linear operator say $D$ on a vector space $V$ of polynomials of degree at most $n$. I mean the first derivative in this case. I can verify that indeed $V$ is a vector space and also $D$ maps the vector space V to a space of polynomials of degree at most $n-1$ which is a subspace of $V$. $Din L(V,V)$.



My concern is with the matrix representation of this linear operator $D$. I read from a textbook that
$$D=left[ {begin{array}{ccccc}
0 & 1 & 0 & 0 &... & 0 \
0 & 0 & 2 &0 &... &0\
0 & 0 & 0& 3 &...& 0\
. & . &. &.&...&.\
. & . &. &.&...&n-1\
0 & 0 & 0& 0 &...& 0
end{array} } right]
$$
I want to know how they came about the matrix $D$. Also
considering a basis for the space $V$, $$B={1,x,x^2,x^3,...,x^n}$$
What is the image of B under the operator D and how we achieve it.






linear-algebra matrices







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I've come across this statement made about the differentiation linear operator say $D$ on a vector space $V$ of polynomials of degree at most $n$. I mean the first derivative in this case. I can verify that indeed $V$ is a vector space and also $D$ maps the vector space V to a space of polynomials of degree at most $n-1$ which is a subspace of $V$. $Din L(V,V)$.



    My concern is with the matrix representation of this linear operator $D$. I read from a textbook that
    $$D=left[ {begin{array}{ccccc}
    0 & 1 & 0 & 0 &... & 0 \
    0 & 0 & 2 &0 &... &0\
    0 & 0 & 0& 3 &...& 0\
    . & . &. &.&...&.\
    . & . &. &.&...&n-1\
    0 & 0 & 0& 0 &...& 0
    end{array} } right]
    $$
    I want to know how they came about the matrix $D$. Also
    considering a basis for the space $V$, $$B={1,x,x^2,x^3,...,x^n}$$
    What is the image of B under the operator D and how we achieve it.






    linear-algebra matrices







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I've come across this statement made about the differentiation linear operator say $D$ on a vector space $V$ of polynomials of degree at most $n$. I mean the first derivative in this case. I can verify that indeed $V$ is a vector space and also $D$ maps the vector space V to a space of polynomials of degree at most $n-1$ which is a subspace of $V$. $Din L(V,V)$.



      My concern is with the matrix representation of this linear operator $D$. I read from a textbook that
      $$D=left[ {begin{array}{ccccc}
      0 & 1 & 0 & 0 &... & 0 \
      0 & 0 & 2 &0 &... &0\
      0 & 0 & 0& 3 &...& 0\
      . & . &. &.&...&.\
      . & . &. &.&...&n-1\
      0 & 0 & 0& 0 &...& 0
      end{array} } right]
      $$
      I want to know how they came about the matrix $D$. Also
      considering a basis for the space $V$, $$B={1,x,x^2,x^3,...,x^n}$$
      What is the image of B under the operator D and how we achieve it.






      linear-algebra matrices







      share|cite|improve this question









      $endgroup$




      I've come across this statement made about the differentiation linear operator say $D$ on a vector space $V$ of polynomials of degree at most $n$. I mean the first derivative in this case. I can verify that indeed $V$ is a vector space and also $D$ maps the vector space V to a space of polynomials of degree at most $n-1$ which is a subspace of $V$. $Din L(V,V)$.



      My concern is with the matrix representation of this linear operator $D$. I read from a textbook that
      $$D=left[ {begin{array}{ccccc}
      0 & 1 & 0 & 0 &... & 0 \
      0 & 0 & 2 &0 &... &0\
      0 & 0 & 0& 3 &...& 0\
      . & . &. &.&...&.\
      . & . &. &.&...&n-1\
      0 & 0 & 0& 0 &...& 0
      end{array} } right]
      $$
      I want to know how they came about the matrix $D$. Also
      considering a basis for the space $V$, $$B={1,x,x^2,x^3,...,x^n}$$
      What is the image of B under the operator D and how we achieve it.






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 2 '18 at 8:03









      J. KyeiJ. Kyei

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          1 Answer
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          $begingroup$

          Note that the polynomials of degree at most $n$ is of dimension $n+1$, so the last column of the matrix $D$ should have entry $n$.



          $D$ is in fact the matrix representation of the differentiation operator under basis $B$.



          That is to say, let $p(x)=a_0+a_1x+dotsc+a_nx^n$ be any polynomial, which we will represent as the vector $vec{p}=[a_0,a_1,dotsc,a_n]^T$.



          Then the derivative of $p(x)$ is simply the vector $Dvec{p}=[a_1,2a_2,dotsc,na_n,0]^T$, which represents $p'(x)=a_1+2a_2x+dotsc+na_nx^{n-1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Exactly. As You pointed out there seems to be a little mistake in the textbook though ($Dinmathbb{R}^{(n+1)times(n+1)}$)
            $endgroup$
            – Peter Melech
            Dec 2 '18 at 10:45










          • $begingroup$
            @PeterMelech, Yes, I think I misquoted that part. thank you.
            $endgroup$
            – J. Kyei
            Dec 3 '18 at 5:41











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          1












          $begingroup$

          Note that the polynomials of degree at most $n$ is of dimension $n+1$, so the last column of the matrix $D$ should have entry $n$.



          $D$ is in fact the matrix representation of the differentiation operator under basis $B$.



          That is to say, let $p(x)=a_0+a_1x+dotsc+a_nx^n$ be any polynomial, which we will represent as the vector $vec{p}=[a_0,a_1,dotsc,a_n]^T$.



          Then the derivative of $p(x)$ is simply the vector $Dvec{p}=[a_1,2a_2,dotsc,na_n,0]^T$, which represents $p'(x)=a_1+2a_2x+dotsc+na_nx^{n-1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Exactly. As You pointed out there seems to be a little mistake in the textbook though ($Dinmathbb{R}^{(n+1)times(n+1)}$)
            $endgroup$
            – Peter Melech
            Dec 2 '18 at 10:45










          • $begingroup$
            @PeterMelech, Yes, I think I misquoted that part. thank you.
            $endgroup$
            – J. Kyei
            Dec 3 '18 at 5:41
















          1












          $begingroup$

          Note that the polynomials of degree at most $n$ is of dimension $n+1$, so the last column of the matrix $D$ should have entry $n$.



          $D$ is in fact the matrix representation of the differentiation operator under basis $B$.



          That is to say, let $p(x)=a_0+a_1x+dotsc+a_nx^n$ be any polynomial, which we will represent as the vector $vec{p}=[a_0,a_1,dotsc,a_n]^T$.



          Then the derivative of $p(x)$ is simply the vector $Dvec{p}=[a_1,2a_2,dotsc,na_n,0]^T$, which represents $p'(x)=a_1+2a_2x+dotsc+na_nx^{n-1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Exactly. As You pointed out there seems to be a little mistake in the textbook though ($Dinmathbb{R}^{(n+1)times(n+1)}$)
            $endgroup$
            – Peter Melech
            Dec 2 '18 at 10:45










          • $begingroup$
            @PeterMelech, Yes, I think I misquoted that part. thank you.
            $endgroup$
            – J. Kyei
            Dec 3 '18 at 5:41














          1












          1








          1





          $begingroup$

          Note that the polynomials of degree at most $n$ is of dimension $n+1$, so the last column of the matrix $D$ should have entry $n$.



          $D$ is in fact the matrix representation of the differentiation operator under basis $B$.



          That is to say, let $p(x)=a_0+a_1x+dotsc+a_nx^n$ be any polynomial, which we will represent as the vector $vec{p}=[a_0,a_1,dotsc,a_n]^T$.



          Then the derivative of $p(x)$ is simply the vector $Dvec{p}=[a_1,2a_2,dotsc,na_n,0]^T$, which represents $p'(x)=a_1+2a_2x+dotsc+na_nx^{n-1}$.






          share|cite|improve this answer









          $endgroup$



          Note that the polynomials of degree at most $n$ is of dimension $n+1$, so the last column of the matrix $D$ should have entry $n$.



          $D$ is in fact the matrix representation of the differentiation operator under basis $B$.



          That is to say, let $p(x)=a_0+a_1x+dotsc+a_nx^n$ be any polynomial, which we will represent as the vector $vec{p}=[a_0,a_1,dotsc,a_n]^T$.



          Then the derivative of $p(x)$ is simply the vector $Dvec{p}=[a_1,2a_2,dotsc,na_n,0]^T$, which represents $p'(x)=a_1+2a_2x+dotsc+na_nx^{n-1}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 8:12









          obscuransobscurans

          1,027311




          1,027311












          • $begingroup$
            Exactly. As You pointed out there seems to be a little mistake in the textbook though ($Dinmathbb{R}^{(n+1)times(n+1)}$)
            $endgroup$
            – Peter Melech
            Dec 2 '18 at 10:45










          • $begingroup$
            @PeterMelech, Yes, I think I misquoted that part. thank you.
            $endgroup$
            – J. Kyei
            Dec 3 '18 at 5:41


















          • $begingroup$
            Exactly. As You pointed out there seems to be a little mistake in the textbook though ($Dinmathbb{R}^{(n+1)times(n+1)}$)
            $endgroup$
            – Peter Melech
            Dec 2 '18 at 10:45










          • $begingroup$
            @PeterMelech, Yes, I think I misquoted that part. thank you.
            $endgroup$
            – J. Kyei
            Dec 3 '18 at 5:41
















          $begingroup$
          Exactly. As You pointed out there seems to be a little mistake in the textbook though ($Dinmathbb{R}^{(n+1)times(n+1)}$)
          $endgroup$
          – Peter Melech
          Dec 2 '18 at 10:45




          $begingroup$
          Exactly. As You pointed out there seems to be a little mistake in the textbook though ($Dinmathbb{R}^{(n+1)times(n+1)}$)
          $endgroup$
          – Peter Melech
          Dec 2 '18 at 10:45












          $begingroup$
          @PeterMelech, Yes, I think I misquoted that part. thank you.
          $endgroup$
          – J. Kyei
          Dec 3 '18 at 5:41




          $begingroup$
          @PeterMelech, Yes, I think I misquoted that part. thank you.
          $endgroup$
          – J. Kyei
          Dec 3 '18 at 5:41


















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