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Kernel, Basis, Rank

  1. Feb 22, 2014 #1
    Please see attached question
    In my opinion this question is conceptional and abstract..

    For part a and b,
    I think dim(Ker(D)) = 1 and Rank(D) = n
    but I do not know how to explain them

    For part c
    What I can think of is if we differentiate f(x) by n+1 times
    then we will get 0

    Can somebody give me some hints, please?
     

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  2. jcsd
  3. Feb 22, 2014 #2

    Dick

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    dim(ker(D))=1, yes. Rank(D)=n, also right. I think you should start by writing down a basis for ##P_n##. Use that to try and explain.
     
    Last edited: Feb 22, 2014
  4. Feb 22, 2014 #3

    PeroK

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    Why is Rank(D) not = n? Pn has dimension n+1.
     
  5. Feb 22, 2014 #4

    Dick

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    It does. I miscounted. Sorry. I edited the original answer.
     
  6. Feb 22, 2014 #5
    I don't think that's a bad thing, is it? :tongue:

    For part (a), what are the only polynomials with derivative zero?

    For part (b), we have a couple options. The easiest way is by our rank-nullity theorem.

    Again, look at the kernel.
     
  7. Feb 24, 2014 #6
    Just figure out what a basis is..

    Basis of P = ##{x^n,x^{n-1},...,x,1}##

    Kernel is everything that gets mapped to 0
    The only polynomial with derivative 0 is ##a_0##, with basis 1
    so basis for kernel D is 1

    Is this a correct explanation?
     
    Last edited: Feb 24, 2014
  8. Feb 24, 2014 #7
    (b) null (D) = 1
    since dim(##P_n##)=n+1
    by rank-nullity theorem, rank (D) = n
    Pretty sure this explanation is correct
     
  9. Feb 24, 2014 #8
    For part (c)

    I think the matrix D is look like (see attached)
    but how does we relate to D^(n+1)?
     

    Attached Files:

  10. Feb 24, 2014 #9

    Dick

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    Best to say the basis for kernel(D) is {1} because it's a set. But yes, correct.

    Also correct.

    What the matrix looks like depends on which components of the vector correspond to which basis elements. Yours is where the topmost element of the vector corresponds to the coefficient of x^n. But, yes, it's fine. D^(n+1) is just that matrix multiplied by itself n+1 times. What's the result?
     
  11. Feb 25, 2014 #10
    Obviously, D^(n+1) = 0
    but how to show it? Is there any suitable method except showing by induction?
     
  12. Feb 25, 2014 #11

    Dick

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    You could argue it's true just by using what you know about derivatives. Also your matrix has zeros along the diagonal and is only nonzero along the first subdiagonal. If you look at D^2 that's zero everywhere except along the second subdiagonal. By the time you get to D^(n+1) you'll run out a subdiagonals. I don't think it needs a formal induction argument.
     
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