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Diagonalizable polynomial

  1. Mar 4, 2014 #1

    Please see attached.

    From part (a), we know that kernel (D) is a constant function, i.e f(x)=c, say
    From part (d), we know that eigenvalue of D is zero

    My question:
    For part (e),
    Is it correct to say that D is diagonalizable if and only if P_n (R) = ker (D) ??

    so the only solution is that n=0, which implies P_n (R) is a constant function
    and a constant function is always diagonalizable
  2. jcsd
  3. Mar 4, 2014 #2


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    The obvious basis of [itex]\ker D[/itex] is the constant function 1.

    Only if all of the eigenvalues vanish.

    You don't diagonalize vectors, you diagonalize matrices. And since a 1x1 matrix is necessarily diagonal, indeed [itex]D: \mathcal{P}_0 \to \mathcal{P}_0[/itex] is diagonalizable.

    That deals with the case [itex]n = 0[/itex]. What about the case when [itex]n > 0[/itex], which you are also asked to deal with?
  4. Mar 4, 2014 #3
    Thanks for your reply

    Yes, I know the basis of Ker (D) = {1}

    For part (e),
    we already know D is diagonalizable for the case n=0, because a 1x1 matrix is obviously diagonal
    So we let the eigenvector of D be v_1, say

    For n>0
    Assume D is diagonalizable
    So D must have at least two eigenvectors
    but the basis of ker(D) is 1
    so it is not diagonalizable because it is not linearly independent

    Is it correct?

    I think basis is a big hint but I do not know how to use it
  5. Mar 4, 2014 #4
    How did you get from "the basis of ker(D) is 1" to "it is not diagonalizable".

    Perhaps your mistake is that you think that al the eigenvectors must be elements of ker(D). This is not true.
  6. Mar 4, 2014 #5
    just guessing...i m a beginner in this topic

    do you mean the eigenvectors must be elements of either Im(D) or ker(D), but not both?
  7. Mar 4, 2014 #6
    Right, there are more eigenvectors than what's in Ker(D). Ker(D) contains exactly the eigenvectors of eigenvalue 0. There might be eigenvectors of another eigenvalue.
  8. Mar 4, 2014 #7
    OK. This is my explanation. Please have a look

    For n=0
    Po(R) = a_0 = ker D , where a_0 is in R
    Basis of Po(R) = {1}, which consists of an eigenvector with eigenvalue 0 for D

    For n≥1
    Pn(R) = {a_0+...+a_n*x^n}
    Dim of Pn(R)=n+1=dim(Ker(D))+dim(Im(D))
    The basis of Pn(R) should consist of n+1 independent eigenvectors, but Ker(D)=1 and rank (D)=n
    which means we cannot find n+1 independent eigenvectors
    so it is not diagonalizable
    Last edited: Mar 4, 2014
  9. Mar 4, 2014 #8
    I follow you so far.

    Why not?

    Why can't we fine n independent eigenvectors in Im(D) and 1 eigenvector in Ker(D)? Together this would give us n+1 independent eigenvectors.
  10. Mar 4, 2014 #9


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    The OP has already established that the only eigenvalue of [itex]D[/itex] is zero, presumably on the grounds that [itex]e^{kx} \notin \mathcal{P}_n(\mathbb{R})[/itex] for [itex]k \neq 0[/itex], or, more generally, on the grounds that an earlier part of the question required him to show that [itex]D^{n+1}[/itex] is the zero map.

    What is not linearly independent?

    The argument here, which you in fact gave in your original post, is that if [itex]D[/itex] is diagonalizable then it has exactly [itex]n+1[/itex] linearly independent eigenvectors. Thus, since the only eigenvalue of [itex]D[/itex] is zero, we should have [itex]\dim \ker D = n+1[/itex] (so that [itex]\ker D = \mathcal{P}_n[/itex]), which if [itex]n \geq 1[/itex] contradicts [itex]\dim \ker D = 1[/itex].
  11. Mar 4, 2014 #10
    I regret studying maths-related degree in university..

    The reason for the eigenvalue is zero implies dim (Ker D) = n+1
    is because n+1-fold composition is the zero map?

    I think I stuck in here so I cannot give a correct explanation
  12. Mar 4, 2014 #11


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    If a linear map is nilpotent (ie, a finite iterate is the zero map), then one can show that the only eigenvalue is zero. That's part (d).

    For part (e): there are a number of arguments one can run, but they all involve a proof by contradiction, in which one assumes that [itex]D[/itex] is diagonalizable and shows that this is inconsistent with, for example, [itex]D(x) = 1[/itex].

    There are various consequences of a matrix [itex]A[/itex] being diagonalizable. Firstly, by definition, there exists an invertible [itex]Q[/itex] such that [itex]QAQ^{-1}[/itex] is diagonal.

    But since the only eigenvalue of [itex]D[/itex] is zero, we have that if it is diagonalizable then there is an invertible [itex]Q[/itex] such that
    QDQ^{-1} = 0
    and it should be obvious that this means that [itex]D = 0[/itex].

    Secondly, if [itex]A[/itex] is diagonalizable then there exists a basis of eigenvectors.

    Again, the only eigenvalue of [itex]D[/itex] is zero, so if [itex]D[/itex] is diagonalizable then [itex]D(v) = 0[/itex] for all [itex]v \in \mathcal{P}_n[/itex] (since then [itex]v[/itex] is a linear combination of eigenvectors whose corresponding eigenvalue is zero). Thus again [itex]D = 0[/itex].

    Either of these arguments is sufficient to establish that if [itex]D[/itex] is diagonalizable then it is the zero map, which is equivalent to saying that [itex]\ker D = \mathcal{P}_n[/itex] or that [itex]\dim \ker D = n+1[/itex].
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