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Let [itex]V= \mathbb{R}_3[x][/itex] be the vector space of polynomials

  1. Oct 20, 2011 #1
    Let [itex]V= \mathbb{R}_3[x][/itex] be the vector space of polynomials with real coefficients with degree at most 3 and let [itex]D:V\to V[/itex] be the linear operator of taking derivatives, [itex]D(f)=f'[/itex].

    I'm trying to check the Rank-nullity theorem for this example but it doesn't seem to hold:

    Since [itex]D[/itex] is not injective (e.g. images of [itex]x^2[/itex] and [itex]x^2+1[/itex] are both [itex]2x[/itex]), [itex]\text{Ker}(D)\neq \{ \bf{0} \}[/itex]. (The kernel is actually all the constant polynomials).

    Since [itex]D[/itex] is surjective (as every polynomial is the derivative of some other polynomial), [itex]\text{Im}(D)=V=\mathbb{R}_3[x][/itex].

    Now the Rank-nullity formula is:

    [itex]\text{dim}(\text{Ker}(D)) + \text{dim}(\text{Im}(D)) = \text{dim}(V)[/itex].

    But [itex]\text{dim}(\text{Im}(D)) = \text{dim}(\mathbb{R}_3[x]) = 4 = \text{dim}(V)[/itex]

    so if the Rank-nullity formula holds this would imply [itex]\text{dim}(\text{Ker}(D))=0[/itex] but it doesn't as [itex]D[/itex] is not injective, so what's going on?

    EDIT: D is not surjective! How do I find a basis for the Kernel and Image?
    Last edited: Oct 20, 2011
  2. jcsd
  3. Oct 20, 2011 #2


    Staff: Mentor

    Re: Rank-nullity

    Your mistake is here. Im(D) != R3. The image of a polynomial of degree 3 is a polynomial of degree 2, so Im(D) does not contain 3rd degree polynomials, hence D is not onto (surjective).

    BTW, the notation used in this problem seems nonstandard to me. The space of real polynomials of degree less than n is often called Pn. So your space would normally be called P4. At least, that's the notation I've seen most often.
  4. Oct 21, 2011 #3
    Re: Rank-nullity

    Yeah I realised that restricting the degrees of the polynomials made D not surjective.

    So with this notation does [itex]\text{Im}(D) = \mathbb{R}_2[x][/itex] (i.e. Polynomials in x with real coefficients with degree at most 2)?

    We know [itex]\text{Ker}(D)=\{ \text{constant polynomials} \}[/itex] so to describe all the bases of [itex]\text{Ker}(D)[/itex] can I just say the set of singletons [itex]\{k\}[/itex] where [itex]k\in\mathbb{R} \setminus \{0\}[/itex] ? So [itex]\text{Ker}(D)[/itex] has dimension 1.

    And [itex]\text{Im}(D) = \mathbb{R}_2[x][/itex] so a basis for the image is [itex]\{1,x,x^2 \}[/itex] with dimension 3. Then 1+3=4=dim(V) so the Rank-nullity formula is satisfied.
  5. Oct 21, 2011 #4


    Staff: Mentor

    Re: Rank-nullity

    Yes, Ker(D) has dimension 1. You don't need to (and shouldn't) exclude the 0 polynomial.
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