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Linear algebra - differentiation operator

  1. Dec 15, 2007 #1
    1. The problem statement, all variables and given/known data
    I have a differentiation operator on P_3, and:

    S = {p \in P_3 | p(0) = 0}.

    I have to show that

    1) D maps P_3 onto the subspace P_2

    2) D : P_3 -> P_2 is NOT one-to-one

    3) D: S -> P_3 is one-to-one

    4) D: S -> P_3 is NOT onto.

    3. The attempt at a solution

    1) Ok, for this I look at the bases. The basis for P_3 is [x^2, x, 1] and since we are dealing with an differentiation operator, the new basis must be [x, 1] - which is the basis for P_2, which we wanted to show.

    2) I have to show that ax^2+bx+c and 2ax+b can have the same values. Do I just choose some values and show that they are equal?

    3 + 4) I have to show that 2ax + b and ax^2+bx+c cannot have same values?

    I hope you can confirm/help me with this.

    Thanks in advance.
  2. jcsd
  3. Dec 15, 2007 #2


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    Yes, since you are showing the "negation" of a general statement (that D is one to one) a counterexample will suffice.

    No! For 3)you want to show that if p and q are distinct members of S, then Dp is NOT equal to Dq.
    For 4), you want to show that there exist some member, q, say, of P_3 (NOT P_2!) that is not of the form Dp for any p in S

  4. Dec 15, 2007 #3
    First, thanks for replying.

    For #3: I have to show that D : S -> P_3 is injective. I'm a little uncertain of this: If p and q are members of S, then they are on the form ax^2+bx. E.g.

    p(x) = q(x) = 2x^2+2x

    D(p(x) != D(q(x)) - aren't they equal?

    For #4: If for example q(x) = 3x^2+x (a member of P_3), then D(p) != q(x)? Is that what I must show?

    I hope you can clarify for me - thanks again!
  5. Dec 15, 2007 #4


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    If you start with p= q, of course, they are! But I specifically said distinct members! If [itex]p\ne q[/itex], that is, if [itex]p= ax^2+ b[/itex] and [itex]q= cx^2+ d[/itex] where either [itex]c\ne a[/itex] or [itex]d\ne b[/itex] or both, is Dp= Dq?

    No, that's not what you must show- it's not true! If [itex]p= x^3+ x^2/2[/itex] then Dp is equal to q. Your particular choice of q(x) is a member of P_2 (which is a subspace of P_3) and I said it was important to note that this problem is about P_3, not P_2.
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