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Basis and Dimension of a Subpace.

  1. May 23, 2012 #1
    1. The problem statement, all variables and given/known data

    V = {p(x) belongs to P3 such that p'(1) + p'(-1) = 0}

    2. Relevant equations

    ...

    3. The attempt at a solution

    Okay, so finding the first derivative of p(x) = ax^3 + bx^2 + cx + d and plugging in the values 1 and -1 (to find p'(1) and p'(-1)), we get c = -3a. Does this make the basis of V = {x^3 - 3x, x^2, 1}. And the dimension hence is 3? I'm fairly new to the subject.

    Also, when wanting to prove a transformation is one-to-one, does finding that the nullspace of T = {0} (i.e, the homogenous system has only one trivial solution which is 0) suffice?
     
    Last edited: May 23, 2012
  2. jcsd
  3. May 25, 2012 #2
    You are exactly correct: Indeed, we can think of [itex] 1,x,x^2,x^3 [/itex] as basis vectors, and the requirement on V is that c=-3a as you have shown. Hence
    [tex] V = \{ ax^3 + bx^2 -3ax + d: a,b,d \in F \} [/tex]
    where F is whatever field in which you are working. Thus there are only three free parameters, so V is three dimensional. You already gave an explicit basis, so everything is fine there.

    You are also correct about showing one-to-one ness: one need only show that the nullspace is trivial. Indeed, having trivial kernel and being injective are equivalent for linear operators on vector spaces.

    Let [itex] T: V \to W [/itex] be a linear map on vector spaces V and W. Suppose first that T is injective and let [itex] x \in \ker T [/itex] so that Tx=0. But T0 = 0 so we have that [itex] Tx = T0 [/itex] and by injectivity this implies that x = 0. Thus the kernel is always trivial.

    Conversely, assume that the kernel is always trivial and consider Tx=Ty. Since the function is linear, we must have that Tx-Ty = T(x-y) = 0 so that [itex] x-y \in \ker T [/itex]. Since the kernel is trivial, this means that x-y=0, so in particular x=y. Thus the map is injective.
     
  4. May 25, 2012 #3
    That's extremely helpful. Thank you for the reply, Kreizhn.
     
  5. May 26, 2012 #4

    HallsofIvy

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    A (very slightly) different way of looking at it is that the condition f'(1)+ f'(-1)= 0 leads to, as you say, c= -3a and so any such polynomial may be written [itex]ax^3+ bx^3+ cx+ d= ax^3+ bx^2- 3ax+ d[/itex][itex]a(x^3- 3x)+ bx^2+ d[/itex] so that any such polynomial can be written as a linear combination of [itex]x^3- 3x[/itex], [itex]x^2[/itex] and [itex]1[/itex]. Those are the "basis vectors".
     
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