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Proof regarding linear functionals

  1. May 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Let V be a vector space over R. let Φ1, Φ2 ∈ V* (the duel space) and suppose σ:V→R, defined by σ(v)=Φ1(v)Φ2(v), also belongs to V*. Show that either Φ1 = 0 or Φ2 = 0.

    2. Relevant equations
    N/A

    3. The attempt at a solution
    Since σ is also an element of the duel space, it is linear, so σ(v+u)=σ(v)+σ(u). Translating both sides into terms of Φ1 and Φ[SUB}2[/SUB], I came up with the equation
    (1) Φ1(v)Φ2(u)+Φ1(u)Φ2(v)=0.

    Doing the same with σ(av+bu) and plugging in equation (1) produced the equation
    (2) a2Φ1(v)Φ2(v)+b2Φ1(u)Φ2(u)=aΦ1(v)Φ2(v)+bΦ1(u)Φ2

    This is where I got stuck, perhaps there is a trick I am not seeing to this equation or perhaps I approached this the wrong way. Any help would be appreciated.
     
  2. jcsd
  3. May 12, 2017 #2

    PeroK

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    You're on the right track. Go back to equation (1) and try ##u = v##.
     
  4. May 13, 2017 #3
    I see, thank you for the help, I solved it.
     
  5. May 13, 2017 #4

    PeroK

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    Are you sure?
     
  6. May 13, 2017 #5
    I think so. Plugging in u=v into Φ1(v)Φ2(u)+Φ1(u)Φ2(v)=0 resulted in Φ1(v)Φ2(v)+Φ1(v)Φ2(v)=0, meaning Φ1(v)Φ2(v)=0. Since these are elements of a field, it means either Φ1(v)=0 or Φ2(v)=0, and since v is arbitrary, this means either Φ1=0 or Φ2=0.
     
  7. May 13, 2017 #6

    PeroK

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    I guessed that is what you had done! What you have shown there is:

    ##\forall \ v: \ \Phi_1(v) = 0## or ##\Phi_2(v) = 0## (Equation 3)

    That means that for some ##v## it could be ##\Phi_1(v) = 0## and for some other ##v## it could be ##\Phi_2(v) = 0##. But, you haven't shown that either ##\Phi_1 = 0## or ##\Phi_2 = 0##.

    So, there is still work to do. Let me help with the next step. Suppose ##\Phi_1 \ne 0##. We have to show that then ##\Phi_2 = 0##.

    Can you use equations (1) and (3) to do this?
     
  8. May 13, 2017 #7
    I see. So let us assume Φ1≠0, Then it means there exists w∈V such that Φ1(w)≠0, let us denote the set of all such vectors with S. since for every vector v∈V Φ1(v) is either 0 or not 0, V=S∪Ker(Φ1).We plug one of these (arbitrary) w into equation 3 and get Φ1(w)Φ2(w)=0, since Φ1(w)≠0, it means Φ2(w)=0. And since this applies to every w∈S, it means Φ2(S)=0.

    Now we must show that Φ2(u)=0 ∀u∈Ker(Φ1). Let us take an arbitrary vector u from Ker(Φ1) and plug it in equation 1 along with another vector w∈S: Φ1(u)Φ2(w)+Φ1(w)Φ2(u)=0. Since u∈Ker(Φ1), Φ1(u)Φ2(w)=0. Thus, Φ1(w)Φ2(u)=0, and since Φ1(w)≠0 by definition, this means Φ2(u)=0 ∀u∈Ker(Φ1).

    Combining these two results give us Φ2(v)=0 ∀v∈V, and so Φ2=0. Doing the same process by assuming Φ2≠0 produces the result Φ1=0.
     
  9. May 13, 2017 #8

    PeroK

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    I can't see any mistakes there, although your solution is more complicated than I was expecting. If I pick up your proof from here:

    Then, I would use equation 3 to note that ##\Phi_2(w) = 0## and hence (from equation 1) we have:

    ##\forall v \in V: \ \Phi_1(w)\Phi_2(v) = 0##

    Hence ##\forall v \in V: \ \Phi_2(v) = 0##

    And the result follows.

    Note also that there is no need to repeat the process for ##\Phi_2##. If ##\Phi_1 = 0##, then we are done. And, if ##\Phi_1 \ne 0## we have shown that ##\Phi_2 = 0##. In either case, one of the functionals must be ##0##. And, there is no need to do any more.
     
  10. May 13, 2017 #9
    Assume that both functions ##\Phi_1,\Phi_2## are not equal to zero identically. Then there exist a vector ##v## such that ##\Phi_1(v)\Phi_2(v)\ne 0##.
    So that ##\sigma(\lambda v)=\lambda^2\Phi_1(v)\Phi_2(v)\Longrightarrow \sigma( v)=\lambda\Phi_1(v)\Phi_2(v),\quad \forall\lambda\ne 0##
    This is possible if only ##\Phi_1(v)\Phi_2(v)=0## Contradiction.
     
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