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Linear functionals: Φ(u)=0 implies Φ(v)=0, then u=kv.

  1. Apr 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose u,v ∈ V and that Φ(u)=0 implies Φ(v)=0 for all Φ ∈ V* (the duel space). Show that v=ku for some scalar k.

    2. Relevant equations
    N/A

    3. The attempt at a solution
    I've managed to solve the problem when V is of finite dimension by assuming u,v are linearly independent, expanding them to a basis of V and defining a function such that Φ(v)=0, Φ(u)=1 and Φ(wi)=0 where wi are all the other basis vectors. This contradicted the original statement in the problem (since Φ(v)=0 but Φ(u)=1) which meant u,v must be linearly dependent, which implies u is a scalar multiple of v.

    However I don't know how to prove it in the general case where V could also be of infinite dimension, the intuitive solution I had was an argument of linear dependence since we have a scalar multiple but I cannot seem to use it in the general case, so this is where I got stuck. Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 12, 2017 #2

    fresh_42

    Staff: Mentor

    What did you need a basis for?
     
  4. Apr 12, 2017 #3
    I assume you mean in my proof for a vector space of finite dimension. I needed to expand u,v to a basis to show that Φ is indeed an element of the duel space by defining its action on a basis.
     
  5. Apr 12, 2017 #4

    fresh_42

    Staff: Mentor

    But you didn't use finiteness, nor did you use the entire basis. All you used, is a complementary space of ##\mathbb{F}\cdot u## on which you defined a functional (that isn't identically zero) to vanish.
     
  6. Apr 12, 2017 #5
    I don't recall defining any complimentary space where Φ vanishes. If V is of dimension n, I merely expanded {v,u} to a basis S={v,u,w1,w2,...,wn-2} and defined a function Φ from V to its field K defined by Φ(v)=0, Φ(u)=1, Φ(w1)=0, Φ(w2)=0,..., Φ(wn-2)=0.
    This function is obviously a linear functional since it's a part of the basis of V* duel to the basis S, and this conflicted with the original statement.

    All I did here was show that if u,v are linearly independent, there exists a function which contradicts the conditions of the question, and so u,v must be linearly dependent.
     
  7. Apr 14, 2017 #6
    There is very important and simple theorem. Let ##X,Y,Z## be vector spaces say over field ##\mathbb{R}## or ##\mathbb{C}##. Perhaps some of these spaces are infinite dimensional. And let ##A:X\to Y,\quad B:X\to Z## be linear operators such that ##A(X)=Y## and ##\mathrm{ker}\,A\subset\mathrm{ker}\,B##.

    Theorem 1. There is a linear operator ##\Lambda:Y\to Z## such that ##B=\Lambda A##.
    This theorem particularly implies thee followin proposition.

    Theorem 2. If ##f,f_1,\ldots, f_n\in X^*## and ##\bigcap_{i=1}^n\mathrm{ker}\, f_i\subseteq \mathrm{ker}\,f## then there exist numbers ##\lambda_1,\ldots,\lambda_n## such that ##f=\lambda_1f_1+\ldots+\lambda_nf_n##.

    By the way this idea is a source of all Lagrange multipliers theorems.

    Now turn to the problem of OP. Define linear functionals ##\tilde u,\tilde v:V^*\to\mathbb{R}## as follows ##\tilde u(\Phi)=\Phi(u)## and the same for ##\tilde v##. From conditions of the OP's problem it follows that ##\mathrm{ker}\,\tilde u\subseteq\mathrm{ker}\,\tilde v##. By Theorem 2 it follows that ##\tilde v=k\tilde u##. That is ##\Phi(v)=k\Phi(u)## for all ##\Phi\in V^*##. Thus ##v=ku##.

    This prompts that there is no need to employ the whole space ##V^*## it is sufficient to use only its total subspace
     
    Last edited: Apr 14, 2017
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