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Dual basis and kernel intersection

  1. Jun 1, 2014 #1
    The problem statement, all variable
    Let ##\phi_1,...,\phi_n \in V^*## all different from the zero functional. Prove that

    ##\{\phi_1,...,\phi_n\}## is basis of ##V^*## if and only if ##\bigcap_{i=1}^n Nu(\phi_i)={0}##.

    The attempt at a solution.

    For ##→##: Let ##\{v_1,...,v_n\}## be a basis of ##V## such that ##\phi_i(v_j)=δ_{ij}##. Now let ##x \in \bigcap_{i=1}^n Nu(\phi_i)={0}##, as ##x \in V## then ##x=α_1v_1+...+α_nv_n##.
    By hypothesis ##\phi_i(x)=0## for all ##1\leq i \leq n##. But this means ##0=\phi_i(x)=\phi_i(α_1v_1+...+α_nv_n)=α_i## for all ##1\leq i \leq n##, it follows ##x=0##.

    I don't know how to show the other implication. For ←, in the exercise is given the suggestion: if ##\{ψ_1,...,ψ_n\}## is a basis of ##V^*##, one can try to show that the matrix obtained from writing ##\phi_1,...,\phi_n## in coordinates with respect to the basis ##\{ψ_1,...,ψ_n\}##, is an invertible matrix, I am stuck and I haven't got a clue on how to use this idea, I would like any hints or another idea to prove the remaining implication.
     
  2. jcsd
  3. Jun 2, 2014 #2

    pasmith

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    Homework Helper


    An alternative approach is to assume that [itex]\Phi = \{\phi_1, \dots, \phi_n\}[/itex] is not a basis for [itex]V^{*}[/itex], and exhibit a non-zero [itex]v \in \bigcap_i \ker(\phi_i)[/itex].

    To that end: let [itex]\Phi' \subset \Phi[/itex] be a linearly-independent subset of [itex]\Phi[/itex] of maximum dimension. This [itex]\Phi'[/itex] can then be extended to a basis [itex]B[/itex] of [itex]V^{*}[/itex]. Now consider the dual basis of [itex]B[/itex].
     
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