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Prove that three functions form a dual basis

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations

    Z95rV0g.png
    3. The attempt at a solution
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    From that point, I don't know what to do. How do I prove linear independence if I have no numerical values? Thank you.
     
  2. jcsd
  3. Feb 17, 2016 #2

    fresh_42

    Staff: Mentor

    I think you made some mistakes with your indices calculating the ##F_i(p)##.
    Can you tell what ##E^*## is? Do you know any basis of it?
    Edit: However, your matrix ##F_i(V_j) = (\frac{1}{j+1} \cdot m_i^{j+1})_{j,i}## seems to be correct. Is it a transformation of basis, i.e. an isomorphism?
     
    Last edited: Feb 17, 2016
  4. Feb 18, 2016 #3
    Yeah, probably got confused with the different sub indices and I'm relatively new to latex, so yeah, sorry. But you said the matrix is correct, so...
    That is the dual space of E, a basis can be easily constructed from n linearly independent functions, in this case, three.
    And is it a transformation matrix from [tex]E \rightarrow E^*[/tex] ?
     
  5. Feb 18, 2016 #4

    fresh_42

    Staff: Mentor

    I've thought it this way. ##E^*## is the vector space of all linear functions ##φ : ℝ^2[t] → ℝ##. The canonical basis of ##E## is ##\{1,t,t^2\}##. So the goal is to write any ##φ∈E^*## as a linear combination of ##F_0,F_1,F_2##, i.e. solve the linear equation ##φ=x_0F_0+x_1F_1+x_2F_2##, i.e. ##φ = (F_i(V_j))_{j,i} \vec{x}##.

    In coordinates it is ##φ(1) = φ(1,0,0), φ(t) = φ(0,1,0), φ(t^2) = φ(0,0,1),## and it's a clearer (to me) to write ##F_i(p)=F_i(a_0,a_1,a_2)## instead.
    To decide whether the ##F_i## form a basis you will probably need another property given at the beginning. You can either solve the linear equation(s) or find an argument why the matrix is regular.

    (For further talk remind me please what is meant by anti-dual basis of ##B^*##. Is it a basis of ##(E^*)^*##?)
     
  6. Feb 18, 2016 #5
    Yeah, but how do I go on about solving the system with no numbers? That is my main problem...
    Also, why is the matrix being regular relevant? Doesn't that just mean that there is a number [tex]n[/tex] such that the entries of [tex]A^n[/tex] are all positive?
    Also, the anti dual basis would be the basis for which [tex]B^*[/tex] is the dual basis. Sorry about the confusing terms, I translated the problem from french.
     
    Last edited: Feb 18, 2016
  7. Feb 18, 2016 #6

    fresh_42

    Staff: Mentor

    Alors. You have numbers, mainly the coordinates. Therefore I wrote e.g ##φ(t)=φ(0,1,0)## and ##F_i(p) = F_i(a_0,a_1,a_2)##. This gives you the equation
    $$φ(t)=φ(0,1,0)=\sum_{i=0}^{i=3} x_i F_i(0,1,0)= \frac{1}{2} x_0 m_0^2 + \frac{1}{2} x_1 m_1^2 + \frac{1}{2} x_2 m_2^2$$ and similar the other two. You can also prove that your matrix above is regular. The three (fixed), and (pairwise) different ##m_i## are given constants. (Another triple would give you another basis, so you have to do the work in dependence of the ##m_i##.)

    Edit: For the anti-dual basis you will have to find three linear independent polynomials ##p_j = ∑ a_i^{(j)}t^i## such that ##F_i(p_j)=F_i(a_0^{(j)},a_1^{(j)},a_2^{(j)})=δ_{ij}##.
     
    Last edited: Feb 18, 2016
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