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Linear Algebra Dual Basis

  1. Feb 19, 2008 #1
    [SOLVED] Linear Algebra Dual Basis

    Let V= R3 and define f1, f2, f3 in V* as follows:
    f1 = x -2y
    f2 = x + y +z
    f3 = y -3z

    part (a): prove that {f1, f2, f3} is a basis for V*

    I did this by using the gauss jordan method and showing that {f1, f2, f3} is linearly independent. Now because dim(V) is finite, I know that dim(V) = dim(V*). Because the set {f1, f2, f3} has exactly three vectors and the dimension of V* is three, by a corollary to the the Replacement Theorem, {f1, f2, f3} is a basis for V*

    part (b) Find a basis for V for which {f1,f2,f3} is the dual basis.

    I know that a for a dual basis, fi(xj) = [tex]\delta[/tex]ij , but I can’t find the x1,x2,x3 for which this works. Any suggestions?
     
  2. jcsd
  3. Feb 19, 2008 #2
    Each basis vector in V that satisfies fi(ej) = [tex]\delta[/tex] ij defines a system of 3 equations in 3 unknowns. Each of these systems gives you one of the basis vectors.
     
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