Linear Algebra Dual Basis

  • Thread starter pezola
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[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?
 

Answers and Replies

  • #2
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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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