What is the Dual Basis for Linear Algebra?

pezola
Messages
11
Reaction score
0
[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) = \deltaij , but I can’t find the x1,x2,x3 for which this works. Any suggestions?
 
Physics news on Phys.org
Each basis vector in V that satisfies fi(ej) = \delta ij defines a system of 3 equations in 3 unknowns. Each of these systems gives you one of the basis vectors.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top