How Are Dual Vector Components Determined by Basis Vectors?

noahcharris
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Hi everyone,

So I'm going through a chapter on dual spaces and I came across this:

"A key property of any dual vector ##f## is that it is entirely determined by its values on basis vectors.

## f_i \equiv f(e_i) ##

which we refer to as the components of ##f## in the basis ##{e_i}##, this is justified by

##e^i(e_j) = \delta^i_j## "

-- This doesn't make any sense to me. How are the components of the dual space only dependent on the basis of V? Wouldn't that mean the components are always the same? I thought the whole point of components was that they varied.

Also, is the second equation assuming a cartesian basis? (not sure what the technical term is) Because I can think of some orthonormal bases for which this doesn't hold. i.e. I can think of some orthonormal basis where 'picking off' the ith component does not yield 1.

I'm obviously confused.

Dual spaces in general just confuse me, I understand that the dual space is the set of linear functionals on V, and that they can be represented as 1-forms, but as far as the details go (coordinates and bases of the dual space) I'm completely lost. Any help would be much appreciated.
 
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Using the Riesz Representation of functionals in an inner-product space, you can express any functional as an inner product with a vector in the space. Implying that you would use the same basis set to express the representation (i.e. vector form).
https://en.wikipedia.org/wiki/Riesz_representation_theorem
 
RUber said:
Using the Riesz Representation of functionals in an inner-product space, you can express any functional as an inner product with a vector in the space. Implying that you would use the same basis set to express the representation (i.e. vector form).
https://en.wikipedia.org/wiki/Riesz_representation_theorem

Could you expand what you mean by 'vector form'? Are you talking about a differential form (covector)?
 
The vector form I was referring to is just the ##f_i = f(e_i)## notation you used.
 
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