Dual Vector Spaces

  • Thread starter Etenim
  • Start date
  • #1
9
0
Greetings,

Slowly I am beginning to think that I must be some sort of retard for not getting this fundamental concept. For this post, I will adapt the bracket notation as introduced by P. Halmos' "Finite-dimensional Vector Spaces". [tex] \left[ \cdot, \cdot \right] : V \times V^* \to K [/tex].

A linear functional on a vector space V is a scalar-valued function, defined for each [tex]v \in V[/tex], mapping vectors into the underlying coefficient field and having the well-known property of linearity. -- Let V be a finite vector space over a field K. V* is defined to be the space of all linear functionals [tex] f : V \to K [/tex], which shall be referred to as the dual space of V.

Once a basis for V is chosen, fixing [tex] x \in V [/tex], for all [tex] f, f^\prime \in V^* [/tex], we have [tex] \left[ x, f \right] = \left[ x, f^\prime \right] \, \Rightarrow \, f = f^\prime [/tex]. Which is obvious, for by choosing a basis, we can show that f must be unique for the expression [tex] \left[ x, f \right] [/tex] to be well-defined. After representing the fixed vector as a linear combination of V's basis vectors [tex] \left( \beta_i \right)_{i=0}^n [/tex], and applying a linear functional f, the term [tex] \left[ \beta_i, f \right] = a_i [/tex] emerges.

That is, given some [tex] a_i \in K [/tex] and [tex] x \in V [/tex], can I find a unique [tex] y^i \in V^* [/tex] such that [tex] \left[ x, y^i \right] = a_i [/tex]?

I interpret this to be a result of our previous definition of the functional to be linear. Conversely, could we give the functional's now known property of uniqueness axiomatically and then deduce that it must be linear on such a foundation? Then that result would not seem so coincidental to me, but be rather a rediscovery of a historical definition made for the very purpose of making the elements of the dual space linear.

Now, I can take the f apart and write it as a linear combination of dual basis vectors, the basis of the dual vector space. Given a basis [tex] \left( \beta_i \right)_{i=0}^n [/tex] for V, we define the elements of the dual basis [tex] \left( \beta^*^i \right)_{i=0}^n [/tex] uniquely by [tex] \left[ \beta_i, \beta^*^j \right] = \delta_i^j [/tex].

Why do we do this? To later make the set of dual basis vectors linearly independent? Is there no other choice for [tex] \left[ \beta_i, \beta^*^j \right] [/tex]'s value to do this feat?

I hope I didn't mess up the indices. This is my first exposure to advanced linear algebra - I would be happy if someone could enlighten me about dual spaces, and what motivates the definitions.

Thanks a lot,

Cheers,
- Etenim.
 

Answers and Replies

  • #2
mathwonk
Science Advisor
Homework Helper
11,063
1,254
a basis represents every vector as a sequence of coordinates (a1,...,an).

then the most natural way to assign a number to such a vector is to choose one of the coordinates.

choosing the ith coordinate, is exactly your definition of the ith dual basis vector.

what other choice could be simpler?
 
  • #3
9
0
Ah. By 'naturally' defining the action of a dual basis vector on an arbitrary vector [tex] v = c^i \beta_i [/tex] to be [tex] \beta^*^j ( c^1 \beta_1 + c_2 \beta_2 + \cdots + c^n \beta_n ) = c_j [/tex], we want, by the dual basis vector's linearity, [tex] \beta^*^j ( \beta_i ) = \delta_i^j [/tex], to "extract" the [tex]c_j[/tex].

'Naturally'. Well, I wonder what understanding feels like. Meh. But, yes, it makes more sense now. Thanks. :)
 
Last edited:

Related Threads on Dual Vector Spaces

  • Last Post
2
Replies
25
Views
7K
Replies
2
Views
4K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
16
Views
3K
  • Last Post
2
Replies
48
Views
4K
  • Last Post
Replies
15
Views
332
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
20
Views
6K
Top