The Dual Space

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Summary:

Trying to get the basics of Covectors into my head.

Main Question or Discussion Point

I understand that the Dual Space is composed of elements that linearly map the elements of the Vector Space onto Real numbers

If my preamble shows that I have understood correctly the basic premise, I have one or two questions that I am trying to work through.

So:
1: Is there a one to one correspondence between the elements of V and V* ? (Do the two sets have the same cardinality?)

2: Are the linear mappings that occur in V* all dot product relationships or are there different linear maps that accomplish the same end?

3: What actually distinguishes V from V* ? They are not identical sets are they?
 

Answers and Replies

  • #2
Math_QED
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A meaningful answer to this question will depend on the following: Is ##\dim_\mathbb{R}(V) < \infty?##

In that case, we have ##V \cong V^*## as ##\mathbb{R}##-vector spaces since if ##\{e_1, \dots, e_n\}## is a basis of ##V##, then ##\{e_1^*, \dots, e_n^*\}## is a basis for ##V^*## where ##e_i^*: V \to \mathbb{R}## is the functional mapping ##e_i \mapsto 1## and ##e_j \mapsto 0, j \neq i## and extending this linearly.

In particular, ##V## and ##V^*## have the same cardinality. They are of course not identical as sets, but have exactly the same vector space structure.

If ##\dim_\mathbb{R}(V) = \infty##, it is another story.
 
  • #3
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They are of course not identical as sets, but have exactly the same vector space structure
By "vector space structure" do you just mean they have the same number of bases (and the same cardinality) or do these two vector spaces share other characteristics that other vector spaces might not necessarily share?

Do they have any other internal orderings in common or is the above description sufficient?
 
  • #4
Math_QED
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By "vector space structure" do you just mean they have the same number of bases (and the same cardinality) or do these two vector spaces share other characteristics that other vector spaces might not necessarily share?

Do they have any other internal orderings in common or is the above description sufficient?
Have the same vector space structure means that they are isomorphic as vector spaces. Basic linear algebra says that this is the same as saying that both spaces have equal dimension (indeed, up to isomorphism the only ##\mathbb{R}##-vector space is ##\mathbb{R}^n##).
 
  • #5
lavinia
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Summary::
2: Are the linear mappings that occur in V* all dot product relationships or are there different linear maps that accomplish the same end?
Elements of ##V^{*}## can be defined without a dot product. If one happens to have a dot product around, then it can be used to translate the linear map into a dot product with a specific fixed vector. A different dot product gives a different vector.

Example: In Calculus one learns that the derivative of a function along a curve is the dot product of the gradient of the function with the tangent vector to the curve. This uses the the Euclidean dot product. But a different dot product on Euclidean space would give a different gradient. Moreover, in order to take the derivative you do not need the gradient but just to differentiate the function along the curve. This defines a linear map from tangent vectors to Euclidean space at a point into the real numbers without a dot product.

As @Math_QED explained in Post #2, given a basis ##e_{i}## for a vector space ##V## then there is a dual basis ##e_{i}^{*}##. One can define a dot product by ##e_{i}⋅e_{j} = δ_{ij}##. Then the dual linear map ##e_{i}^{*}## is the dot product with ##e_{i}##.
 
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  • #6
mathwonk
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You are familiar with functions evaluated on points, such as f(p). In this setting p is like the vector and f is like the covector or dual vector, i.e. f is a function on p. Obviously f and p are not the same. Each one can be viewed as a function on the other. I.e. given f, each p yields a number f(p), and given p, each f yields a number, f(p). Thus f and p are "dual" to each other.


suppose we think in the real plane R^2. Let's denote points by letters like p, and we may think of them as vectors, namely the arrow from (0,0) to p. If we want to describe a vector p by its coordinates, then the x and y coordinates are numbers, x(p) and y(p). In these expressions, x and y are functions and p is the argument. Moreover, if we know how to add vectors p and q, by the parallelogram law for their associated arrows, then we know that the x coordinate of p+q is the sum of the x coordinates of p and of q. I.e. x, and also y, are "linear" functions of p.

Thus the coordinate functions x and y, on the plane, are covectors. I.e. they are linear functions from the plane to the real numbers.

Now a dot product is a function of two arguments, given two vectors p and q, it gives a number <p,q>. Looked at asymmetrically, if we fix one vector p, then we have <p, >, which assigns a number to one other vector q. I.e. given a dot product < , >, p determines <p, >, which is a function of q. Moreover since by hypotheses a dot product is linear in each variable, this function <p, > is linear in q. I.e. the expression <p, > is a covector.

Thus a dot product defines a map from each vector p to a covector <p, >. Since the dot product is also linear in the first variable, this map from p to <p, > is also linear. and this is reversible. So essentially a dot product can be viewed as a linear map from vectors to covectors.

Thus V* is the space of linear functions on V. Since we can add functions, V* is also a vector space, the "dual" space to V, and a dot product is a linear map from V to V*. We usually make some assumptions about this map as well, e.g. in finite dimensions, a dot product on V can be defined as a linear isomorphism from V to V*.

E.g. in the plane, the x coordinate can be realized as a dot product with the positive unit vector e1 on the x axis. Namely given a vector p in the plane, to dot it with e1 means to project the arrow 0p orthogonally onto the x axis, and then ask what multiple the projected vector is of e1. I.e. <e1,p> = how far does 0p reach in the x direction?

ok.hope this is food for thought.
 
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