B Understanding Dual Space: Mapping Vector Space to Real Numbers

[fresh_42]

So what is the connection between the vector space and it's dual space?

One is a set of arrows and the other a set of functions, which attach a number to the arrows.

Do they share the same origin ?

No. Only the dimension. $(x,y)$ is a vector, $(x,y)\longrightarrow f(x,y)=\alpha x+\beta y$ is a dual vector.

If ,as a simple example the vector space is the set of all real number pairs is perhapd its Dual Space the set of all number pairs where one of the pairs is zero? (meaning [x,y] would be mapped to either [x,0] or [0,y]

No. These are all vectors. But if you define a function $f\, : \,(x,y) \longmapsto x$ then you have a dual vector, namely $f$ - without a second component, zero or not, none.

And ,if the origin is shared by both spaces (the vector and its dual) is it possible to take any vector and ,starting from the origin go to it by either adding vectors or (the scenic route ) by adding covectors?

No. That's confusing. You confuse a domain (vectors) with the functions on that domain (covectors).

 

[geordief]

Suppose we have any vector in the vector space; say (3,4) just to have a concrete example to keep it simple for me...

If a covector acts on (3,4) are there an infinite number of possible results of a covector acting on (3,4)?

Can I result be the number 88?

Would another covector give a different number ,say 56?

 

[PeterDonis]

If a covector acts on (3,4) are there an infinite number of possible results of a covector acting on (3,4)?

Since a covector maps pairs (x, y) to numbers, and since there are an infinite number of numbers, the answer to this would be yes.

Can I result be the number 88?

Sure, why not?

Would another covector give a different number ,say 56?

Meaning, another covector acting on the same pair (3, 4)? Yes, there will be a covector that gives 56 acting on that pair.

 

[fresh_42]

Sure, no problem. If we define $f(3,4)=\alpha\cdot 3+ \beta \cdot 4 = 88$ then we still have infinitely many possibilities to choose $\alpha,\beta$.

On the risk of confusing you even more, let me explain where the names come from. Let us choose $(\alpha,\beta) = (22, 5.5)$. Then $$f(3,4)=22\cdot 3+ 5.5 \cdot 4 = (22,5.5) \cdot \begin{bmatrix}3\\ 4 \end{bmatrix} = 88$$ Hence the covector $(22,5.5)$ maps the vector $(3,4)$ onto the real number $88$. The same is true for the covector $(0,22)$ which results in $88$, too: $$g(3,4)=0\cdot 3+ 22 \cdot 4 = (0,22) \cdot \begin{bmatrix}3\\ 4 \end{bmatrix} = 88$$

So in a way our functions are represented by the covectors $(\alpha,\beta)=(22,5.5)$ for the function $f$ and by $(\alpha,\beta)=(0,22)$ for the function $g$. Now both, $(3,4)$ as well as $(\alpha,\beta)$ are written as vectors. However, there is a small but important difference:

Vector $(x,y) = (3,4)$ is an arrow in the real Euclidean plane, starting at the origin and ending in $(3,4)$. We only write the endpoint, because it is more convenient and indicates the direction, so that we can attach this arrow on any other point than the origin, too.

Covector $(\alpha,\beta)=(22,5.5)$ is also just a brief notation, but this time it abbreviates something different, namely an instruction rather than an arrow. And the instruction reads: Multiply any vector $(x,y)$ by $(22,5.5)$. The entire instruction with the order to multiply is now the covector. That it can be written like a vector is only a notation.

 

[geordief]

Thanks for your help and patience.

Yes ,it is confusing, but I feel I am laying the ground for removing some of the preconceptions I have in this area.

It will take a while for these lessons to sink in and I will have to familiarize myself gradually with this new territory :)

 

[robphy]

This might be useful:


Covectors can be visualized as a stack of equally spaced planes,
which maps a vector to the number of planes pierced by the vector.

A physical example is the electric field as a field of covectors, which could be thought of as linear approximations to the equipotential surfaces. Infinitesimal displacement vectors are mapped to voltage differences.

 

[mathwonk]

what lavinia says is of course correct, but some of us stretch the meaning of the word "orthogonal" so it applies to a relation between elements of a given vector space, and elements of its dual space. I.e. given a veftor space V and its dual space V*, we say that a linear functional f in V* is orthogonal to a vector x in V, precisely when f(x) = 0. Then if we are also given an inner product on V, which can be viewed as a map from V to V*, it turns out that a vector y in V is orthogonal to x in V, in the sense that their dot product is zero, precisely when the image fy of y in V*, is orthogonal to x as above, i.e. when fy(x) = 0. This is obvious since by defini†ion, fy(x) = <y,x>.

In this setting, even without an inner product on V, a subspace W of V has an orthogonal complement Wperp in V*, where Wperp consists of those functionals in V* which vanish identically on W. I.e. the orthocomplement of a subspace of V, is intrinsically defined as a subspace of V*. But maybe the books you are using don't use this abstraction.

 

[geordief]

This might be useful:


Covectors can be visualized as a stack of equally spaced planes,
which maps a vector to the number of planes pierced by the vector.

A physical example is the electric field as a field of covectors, which could be thought of as linear approximations to the equipotential surfaces. Infinitesimal displacement vectors are mapped to voltage differences.

Is there a way that it can be said that a vector can give a real number when acting on a covector?

Can a vector also be considered as a function acing on a covector?

Does the same visualization apply (in reverse?)

 

[lavinia]

Is there a way that it can be said that a vector can give a real number when acting on a covector?

Can a vector also be considered as a function acing on a covector?

Does the same visualization apply (in reverse?)

If $v$ is a vector and $l$ is a covector then $l(v)$ assigns a number to $l$. $v$ is a cocovector.

 

[geordief]

If $v$ is a vector and $l$ is a covector then $l(v)$ assigns a number to $l$. $v$ is a cocovector.

So ,are we saying that all the number pairs in a given vector space can be treated entirely interchangeably as either a vector or a covector ,depending entirely on the way they are used?

If the number pair are both ,say(3,4) can we use (3,4) as a covector to operate on the vector (3,4) and ,if we use the vector (3,4) to operate on the covector (3,4) then we say that the vector (3,4) is now being used as a covector and so is a covector and no longer a vector?

 

[fresh_42]

So ,are we saying that all the number pairs in a given vector space can be treated entirely interchangeably as either a vector or a covector ,depending entirely on the way they are used?

Yes, but this is misleading. Better is to say: A tuple of numbers represent a vector according to some coordinate system, i.e. choice of basis. Now vector spaces can consist of various types of objects: the commonly thought of arrows, functions, covectors, operators, differential forms, or whatever can be added and stretched.

It is not the pair which is used differently, it is the pair which is meant / defined differently. Usage comes from context, not the other way around.

If the number pair are both ,say(3,4) can we use (3,4) as a covector to operate on the vector (3,4) and ,if we use the vector (3,4) to operate on the covector (3,4) then we say that the vector (3,4) is now being used as a covector and so is a covector and no longer a vector?

Covectors are elements of the dual vector space and as such they are again vectors. For finite dimensional vector spaces $V$ we have $V\cong V^*$ and $(V^*)^*=V$. But you should first learn what a vecttor space is! Take continuous functions as example: $V=C^0(\mathbb{R})$.

 

[PeterDonis]

If the number pair are both ,say(3,4) can we use (3,4) as a covector to operate on the vector (3,4) and ,if we use the vector (3,4) to operate on the covector (3,4) then we say that the vector (3,4) is now being used as a covector and so is a covector and no longer a vector?

You need to stop thinking in terms of "number pairs" and start thinking of what the number pairs represent.

First, as several of us have told you, you need to understand what a vector space is. As I said in post #30, a vector space is anything that satisfies the vector space axioms. What are those axioms? Different sources might organize them differently, but in a nutshell, you have the following:

(1) A field, which for this discussion we will take to be the real numbers, $\mathbb{R}$. Elements of the field are called "scalars".

(2) A set defined over this field. In this discussion, we have been using the set of ordered pairs of reals, i.e., $\mathbb{R}^2$. Elements of the set are called "vectors".

(3) Two operations on the set, called "addition" and "scalar multiplication". For $\mathbb{R}^2$, these operations are obvious: addition just adds the pairs, so $(a, b) + (c, d) = (a + c, b + d)$, and scalar multiplication just multiplies each number in the pair by the scalar, so $m (a, b) = (ma, mb)$.

(4) A set of properties that the operations must satisfy: we don't need to delve too deeply into this here, but they are basically the obvious properties that we expect addition and scalar multiplication to have for our examples, e.g., associativity, identity, inverse, commutativity of addition, multiplication distributive over addition, etc.

The Wikipedia article on vector spaces [1] discusses all this in more detail.

Now, given the above definition of a vector space, what is a "covector"? A covector is a linear map from a vector space into its underlying field. So in the case of the example we have been using, it is a linear map from $\mathbb{R}^2$ into $\mathbb{R}$. Now, it should be clear that, as has already been said in this discussion, any such linear map can be written as follows: $(x, y) \rightarrow \alpha x + \beta y$. Notice that we have just characterized the linear map by an ordered pair of real numbers. In other words, if our vector space is the set $\mathbb{R}^2$, then the space of all covectors--all linear maps from our vector space into its underlying field--is also the set $\mathbb{R}^2$. What's more, if we think about what it means to add two linear maps, or multiply a linear map by a real number, we will see that the space of all linear maps from $\mathbb{R}^2$ into $\mathbb{R}$ satisfies all the axioms of a vector space. So the space of all covectors is also a vector space.

What all this means is that the set $\mathbb{R}^2$, considered as a vector space, can be interpreted in two different ways: it can be interpreted as a set of ordered pairs $(x, y)$ that describe the locations of points in a plane, given an origin; or it can be interpreted as a set of linear maps $\alpha x + \beta y$ from ordered pairs $(x, y)$ to real numbers. So if we are talking about vector spaces, we can't just talk about $\mathbb{R}^2$ as a set of "number pairs". We have to be clear about whether we are using the number pairs to represent points, or linear maps.

And we can go even further. Suppose we take $\mathbb{R}^2$ to represent the set of linear maps $\alpha x + \beta y$ from ordered pairs to real numbers; i.e., each member of $\mathbb{R}^2$ is interpreted as the pair $(\alpha, \beta)$ that defines a linear map. Now pick some ordered pair $(x, y)$. This ordered pair will give us a real number $\alpha x + \beta y$ for every pair $(\alpha, \beta)$. In fact, since multiplication of reals is commutative, we could just as well write this number as $x \alpha + y \beta$, and we could write the linear map as $(\alpha, \beta) \rightarrow x \alpha + y \beta$. This looks just like the covector definition we gave above! All that has changed is that we have switched $(x, y)$ and $(\alpha,\beta)$. In other words, we have now used the ordered pair $(x, y)$ to define a linear map from the space of linear maps $(\alpha, \beta)$ to the real numbers! In other words, the set of ordered pairs $(x, y)$ can be viewed as the set of covectors of the vector space $(\alpha, \beta)$. This is what @lavinia was talking about in post #39.

What this is telling us is that, if we have two interpretations of $\mathbb{R}^2$ as a vector space, which interpretation we call "vectors" and which interpretation we call "covectors" is a matter of choice. Each interpretation--ordered pairs describing points, and ordered pairs describing linear maps--is "dual" to the other, and both satisfy all the vector space axioms so each one is a vector space, and each one is a covector space with respect to the other one. There is no "fact of the matter" about which one is the "real" vector space and which one is the "real" covector space. It all depends on what specific problem you are trying to solve and how you want to use these spaces and interpretations to solve it.

[1] https://en.wikipedia.org/wiki/Vector_space#Definition

 

[WWGD]

Ultimately, as a meta -comment , @geordief , you cannot always readily visualize Mathematical concepts and sometimes just try to deal with abstractions best way you van until they hopefully sink in some day. These concepts are not simple nor readily comparable to things you may be familiar with.

 

[geordief]

You need to stop thinking in terms of "number pairs" and start thinking of what the number pairs represent.

First, as several of us have told you, you need to understand what a vector space is. As I said in post #30, a vector space is anything that satisfies the vector space axioms. What are those axioms? Different sources might organize them differently, but in a nutshell, you have the following:

(1) A field, which for this discussion we will take to be the real numbers, $\mathbb{R}$. Elements of the field are called "scalars".

(2) A set defined over this field. In this discussion, we have been using the set of ordered pairs of reals, i.e., $\mathbb{R}^2$. Elements of the set are called "vectors".

(3) Two operations on the set, called "addition" and "scalar multiplication". For $\mathbb{R}^2$, these operations are obvious: addition just adds the pairs, so $(a, b) + (c, d) = (a + c, b + d)$, and scalar multiplication just multiplies each number in the pair by the scalar, so $m (a, b) = (ma, mb)$.

(4) A set of properties that the operations must satisfy: we don't need to delve too deeply into this here, but they are basically the obvious properties that we expect addition and scalar multiplication to have for our examples, e.g., associativity, identity, inverse, commutativity of addition, multiplication distributive over addition, etc.

The Wikipedia article on vector spaces [1] discusses all this in more detail.

Now, given the above definition of a vector space, what is a "covector"? A covector is a linear map from a vector space into its underlying field. So in the case of the example we have been using, it is a linear map from $\mathbb{R}^2$ into $\mathbb{R}$. Now, it should be clear that, as has already been said in this discussion, any such linear map can be written as follows: $(x, y) \rightarrow \alpha x + \beta y$. Notice that we have just characterized the linear map by an ordered pair of real numbers. In other words, if our vector space is the set $\mathbb{R}^2$, then the space of all covectors--all linear maps from our vector space into its underlying field--is also the set $\mathbb{R}^2$. What's more, if we think about what it means to add two linear maps, or multiply a linear map by a real number, we will see that the space of all linear maps from $\mathbb{R}^2$ into $\mathbb{R}$ satisfies all the axioms of a vector space. So the space of all covectors is also a vector space.

What all this means is that the set $\mathbb{R}^2$, considered as a vector space, can be interpreted in two different ways: it can be interpreted as a set of ordered pairs $(x, y)$ that describe the locations of points in a plane, given an origin; or it can be interpreted as a set of linear maps $\alpha x + \beta y$ from ordered pairs $(x, y)$ to real numbers. So if we are talking about vector spaces, we can't just talk about $\mathbb{R}^2$ as a set of "number pairs". We have to be clear about whether we are using the number pairs to represent points, or linear maps.

And we can go even further. Suppose we take $\mathbb{R}^2$ to represent the set of linear maps $\alpha x + \beta y$ from ordered pairs to real numbers; i.e., each member of $\mathbb{R}^2$ is interpreted as the pair $(\alpha, \beta)$ that defines a linear map. Now pick some ordered pair $(x, y)$. This ordered pair will give us a real number $\alpha x + \beta y$ for every pair $(\alpha, \beta)$. In fact, since multiplication of reals is commutative, we could just as well write this number as $x \alpha + y \beta$, and we could write the linear map as $(\alpha, \beta) \rightarrow x \alpha + y \beta$. This looks just like the covector definition we gave above! All that has changed is that we have switched $(x, y)$ and $(\alpha,\beta)$. In other words, we have now used the ordered pair $(x, y)$ to define a linear map from the space of linear maps $(\alpha, \beta)$ to the real numbers! In other words, the set of ordered pairs $(x, y)$ can be viewed as the set of covectors of the vector space $(\alpha, \beta)$. This is what @lavinia was talking about in post #39.

What this is telling us is that, if we have two interpretations of $\mathbb{R}^2$ as a vector space, which interpretation we call "vectors" and which interpretation we call "covectors" is a matter of choice. Each interpretation--ordered pairs describing points, and ordered pairs describing linear maps--is "dual" to the other, and both satisfy all the vector space axioms so each one is a vector space, and each one is a covector space with respect to the other one. There is no "fact of the matter" about which one is the "real" vector space and which one is the "real" covector space. It all depends on what specific problem you are trying to solve and how you want to use these spaces and interpretations to solve it.

[1] https://en.wikipedia.org/wiki/Vector_space#Definition

Thanks,you have gone to a lot of trouble.I think I am getting there.

 

[geordief]

Ultimately, as a meta -comment , @geordief , you cannot always readily visualize Mathematical concepts and sometimes just try to deal with abstractions best way you van until they hopefully sink in some day. These concepts are not simple nor readily comparable to things you may be familiar with.

Yes,I have read that mathematics is often developed without a real world application and that that application ,if it shows up may be a long time after the mathematics were first formulated.

This covector /vector formulation does appear to me to have real world applications and I think that is what makes me dissatisfied if I cannot connect the mathematics with its application in as direct a way as possible (visually would be ideal but I appreciate it may not always be possible)

And of course I also understand that it can be a very gradual process for an understanding of these concepts to sink in and become something like second nature.

I still feel I have made some progress ,with everyone's help.

 

[lavinia]

Yes,I have read that mathematics is often developed without a real world application and that that application ,if it shows up may be a long time after the mathematics were first formulated.

This covector /vector formulation does appear to me to have real world applications and I think that is what makes me dissatisfied if I cannot connect the mathematics with its application in as direct a way as possible (visually would be ideal but I appreciate it may not always be possible)

Vectors and covectors appear naturally in calculus and through calculus are instrumental in physics, engineering, and many other disciplines.

 

[Stephen Tashi]

Well ,just to take the example I gave (3,4) in the original vector space

We should begin by making a distinction between a vector in a vector space versus a particular representation of that vector. It appears your concept of a 2-D vector is that it is a pair of real numbers like (3,4). A cartesian representation of a 2-D vector (such as (3,4)) is very useful. However, there are other representations, such as polar coordinates.

If $L$ is a linear mapping of 2-D vectors to the real numbers, we could choose to represent a particular $L$ by $(3,4)$ with the understanding that this representation implies, in cartesian coordinates, $L((x,y)) = 3x + 4y$. However, the common use of the notation "$(3,4)$" for a vector in a vector space and also for a function $L$ whose domain is that vector space does not imply that the vector $(3,4)$ and the linear function represented by $(3,4)$ are elements of the same set of vectors.

The notation $(3,4)$ for a linear functional does show we can think of the vector space of linear functions on a 2-D real vector space as a set of 2-D vectors in a different 2-D vector space. And there is a sense in which each of those vector spaces is "the same" as a vector space representing locations on a map or some other 2-D physical situation.

In mathematics, there are many different notions for what it means for two things to be "the same" or "equal". For example, "A=B" means one thing for two numbers A,B and it means something different for two sets A,B. Defining the meaning of "=" in a particular context requires defining the "equivalence relation that it symbolizes. Two 2-D vector spaces over the real numbers are "the same" in a certain technical sense. They are said to be "isomorphic". You can look up the technical definitions of "isomorphic". There are different definitions for "isomorphic" for various different mathematical structures - groups, rings, fields, vector spaces etc. (Technically the definition of a vector space does not say it must have an inner product ( a dot product). So we need the definition of isomorphic for a "vector space with inner product") As an intuition, the definition of "isomorphic" is in the same spirit as the definition of congruent triangles. Two congruent triangles aren't (necessarily) the same triangle in the common language sense of the word "same" , but they are the same in certain respects.

,wouldn't there be a (3,4) in the rotated space too

How are you thinking of "the rotated space". A transformation, such as a rotation can be thought about in two different ways. In the active context we think of the vectors as representing physical objects and we rotate them in space. In the passive context, we think of a coordinate system for a set of points in space. We define a different coordinate system for the same points by rotating the axes of the original coordinate system. The representation of a vector in the rotated coordinate system is different than its rotation in the original coordinate system, but the vector is not a different vector.

and wouldn't there be equivalent linear maps to the real numbers in that rotated vector space?

You have to define what you mean by "equivalent". It's another example of the difficulty of defining notions of equality and sameness in mathematics.

 

[Stephen Tashi]

So ,are we saying that all the number pairs in a given vector space can be treated entirely interchangeably as either a vector or a covector ,depending entirely on the way they are used?

It's tempting to answer "yes". However, the language of your question has some problems, as others have pointed out. In the first place, there need not be number pairs "in" a given vector space. A vector in a 2-D real vector space can be represented by a pair of real numbers. However, the definition of a vector space might be that it is some collection of physical things such a forces. A vector in a vector space might be a mathematical structure, such as a polynomial. By analogy, we would not say that social security numbers are "in" the population of the USA. Social security numbers represent members of the population.

Having said that, it is correct that pairs of real numbers can be used to define one example of a 2-D vector space. That's a special case of defining a vector to be a mathematical structure.

If the number pair are both ,say(3,4) can we use (3,4) as a covector to operate on the vector (3,4) and ,if we use the vector (3,4) to operate on the covector (3,4) then we say that the vector (3,4) is now being used as a covector and so is a covector and no longer a vector?

This is like asking if the same person can be both a son and a grandfather. The answer is "Yes" with the understanding that a person can be a son with respect to one person ( his father) and a grandfather with respect to a different person (a grandchild).

The usual presentation of vectors and covectors begins by assuming we have a vector space $V$. A "vector" is understood to be a "vector" with respect to $V$. The set of covectors of $V$ is another vector space $W$. A covector in $W$ is a "vector" with respect to the space $W$, but it is a "covector" with respect to the space $V$.

Likewise, if we take a transformation of a basis for $V$ to another basis for $V$, the covectors in $W$ are covariant with respect to that transformation. They need not be covariant with respect to transformations on a different vector space.

As a physical example of vectors and co-vectors, take $V$ to be the vector space of 2-D displacements. Let an element of $W$ be a function $w$ that calculates the work done to displace an object against a constant force field whose components are $(x_w, y_w)$ at every point in space. Different $w$'s are defined by different constant forces.

 

[zinq]

As long as you start with a basis B = {v_j | j ∈ J} for a vector space V (where J is any index set having the cardinality of the basis), then a dual vector is completely determined by which element of the underlying field F that it takes each basis vector to. It's easy to check that, since 1) each vector v ∈ V is a finite linear combination of basis elements, and since 2) a dual vector (call it w*) is a linear map to F, it follows that the value w*(v) that w* takes when you plug in v is completely determined. (This is a good, easy exercise if you haven't tried it.)

So given a basis B for V, you can think of a dual vector as (determined by) a function taking each basis vector to an element of the field.