Understanding Dual Space: Mapping Vector Space to Real Numbers

[geordief]

TL;DR Summary

Trying to get the basics of Covectors into my head.

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?

 

[member 587159]

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.

 

[geordief]

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?

 

[member 587159]

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 $\mathbb{R}^n$).

 

[lavinia]

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 $e_{i}$.

 

[mathwonk]

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.

 

[geordief]

I found those last 2 posts a bit hard .

But can I ask another question?

Suppose we have ,as an example an earthlike sphere that is lumpy and not completely round ,can I specify any point on the surface by two numbers (latitude and longitude) and associate each point with two other numbers that specify its max/min curvature at that point?

Do we have two vector spaces and are they by any chance dual to each other?

 

[mathwonk]

If you found my post difficult, I recommend you spend some time reading and studying a well written book on linear algebra, such as this free one from Sergei Treil at Brown:

https://www.math.brown.edu/~treil/papers/LADW/LADW-2014-09.pdf

 

[geordief]

If you found my post difficult, I recommend you spend some time reading and studying a well written book on linear algebra, such as this free one from Sergei Treil at Brown:

https://www.math.brown.edu/~treil/papers/LADW/LADW-2014-09.pdf

Thanks,I will give it a look.

 

[mathwonk]

since the book is long, you might start by just reading the introduction to the chapter on dual spaces. but it sounds to me as if you really need to think about the difference between a function and a point of its domain, since a dual space is made up of functions. or maybe you could just think about why a coordinate, like x, is a function on plane vectors.

 

[WWGD]

I found those last 2 posts a bit hard .

But can I ask another question?

Suppose we have ,as an example an earthlike sphere that is lumpy and not completely round ,can I specify any point on the surface by two numbers (latitude and longitude) and associate each point with two other numbers that specify its max/min curvature at that point?

Do we have two vector spaces and are they by any chance dual to each other?

Im not sure where you're heading with this but , just in case, a sphere is not a vector space; it does not necessarily contain the origin and it is not closed under sum, i.e., if (p,q) and (r,s) are in the sphere, then (p+q, r+s) is not necessarily in it. I wonder if one can show there is no way of defining a sum that would have this property.

 

[geordief]

Im not sure where you're heading with this but , just in case, a sphere is not a vector space; it does not necessarily contain the origin and it is not closed under sum, i.e., if (p,q) and (r,s) are in the sphere, then (p+q, r+s) is not necessarily in it. I wonder if one can show there is no way of defining a sum that would have this property.

Oh,I see. A sphere is not a vector space. I stand very much corrected.Thanks

 

[member 587159]

Im not sure where you're heading with this but , just in case, a sphere is not a vector space; it does not necessarily contain the origin and it is not closed under sum, i.e., if (p,q) and (r,s) are in the sphere, then (p+q, r+s) is not necessarily in it. I wonder if one can show there is no way of defining a sum that would have this property.

Well it is not possible to define an $\mathbb{R}$- scalar multiplication on the sphere . Indeed, suppose there is one. Then if $p\in S^n$, we get $2p\in S^n$. Taking Euclidean norms, this means that $1=\Vert 2 p\Vert =2\Vert p\Vert=2$, impossible.

Thus the sphere does not admit a vector space structure.

Note that $S^1$ and $S^3$ do admit Lie group structures, so a priori it might be possible to define abelian group structures on (some) spheres, but we cannot make these compatible with a scalar multiplication.

 

[WWGD]

Well it is not possible to define an $\mathbb{R}$- scalar multiplication on the sphere . Indeed, suppose there is one. Then if $p\in S^n$, we get $2p\in S^n$. Taking Euclidean norms, this means that $1=\Vert 2 p\Vert =2\Vert p\Vert=2$, impossible.

Thus the sphere does not admit a vector space structure.

Note that $S^1$ and $S^3$ do admit Lie group structures, so a priori it might be possible to define abelian group structures on (some) spheres, but we cannot make these compatible with a scalar multiplication.

Sure, it cannot be done over the Reals but I was wondering ; I think I remember hearing ( too tired to think at the moment) that by restricting the scalars somehow to those having modulus/norm 1, correlation could ne made into an inner-product in the space of RVs.

 

[geordief]

Is this an example of a vector space and it's dual?

1:) The vector space is just the "normal" 2D x,y plane where each point ,is a pairing of x and y values - So all the possible values of (x,y)

2:) Its dual would be the same co ordinate system rotated around the origin at an angle of choice and containing corresponding pairs of (x,y) whose position has been moved.*Would those 2D vectors spaces be Dual to each other?

If they are Duals to each other ,would that mean they can be used to create a Tensor Space?

* ie (3,4) in the first vector space might correspond to ,just guessing (-3,4) in the second,Dual space...and so on.

 

[WWGD]

Is this an example of a vector space and it's dual?

1:) The vector space is just the "normal" 2D x,y plane where each point ,is a pairing of x and y values - So all the possible values of (x,y)

2:) Its dual would be the same co ordinate system rotated around the origin at an angle of choice and containing corresponding pairs of (x,y) whose position has been moved.*Would those 2D vectors spaces be Dual to each other?

If they are Duals to each other ,would that mean they can be used to create a Tensor Space?

* ie (3,4) in the first vector space might correspond to ,just guessing (-3,4) in the second,Dual space...and so on.

In what sense would they be dual? Does the rotated copy contain all linear maps from the initial copy of the plane into the Reals? How so? Moreover, you can tensor any two vector spaces ( over the same field) , not just a vector space and its dual.

 

[geordief]

E

In what sense would they be dual? Does the rotated copy contain all linear maps from the initial copy of the plane into the Reals? How so?

Well ,just to take the example I gave (3,4) in the original vector space ,wouldn't there be a (3,4) in the rotated space too and wouldn't there be equivalent linear maps to the real numbers in that rotated vector space?

I appreciate that I am surely badly wrong but that is my reasoning.

I think you must be telling me that the Vector Space and the Dual Space inhabit each other's Space quite completely (no "overlapping")

 

[WWGD]

E
Well ,just to take the example I gave (3,4) in the original vector space ,wouldn't there be a (3,4) in the rotated space too and wouldn't there be equivalent linear maps to the real numbers in that rotated vector space?

I appreciate that I am surely badly wrong but that is my reasoning.

I think you must be telling me that the Vector Space and the Dual Space inhabit each other's Space quite completely (no "overlapping")

There is no condition of inhabiting or being contained in the definition of vector space and their duals. These can be abstract spaces without geometric models. The one thing needed is that the dual space has to somehow be a space containing all linear maps from the original space into the base field.

 

[PeterDonis]

Is this an example of a vector space and it's dual?

Before looking for examples, you should first be sure you understand what a vector space is. There are certain axioms that every vector space must satisfy. That is how you tell whether something is a vector space--by seeing if it satisfies the axioms.

Similarly, to see whether something is a covector space, you need to understand what axioms a covector space (or dual vector space) satisfies.

The vector space is just the "normal" 2D x,y plane where each point ,is a pairing of x and y values - So all the possible values of (x,y)

How do you know this is a vector space?

2:) Its dual would be the same co ordinate system rotated around the origin

Why do you think this would be a dual vector space? How is it a space of functions from a vector space into the reals?

 

[geordief]

There is no condition of inhabiting or being contained in the definition of vector space and their duals. These can be abstract spaces without geometric models. The one thing needed is that the dual space has to somehow be a space containing all linear maps from the original space into the base field.

I see. Are there examples of Vector Spaces (and their Dual spaces) that can be represented geometrical (and in 2d) for purely pedagogic purposes?

 

[fresh_42]

You live in a three dimensional real vector space: height, length and width.

Dual spaces are harder to visualize, because they are functions. Real valued functions which map vectors in a certain way into numbers.

I hesitate to give you examples as long as you don't understand the basics. They could do more bad than good, will say confuse you even more. The only example I dare to give is a projection onto a certain coordinate, say the $x-$ coordinate. So $(x,y)=(4,2)$ is a vector, and the function $(x,y) \longmapsto x\, , \,(4,2) \longmapsto 4$ is an element of the dual space $(\mathbb{R}^2)^*$ of $\mathbb{R}^2$. Both have are completely different things: one space are arrows, the other one linear functionals, which means linear mappings with the real numbers as range.

 

[geordief]

You live in a three dimensional real vector space: height, length and width.

Dual spaces are harder to visualize, because they are functions. Real valued functions which map vectors in a certain way into numbers.

I hesitate to give you examples as long as you don't understand the basics. They could do more bad than good, will say confuse you even more. The only example I dare to give is a projection onto a certain coordinate, say the $x-$ coordinate. So $(x,y)=(4,2)$ is a vector, and the function $(x,y) \longmapsto x\, , \,(4,2) \longmapsto 4$ is an element of the dual space $(\mathbb{R}^2)^*$ of $\mathbb{R}^2$. Both have are completely different things: one space are arrows, the other one linear functionals, which means linear mappings with the real numbers as range.

Can I also map (4,2) to the y-coordinate? The number 2...

Does (4,2) map only to those 2 numbers in the dual space?

Or can (4,2) project onto any other vector in the vector space say and give a real number ( say onto (4,1) )

 

[fresh_42]

Can I also map (4,2) to the y-coordinate? The number 2...

Yes.

Does (4,2) map only to those 2 numbers in the dual space?

No. However, one can prove that these two functions span the dual space. But that's not surprising as the dual space here is two dimensional and these two functions are linearly independent.

Or can (4,2) project onto any other vector in the vector space say and give a real number ( say onto (4,1) )

This depends on how you setup your function. If it is a linear combination of the two projections, then yes. The resulting function has still to be linear.

 

[geordief]

Can I project (1,1) onto (1,0) and then add the vector which projects (1,0) onto(1/2,1/2) and so on ** continuously getting back to (0,0) as the limit?

Would those be linear combinations in the Dual Space?

** next would be projecting (1/2, 1/2) to (1/2, 0)...

 

[fresh_42]

Can I project (1,1) onto (1,0) and then add the vector which projects (1,0) onto(1/2,1/2) and so on ** continuously getting back to (0,0) as the limit?

Would those be linear combinations in the Dual Space?

** next would be projecting (1/2, 1/2) to (1/2, 0)...

This confuses me. Say we have $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$. Then every function $\varphi = \alpha \pi_1+\beta \pi_2$ is a vector of the dual space. So we get all functions $\varphi(x,y)=\alpha \cdot x + \beta \cdot y$. You can now set any $\varphi(x,y)=(a,b)$ and see whether you can find values $\alpha,\beta$ which will do.

 

[geordief]

This confuses me. Say we have $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$. Then every function $\varphi = \alpha \pi_1+\beta \pi_2$ is a vector of the dual space. So we get all functions $\varphi(x,y)=\alpha \cdot x + \beta \cdot y$. You can now set any $\varphi(x,y)=(a,b)$ and see whether you can find values $\alpha,\beta$ which will do.

Yes you are clearer than my attempt. Thanks I am seeing it a little clearer now.

 

[geordief]

Can I keep this thread going? (I feel I have made some progress)

So the dual vector bases are chosen (from countless possibilities) precisely for the property that they are orthogonal to the vector bases?

And what about spaces with higher dimensions...are all the dual bases likewise orthogonal to or does it get more complex?(is that where the Kronecker Delta function applies?)

 

[lavinia]

Can I keep this thread going? (I feel I have made some progress)

So the dual vector bases are chosen (from countless possibilities) precisely for the property that they are orthogonal to the vector bases?

And what about spaces with higher dimensions...are all the dual bases likewise orthogonal to or does it get more complex?(is that where the Kronecker Delta function applies?)

Dual spaces are not orthogonal to the underlying vector space. This in fact makes no sense since they are different vector spaces. It only makes sense to say that two vectors are orthogonal if they are in the same vector space and this vector space has an inner product.

Dual spaces are simply the space of linear maps from a vector space into the field of scalars e.g. the real numbers in the case of real vector spaces. They do not have inner products by themselves so there is no intrinsic idea of angle or length. An inner product must be chosen as an additional feature. It is extra.

If one chooses a basis for a vector space then there is a dual basis whose values on the basis of the vector space can be described by the Kronecker delta. But this use of Kronecker delta does not come from an inner product. It is merely a short hand way of describing the values of the dual basis on the underlying basis vector.

However, given a basis one can define an inner product on the vector space by declaring the basis vectors to be orthogonal and of length 1. So every choice of a basis determines an inner product. But this inner product is on the vector space not on the dual space. However, one can uses this to define a second inner product on the dual space by declaring the dual basis to be orthogonal and of length 1.

- In general given two vector spaces, one can consider the space of all linear maps from one of the vector spaces into the other. These linear maps themselves form a vector space since one can add them and multiply them by scalars and still get linear maps. The dual space is a special case where the target vector space is the one dimensional vector space of scalars.

 

[geordief]

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

Do they share the same origin ?

If ,as a simple example the vector space is the set of all real number pairs is perhaps 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]

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?

 

[PeterDonis]

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

The definition of the dual space.

You appear to be confused by the fact that the dual space is also a vector space. But I think that's because you are confused about what a "vector space" is, mathematically. Mathematically, a vector space is not "a space of little arrows". It is anything that satisfies the axioms of a vector space.

The dual space to a vector space is a space of linear maps. And it just so happens that a space of linear maps satisfies the vector space axioms, so it is a vector space. But that doesn't mean a space of linear maps is a space of little arrows, or points, or that it is somehow "the same space" as the vector space it's a dual of.

With that in mind, let's look at the questions you are asking:

Do they share the same origin ?

What does this even mean? Suppose the original vector space is $\mathbb{R}^2$--roughly speaking, the set of points in a plane viewed as a vector space (so each vector can be thought of as an arrow from the origin to a given point). (Your "the set of all number pairs" is basically the same thing.) The dual space is then the space of linear maps from $\mathbb{R}^2$ to $\mathbb{R}$. The "origin" of the dual space, as a member of the dual space, is then a linear map, not a point in $\mathbb{R}^2$. So what does it even mean to ask whether it is "the same" as the origin of the original vector space?

is perhapd its Dual Space the set of all number pairs where one of the pairs is zero?

No. See above.

(meaning [x,y] would be mapped to either [x,0] or [0,y]

Neither. See above.

 

[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 :)