Vectors and Dual Vectors

In summary: One colloquial defining property of dual vectors is that they "eat vectors and produce a scalar." An example is -Eμdxμ = work done. But one can take any two regular vectors and form the sum FμGμ which would also be a scalar. So how are dual vectors different in this regard?Dual vectors are different because they are defined as linear maps from vectors to scalars, while regular vectors are simply elements of a vector space. So while you can form a scalar from two regular vectors by taking the dot product, you can also form a scalar from a dual vector and a regular vector by applying the
  • #1
pixel
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I've been looking at various online sources for relativity and have some confusion about "dual vectors." I'm hoping for some very basic information/examples from physics, not abstract mathematical concepts from the field of vector spaces.

1. In addition to the vector/dual vector distinction, there is also the distinction between contravariant/covariant forms of vectors. I believe covariant vectors and dual vectors transform the same way. Are they essentially the same thing?

2. Someone gave the electric field as an example of a dual vector because it is found as the gradient of the potential field and transforms as a covariant vector. One colloquial defining property of dual vectors is that they "eat vectors and produce a scalar." An example is -Eμdxμ = work done. But one can take any two regular vectors and form the sum FμGμ which would also be a scalar. So how are dual vectors different in this regard?
 
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  • #2
Possibly useful:
https://www.physicsforums.com/threa...ion-and-operations-of-gr.968301/#post-6149583

To tie in some physics, it is helpful to think of the electric field 1-form [dual vector, covector]
as a map from displacements to voltages. The units of electric field are "Volts/meter".
While an electrostatic field---as the gradient of a scalar potential---is a good example,
the electric field need not be electrostatic and still retain its identity as a 1-form.
(In Gauss's law, the electric field is best thought of as a two-form.)

Similarly, a [conservative] force maps displacements to work-done and thus has units of "Joules/meter".
Momentum can be thought of as something that maps displacements to "actions" and this has units [itex]\rm J\cdot s/m [/itex]
Up to unit conventions, the magnetic field in Ampere's law can be thought of as something that maps displacements (along a closed curve) to currents (enclosed).
The wave-vector [itex]\vec k[/itex] is better thought of as the gradient of a phase, which maps displacements to phase-changes and thus has units of [itex]\rm rad/m[/itex].
 
  • #3
Abstractly, vectors are "vector spaces", and represent anything that can be multiplied by numbers (scalars), and added together. The additive operator must commute, so ##\vec{a}+\vec{b}=\vec{b}+\vec{a}##, the scalar multiplication must be distributive, so that ##a(\vec{b}+\vec{c}) = a\vec{b} + a\vec{c}##, and the whole space has to be closed.

The duality principle says that a linear map from a vector (space) to a scalar is another vector space of the same dimension as the original vector space. It's called the "dual space". The duality principle also says that the dual of a dual space is the same as the original space. While the dual space has the same dimension as the original space, but it's not the "same space".

Proofs of this can be found in most linear algebra textbooks, but it's all rather abstract and I've long since forgotten the details.

Covariant and contravariant vectors are duals of one another. The concept also comes up in pre-tensor analysis as "row vectors" and "column vectors".

Because dual vectors are maps from vectors to scalars, the pairing of a dual vector (a map from a vector to a scalar) and a vector gives a scalar. This is called the "natural pariing".

An application to the previously mentioned row and column vectors. If you have a row vector and a column vector, the matrix product of the row vector with the column vector gives you a scalar. This is all inear, so this makes a row vector a linear map from a column vector to a scalar, and hence the row vector is the dual of the column vector. The reverse is also true.

Graphically, my favorite technique is to represent vectors by little lines with arrows, and dual vectors as "stacks of plates". The number of plates that the arrow "pierces" is the scalar assocaited with the natural pairing of the dual vector and the vector. This is used in textbooks such as MTW's "Gravitation", which is a GR textbook.I
 
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  • #4
Can I ask a question in this thread as I have been looking into this also over the past few weeks,..?

Does the distinction between contravariant and covariant vectors only apply when the base vectors are not orthogonal (ie are they identical otherwise?)

If the bases are not orthogonal what happens when a covariant vector is added to a contravariant vector?

Is the result the same irrespective of the order of the operation?
 
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  • #5
geordief said:
Does the distinction between contravariant and covariant vectors only apply when the base vectors are not orthogonal (ie are they identical otherwise?
Covariant and contravariant vectors generally belong do different vector spaces. Only when you have a non-degenerate 2-form is there a connection between them.

geordief said:
If the bases are not orthogonal what happens when a covariant vector is added to a contravariant vector?
You cannot add a covariant and a contravariant vector. They belong to different vector spaces.
 
  • #6
Orodruin said:
Covariant and contravariant vectors generally belong do different vector spaces. Only when you have a non-degenerate 2-form is there a connection between them.You cannot add a covariant and a contravariant vector. They belong to different vector spaces.

Thanks,you are helpful. Back to the drawing board:smile:
 
  • #7
pixel said:
I've been looking at various online sources for relativity and have some confusion about "dual vectors." I'm hoping for some very basic information/examples from physics, not abstract mathematical concepts from the field of vector spaces.

1. In addition to the vector/dual vector distinction, there is also the distinction between contravariant/covariant forms of vectors. I believe covariant vectors and dual vectors transform the same way. Are they essentially the same thing?

That's exactly the way to think about dual vectors: They are functions that "eat vectors and produce a scalar". If you have a "dot" product, then you can convert any vector into a dual vector:

If ##u## is a vector, then you define a corresponding function ##f_u## that acts on vectors this way:

##f_u(v) = u \cdot v##

So the covariant form of a contravariant vector always implicitly involves the dot product, which is more properly called the "metric tensor". If ##u## and ##v## are vectors, and ##g## is the metric tensor, then ##u \cdot v = g(u,v)##. In terms of components, ##g(u,v) = \sum_{i j} g_{ij} u^i v^j##

2. Someone gave the electric field as an example of a dual vector because it is found as the gradient of the potential field and transforms as a covariant vector. One colloquial defining property of dual vectors is that they "eat vectors and produce a scalar." An example is -Eμdxμ = work done. But one can take any two regular vectors and form the sum FμGμ which would also be a scalar. So how are dual vectors different in this regard?

No, you cannot take two vectors ##F^\mu## and ##G^\mu## (by convention, raised indices indicate vectors and lowered indices indicate dual vectors) and form the sum ##\sum_\mu F^\mu G^\mu##. The latter sum is not a meaningful quantity. What is meaningful, as I said, is ##\sum_{\mu \nu} g_{\mu \nu} F^\mu G^\nu##. You need the metric tensor ##g## to convert a vector into a dual vector.
 
  • #8
stevendaryl said:
If you have a "dot" product, then you can convert any vector into a dual vector
It is not necessary that this is an inner product. All that is required is a non-degenerate 2-form. For example, in symplectic geometry you use a symplectic form rather than a metric. This is used in Hamiltonian mechanics to relate Hamiltonian vector fields to the exterior derivative of the Hamiltonian function.
 
  • #9
stevendaryl said:
If you have a "dot" product, then you can convert any vector into a dual vector
So one can convert any vector into a covector (and vice versa?) but the two vectors (in their different spaces) cannot act on each other directly ?(indirectly?)
 
  • #10
geordief said:
So one can convert any vector into a covector (and vice versa?) but the two vectors (in their different spaces) cannot act on each other directly ?(indirectly?)
What do you mean by "act on"? By definition a dual vector acts on a tangent vector to produce a scalar. What you cannot do is to add a tangent vector and a dual vector.
 
  • #11
Orodruin said:
What do you mean by "act on"?
By "act on" I had in mind "add to" .

Still ,if one converts a covector to a vector as stevendaryl suggested ,is it then possible to add this vector to one of the vectors in the covectors' dual space (the original vector space) ?

Can you "switch" between the vector space and its dual space in this way?
 
  • #12
geordief said:
So one can convert any vector into a covector (and vice versa?) but the two vectors (in their different spaces) cannot act on each other directly ?(indirectly?)

Right, two vectors (or two dual vectors) can only act on each other if you have a metric.

I think it's worthwhile to think about what it's like to have a space without a metric. That to me illustrates the importance of metrics and the importance of distinguishing vectors from dual vectors.

Here's a simple example of such a space, from thermodynamics. If you have a quantity of a gas, then the state of that gas can be characterized by two thermodynamic variables: ##V## (volume) and ##T## (temperature) (or we could have chosen volume and pressure, or some other combination of quantities). The state is then a point on a two-dimensional space. In this two-dimensional space, we can define two different notions of vector-like quantities.

Tangent vectors:

If the state ##S## of the system is changing as a function of time, then you can represent the rate of change by a vector: ##\overrightarrow{v} = \frac{dS}{dt}##. This vector has two components: ##v^V = \frac{dV}{dt}## and ##v^T = \frac{dT}{dt}##. Note that the units of the two components are different. ##v^V## has units of "volume per unit time" and ##v^T## has units of "temperature per unit time". It would be completely meaningless to take the "dot" product of ##\overrightarrow{v}## with itself: ##\overrightarrow{v} \cdot \overrightarrow{v} = ## some combination of terms of the form temperature-squared per time-squared and volume-squared per time-squared. That quantity doesn't have consistent units.

Dual vectors:

We can construct a different kind of vector, as well. If we have a function such as ##P## (pressure) that depends on both ##V## and ##T##, then we can construct a different kind of vector that characterizes how ##P## changes with "location" in the state space: ##\overrightarrow{\omega} = \nabla P##, with components ##\omega_V = \frac{\partial P}{\partial V}## and ##\omega_T = \frac{\partial P}{\partial T}##. Note again that the components do not have the same units, so it makes no sense to take the dot product of such a vector with itself.

However, it does make sense to take the dot-product of a vector and a dual-vector:

##\nabla P \cdot \frac{dS}{dt} = \frac{\partial P}{\partial V} \frac{dV}{dt} + \frac{\partial P}{\partial T} \frac{dT}{dt}##

This combination does have consistent units: pressure per unit time. It gives the time rate of change of ##P## as the state changes.

Note: I probably shouldn't use the same symbol, ##\cdot## to mean the action of a dual vector on a vector and to mean the combination of two vectors.
 
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So a metric is a mathematical object that allows one to work with two recalcitrant quantities ,like a kind of bridge?

It comes with a bit of extra information that slots into either side of the question and let's them "talk to each other mathematically"?
 
  • #14
geordief said:
So a metric is a mathematical object that allows one to work with two recalcitrant quantities ,like a kind of bridge?

It comes with a bit of extra information that slots into either side of the question and let's them "talk to each other mathematically"?

Vectors and dual vectors talk to each other just fine. It's two vectors that can't talk without help.

You can think of it like the male and female ends of a garden hose. You can't connect two male ends or two female ends without some kind of adapter.
 
  • #15
stevendaryl said:
Vectors and dual vectors talk to each other just fine. It's two vectors that can't talk without help.
This depends what you mean by "talk to". With two vectors in the same vector space you can always add them (by definition of "vector space"). What you cannot do is map them linearly to scalars, for this you need a vector and a dual vector.
 
  • #16
stevendaryl said:
I think it's worthwhile to think about what it's like to have a space without a metric. That to me illustrates the importance of metrics and the importance of distinguishing vectors from dual vectors.

What is the metric in your example?
 
  • #17
stevendaryl said:
Vectors and dual vectors talk to each other just fine. It's two vectors that can't talk without help.

You can think of it like the male and female ends of a garden hose. You can't connect two male ends or two female ends without some kind of adapter.
Thanks . I feel like I have learned quite a bit.

Gooseberries can be useful
(https://dictionary.cambridge.org/dictionary/english/play-gooseberry)
:wink
:
 
  • #18
pixel said:
What is the metric in your example?
The entire point was that he was looking at a space without a metric where there is no direct way to relate tangent to dual vectors.
 
  • #19
Orodruin said:
The entire point was that he was looking at a space without a metric where there is no direct way to relate tangent to dual vectors.
I don't understand this. A dual vector is an element of ##V^*## the vector space of linear functionals of ##V##. They automatically relate to vectors via ##(v^*,u)\longmapsto v^*(u)##. There is no need for any metric, and they are still directly related.
 
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  • #20
fresh_42 said:
I don't understand this. A dual vector is an element of ##V^*## the vector space of linear functionals of ##V##. They automatically relate to vectors via ##(v^*,u)\longmapsto v^*(u)##. There is no need for any metric, and they are still directly related.
The problem here is the imprecision in that word "relate"... Without a metric there's no way of raising and lowering indices, no dual vector ##V_\mu## that we can say is "related to" the vector ##V^\mu##.
 
  • #21
fresh_42 said:
I don't understand this. A dual vector is an element of ##V^*## the vector space of linear functionals of ##V##. They automatically relate to vectors via ##(v^*,u)\longmapsto v^*(u)##. There is no need for any metric, and they are still directly related.
Relating here does not imply ”mapping a vector and dual vector to a scalar”. It implies a bijection between a vector space and its dual.
 
  • #22
Orodruin said:
Relating here does not imply ”mapping a vector and dual vector to a scalar”. It implies a bijection between a vector space and its dual.

Natural, basis-independent linear bijection.

Since, for finite-dimensional ##V##, ##V## and ##V^*## have the same dimension, it is always possible to construct a linear bijection between ##V## and ##V^*##, even without a "metric".
 
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  • #23
Orodruin said:
The entire point was that he was looking at a space without a metric where there is no direct way to relate tangent to dual vectors.

I asked because he gave an example of a vector and a dual vector but I think I understand what you are saying is that the latter was not the dual of the former.

To put things in concrete terms, can you give an example of a vector representing a physical quantity and its corresponding dual vector, where the latter was derived by applying the metric to the former?
 
  • #24
Nugatory said:
The problem here is the imprecision in that word "relate"... Without a metric there's no way of raising and lowering indices, no dual vector ##V_\mu## that we can say is "related to" the vector ##V^\mu##.
Even this makes me wonder.
Let's say we have ##\varphi_1\in V^*-\{\,0\,\}##. Then there is a vector ##v_1\in V-\{\,0\,\}## such that ##\varphi(v_1)=1##. Now we can split ##V=\mathbb{F}\cdot v_1 \oplus \operatorname{ker}\varphi_1## and continue with ##V_2:= \operatorname{ker}\varphi_1##. Where is the problem?
 
  • #25
pixel said:
What is the metric in your example?

You included the following quote from me:
I think it's worthwhile to think about what it's like to have a space without a metric.

The ##V,T## space is a space without a metric.
 
  • #26
pixel said:
I asked because he gave an example of a vector and a dual vector but I think I understand what you are saying is that the latter was not the dual of the former.

The two types of vectors ("velocity" vectors with components ##\frac{dV}{dt}## and ##\frac{dT}{dt}## and "gradient" vectors with components ##\frac{\partial P}{\partial V}## and ##\frac{\partial P}{\partial T}##) are duals of each other. But there is no metric, so you can't convert the one type of vector into the other type.

To put things in concrete terms, can you give an example of a vector representing a physical quantity and its corresponding dual vector, where the latter was derived by applying the metric to the former?

Any time you compute the "length" of a vector, that's what you're doing. You have a velocity vector ##\overrightarrow{v}##. When you write ##|v| = \sqrt{\overrightarrow{v} \cdot \overrightarrow{v}}## you're using the metric to convert one of the ##\overrightarrow{v}## into a dual vector.

##|v| = \sqrt{\overrightarrow{v} \cdot \overrightarrow{v}} = \sqrt{g_{\mu \nu} v^\mu v^\nu}##

The quantity ##g_{\mu \nu} v^\mu## is the dual vector ##v_\nu##.

You might be asking what's a natural interpretation of ##v_\nu## as a dual vector. Dual vectors are most naturally created as a kind of "gradient" of a scalar function: ##\omega_\mu = \frac{\partial \phi}{\partial x^\mu}##. You can certainly reverse engineer a scalar function ##\phi## such that ##v_\mu## are the components of its gradient, but I guess that's not very interesting.
 
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  • #27
fresh_42 said:
Even this makes me wonder.
Let's say we have ##\varphi_1\in V^*-\{\,0\,\}##. Then there is a vector ##v_1\in V-\{\,0\,\}## such that ##\varphi(v_1)=1##. Now we can split ##V=\mathbb{F}\cdot v_1 \oplus \operatorname{ker}\varphi_1## and continue with ##V_2:= \operatorname{ker}\varphi_1##. Where is the problem?

Let's consider my example. We have a 2-D state space for the thermodynamic state of a quantity of a gas, which can be characterized by the coordinate system ##V,T## (volume, temperature). We can use the rate of change of the state as an example tangent vector with components ##v^V = \frac{dV}{dt}## and ##v^T = \frac{dT}{dt}##. The point is that there is no meaningful notion of ##v \cdot v##. You can certainly make up a way to make sense of that, but it's not likely to be physically meaningful.
 
  • #28
stevendaryl said:
Let's consider my example. We have a 2-D state space for the thermodynamic state of a quantity of a gas, which can be characterized by the coordinate system ##V,T## (volume, temperature). We can use the rate of change of the state as an example tangent vector with components ##v^V = \frac{dV}{dt}## and ##v^T = \frac{dT}{dt}##. The point is that there is no meaningful notion of ##v \cdot v##. You can certainly make up a way to make sense of that, but it's not likely to be physically meaningful.
Yes, but there is a major difference between physically meaningful and the requirement of a metric or even impossible. The algorithm I mentioned allows a unique identification ##\varphi_i \longleftrightarrow v_i## of basis vectors with the property ##\varphi_j(v_i)=\delta_{ij}##. No metric anywhere near, only fundamental properties of the category of vector spaces.
 
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  • #29
fresh_42 said:
Even this makes me wonder.
Let's say we have ##\varphi_1\in V^*-\{\,0\,\}##. Then there is a vector ##v_1\in V-\{\,0\,\}## such that ##\varphi(v_1)=1##. Now we can split ##V=\mathbb{F}\cdot v_1 \oplus \operatorname{ker}\varphi_1## and continue with ##V_2:= \operatorname{ker}\varphi_1##. Where is the problem?

Requiring that the "relation" between ##V## and ##V^*## be a vector space isomorphism (linear bijection) would mean that
$$\psi_1 \sim v \Rightarrow c \psi_1 \sim c v$$
for ##c\neq0## in ##\mathbb{F}##.

Now, let ##\phi_1 = c\psi_1##, and run the procedure again. When this is done, ##\phi_1 = c \psi_1 \sim \frac{1}{c} v##. Consequently, this procedure does not lead to a natural isomorphism.
 
  • #30
George Jones said:
Requiring that the "relation" between ##V## and ##V^*## be a vector space isomorphism (linear bijection) would mean that
$$\psi_1 \sim v \Rightarrow c \psi_1 \sim c v$$
for ##c\neq0## in ##\mathbb{F}##.

Now, let ##\phi_1 = c\psi_1##, and run the procedure again. When this is done, ##\phi_1 = c \psi_1 \sim \frac{1}{c} v##. Consequently, this procedure does not lead to a natural isomorphism.
I do not understand this .

I agree that the isomorphism isn't natural, because certain correspondences are deliberately chosen, but whether I have ##\varphi_i(v_j)=\delta_{ij}## or ##\langle \varphi_i , v_j \rangle =\delta_{ij}## doesn't make a difference. If you stretch both, vector and covector with the same factor, you will get a squared factor anyway: ##\begin{bmatrix}2&0\end{bmatrix}\cdot\begin{bmatrix}2\\0\end{bmatrix}=4##, even in the ordinary plane.
 
  • #31
fresh_42 said:
Yes, but there is a major difference between physically meaningful and the requirement of a metric or even impossible.

But I think for the purposes of getting an intuition about difference between tangent vectors and their duals, it helps to consider a case where there is no obvious physically meaningful way to convert one to the other.
 
  • #32
George Jones said:
Natural, basis-independent linear bijection.
"Natural" is a matter of definition. I don't understand your need to put "basis-independent" in there. Nowhere did we mention a basis. You can also use any basis to define a basis-independent linear bijection - it just will not have the same components. As soon as you define your non-degenerate linear map between ##V## and ##V^*## you have defined your non-degenerate 2-form that determines what "natural" means.

pixel said:
To put things in concrete terms, can you give an example of a vector representing a physical quantity and its corresponding dual vector, where the latter was derived by applying the metric to the former?
The phase differential ##d\phi## and the wave vector ##N## of a plane wave in Minkowski space (or any other 3+1-spacetime).

fresh_42 said:
Yes, but there is a major difference between physically meaningful and the requirement of a metric or even impossible. The algorithm I mentioned allows a unique identification ##\varphi_i \longleftrightarrow v_i## of basis vectors with the property ##\varphi_j(v_i)=\delta_{ij}##. No metric anywhere near, only fundamental properties of the category of vector spaces.
I do not agree with the bold part. You just defined a type (0,2) tensor in your bijection between ##V## and ##V^*## with exactly the properties of a metric so your bijection in itself is an actual metric.

George Jones said:
Now, let ##\phi_1 = c\psi_1##, and run the procedure again. When this is done, ##\phi_1 = c \psi_1 \sim \frac{1}{c} v##. Consequently, this procedure does not lead to a natural isomorphism.
Simply because this defines a different metric on your vector space ##V##. By choosing that your particular bases should be orthonormal you have imposed a metric on the space. If you choose another basis to be orthonormal you will obviously get a different metric, but it will still define a metric. Whether or not the metric is "natural" is a matter of semantics.
 
  • #33
Orodruin said:
I do not agree with the bold part. You just defined a type (0,2) tensor in your bijection between VVV and V∗V∗V^* with exactly the properties of a metric so your bijection in itself is an actual metric.
But I do not require one. Only that I can write a vector space as a direct sum of a kernel and a non-zero vector (line). I can do this as soon as I defined ##V^*##, no matter whether there is a given metric or not, or finite dimensionality. Only thing is, that there is no unique (or canonical) way to do it. Once a metric is given, one would adjust the algorithm appropriately.

But the debate doesn't fit very well into this physical forum. (I recognized too late that it is SR/GR where metrics are crucial parts.)
 
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  • #34
fresh_42 said:
But I do not require one.
No, you do not require one to define the map, but as soon as you do you have one in the map you defined. So why did you define it if you did not require one?

fresh_42 said:
Only thing is, that there is no unique (or canonical) way to do it.
Agreed. But once you have defined your map, that in itself provides you with a metric.
 
  • #35
geordief said:
Can I ask a question in this thread as I have been looking into this also over the past few weeks,..?

Does the distinction between contravariant and covariant vectors only apply when the base vectors are not orthogonal (ie are they identical otherwise?)

If the bases are not orthogonal what happens when a covariant vector is added to a contravariant vector?

Is the result the same irrespective of the order of the operation?
A lot of the mysteries of tensor analysis immediately goes away when you are just slightly more precise with the language (which physicists usually never are, which is a pity, but you can't help it :-(): Vektors and dual vectors are not "variant" at all but independent of any chosen basis/reference frame. The components are what changes when changing from one basis to another.

Now, if you just have a finite-dimensional real vector space, there are first the vectors ##\vec{v}## and bases of the vector space ##\vec{b}_j##. Any vector can be uniquely represented by its components with respect to this basis,
$$\vec{v}=v^j \vec{b}_j.$$
Here and in the following the Einstein summation convention is used, i.e., one has to sum over repeated indices.

Now the change from one basis to another is defined by an invertible matrix,
$$\vec{b}_j={T^k}_j \vec{b}_k'.$$
The transformation of the vector components is immediately derived by the invariance of the vector, i.e.,
$$\vec{v}=v^j \vec{b}_j = v^j {T^k}_j \vec{b}_k' \; \Rightarrow \; v^{\prime k}={T^k}_j v^j.$$

The next very natural notion are the linear maps from vectors to the real numbers, socalled linear forms (also called 1-forms, or tensors of rank 1), mapping any vector ##\vec{v} \mapsto M(\vec{v}) \in \mathbb{R}## being linear. Because of the linearity you only need to know, how the basis vectors map, i.e.,
$$M(\vec{b}_j)=M_j,$$
because then you know how all vectors map given their components with respect to the basis
$$M(\vec{v})=M(v^j \vec{b}_j) = v^j M(\vec{b}_j)=v^j M_j.$$
Now the linear form must not change by just changing the basis. This again uniquely defines the transformation properties of its components
$$M_j=M(\vec{b}_j) = M({T^k}_j \vec{b}_k') = {T^k}_j M(\vec{b}_j')={T^k}_j M_j',$$
or using the inverse matrix ##{U^j}_k={(T^{-1})^j}_k##,
$$M_k' = {U^{j}}_{k} M_j.$$
one says the components of a linear form transform contragrediently to the vector components.

It's clear that the linear forms also build a vector space, called the dual space.
 
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<h2>1. What is the difference between a vector and a dual vector?</h2><p>A vector is a mathematical object that represents magnitude and direction. It can be represented by an array of numbers or symbols. A dual vector, also known as a covector, is a mathematical object that represents a linear function on vectors. In other words, it is a mapping from vectors to scalars.</p><h2>2. How are vectors and dual vectors related?</h2><p>Vectors and dual vectors are related through a mathematical operation called the dot product. The dot product of a vector and a dual vector results in a scalar value. This relationship is often used in physics and engineering to calculate work, energy, and other physical quantities.</p><h2>3. Can vectors and dual vectors exist in different dimensions?</h2><p>Yes, vectors and dual vectors can exist in different dimensions. Vectors are typically represented as n-tuples, where n is the number of dimensions. Dual vectors, on the other hand, are represented as row vectors and can exist in a different number of dimensions than the corresponding vector.</p><h2>4. What is the geometric interpretation of a dual vector?</h2><p>The geometric interpretation of a dual vector is as a hyperplane in the vector space. In two dimensions, a dual vector can be thought of as a line perpendicular to the vector it is dual to. In three dimensions, a dual vector can be thought of as a plane perpendicular to the vector it is dual to.</p><h2>5. How are vectors and dual vectors used in machine learning?</h2><p>Vectors and dual vectors are used in machine learning to represent data and parameters. For example, vectors can be used to represent features of a dataset, while dual vectors can be used to represent the weights of a model. The dot product between a vector and a dual vector can be used to make predictions or update the model's parameters during training.</p>

1. What is the difference between a vector and a dual vector?

A vector is a mathematical object that represents magnitude and direction. It can be represented by an array of numbers or symbols. A dual vector, also known as a covector, is a mathematical object that represents a linear function on vectors. In other words, it is a mapping from vectors to scalars.

2. How are vectors and dual vectors related?

Vectors and dual vectors are related through a mathematical operation called the dot product. The dot product of a vector and a dual vector results in a scalar value. This relationship is often used in physics and engineering to calculate work, energy, and other physical quantities.

3. Can vectors and dual vectors exist in different dimensions?

Yes, vectors and dual vectors can exist in different dimensions. Vectors are typically represented as n-tuples, where n is the number of dimensions. Dual vectors, on the other hand, are represented as row vectors and can exist in a different number of dimensions than the corresponding vector.

4. What is the geometric interpretation of a dual vector?

The geometric interpretation of a dual vector is as a hyperplane in the vector space. In two dimensions, a dual vector can be thought of as a line perpendicular to the vector it is dual to. In three dimensions, a dual vector can be thought of as a plane perpendicular to the vector it is dual to.

5. How are vectors and dual vectors used in machine learning?

Vectors and dual vectors are used in machine learning to represent data and parameters. For example, vectors can be used to represent features of a dataset, while dual vectors can be used to represent the weights of a model. The dot product between a vector and a dual vector can be used to make predictions or update the model's parameters during training.

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