How Are Dual Vectors and Covectors Related in Vector Spaces?

[Kaguro]

TL;DR

I want to know how are linear functionals same as covectors.

I understand that a vector space is a set of objects closed under addition and scalar multiplication and satisfies several properties.

A functional is a map that takes a vector and produces a scalar. A functional is also called a dual vector.

A covector is an object which transforms via the same matrix that the basis vectors use to transform under a change of coordinates(while the contravectors use its inverse).

How are these two related?

 

[PeroK]

Summary:: I want to know how are linear functionals same as covectors.

I understand that a vector space is a set of objects closed under addition and scalar multiplication and satisfies several properties.

A functional is a map that takes a vector and produces a scalar. A functional is also called a dual vector.

A covector is an object which transforms via the same matrix that the basis vectors use to transform under a change of coordinates(while the contravectors use its inverse).

How are these two related?

Is this not covered in your course or textbook? You need to find a source that presents this topic fully.

 

[Kaguro]

I have studied Linear Algebra in last year from the book of S. Andrili.
This semester,I have tensor algebra and calculus with some intro to vector spaces. This we're studying from a wide variety of books,notes and YT lectures. The only book I read formally is Arfken,Weber,Harris.

So I don't know of a good resource to describe this.

 

[PeroK]

I have studied Linear Algebra in last year from the book of S. Andrili.
This semester,I have tensor algebra and calculus with some intro to vector spaces. This we're studying from a wide variety of books,notes and YT lectures. The only book I read formally is Arfken,Weber,Harris.

So I don't know of a good resource to describe this.

Is this for physics or mathematics?

 

[Kaguro]

I am a student of Physics. And I have taken tensors as an elective. That course mainly deals with Cartesian tensors,and very minimal amount of general tensors.

 

[Kaguro]

PHYSICS-DSE: ADVANCED MATHEMATICAL PHYSICS-I (Credits: Theory-04, Practicals-02) Theory: 60 Lectures The emphasis of the course is on applications in solving problems of interest to physicists. Students are to be examined on the basis of problems, seen and unseen. Linear Vector Spaces Abstract Systems. Binary Operations and Relations. Introduction to Groups and Fields. Vector Spaces and Subspaces. Linear Independence and Dependence of Vectors. Basis and Dimensions of a Vector Space. Change of basis. Homomorphism and Isomorphism of Vector Spaces. Linear Transformations. Algebra of Linear Transformations. Non- singular Transformations. Representation of Linear Transformations by Matrices. (12 Lectures) Matrices Addition and Multiplication of Matrices. Null Matrices. Diagonal, Scalar and Unit Matrices. Upper-Triangular and Lower-Triangular Matrices. Transpose of a Matrix. Symmetric and Skew-Symmetric Matrices. Conjugate of a Matrix. Hermitian and Skew- Hermitian Matrices. Singular and Non-Singular matrices. Orthogonal and Unitary Matrices. Trace of a Matrix. Inner Product. (8 Lectures) Eigen-values and Eigenvectors. Cayley- Hamiliton Theorem. Diagonalization of Matrices. Solutions of Coupled Linear Ordinary Differential Equations. Functions of a Matrix. (10 Lectures) Cartesian Tensors Transformation of Co-ordinates. Einstein’s Summation Convention. Relation between Direction Cosines. Tensors. Algebra of Tensors. Sum, Difference and Product of Two Tensors. Contraction. Quotient Law of Tensors. Symmetric and Anti-symmetric Tensors. Invariant Tensors : Kronecker and Alternating Tensors. Association of Antisymmetric Tensor of Order Two and Vectors. Vector Algebra and Calculus using Cartesian Tensors : Scalar and Vector Products, Scalar and Vector Triple Products. Differentiation. Gradient, Divergence and Curl of Tensor Fields. Vector Identities. (20 lectures) General Tensors Transformation of Co-ordinates. Minkowski Space. Contravariant & Covariant Vectors. Contravariant, Covariant and Mixed Tensors. Kronecker Delta and Permutation Tensors. Algebra of Tensors. Sum, Difference & Product of Two Tensors. Contraction. Quotient Law of Tensors. Symmetric and Anti-symmetric Tensors. Metric Tensor. (10 Lectures)

This is my syllabus.

 

[PeroK]

Summary:: I want to know how are linear functionals same as covectors.

I understand that a vector space is a set of objects closed under addition and scalar multiplication and satisfies several properties.

A functional is a map that takes a vector and produces a scalar. A functional is also called a dual vector.

A covector is an object which transforms via the same matrix that the basis vectors use to transform under a change of coordinates(while the contravectors use its inverse).

How are these two related?

The set of linear functionals (also called covectors, dual vectors or one-forms) is a vector space that can fairly simply be shown, directly from the definition, to have transformation rules that are related to the vector transformation rules.

Have you not seen that? It's a few lines of algebra to prove - although conceptually it may take some time and practice to digest.

 

[Kaguro]

Yes,yes, the set of all linear functionals itself is a vector space. True. But why is a functional(a row vector) same as a covector?

 

[PeroK]

Yes,yes, the set of all linear functionals itself is a vector space. True. But why is a functional(a row vector) same as a covector?

By definition. They are just different names for the same thing.

One thing you need to prove, of course, is that not only do the linear functionals have the required transformation properties, but anything that has these transformation properties can be associated with a linear functional.

 

[PeroK]

PS I found this in a few seconds:

http://idv.sinica.edu.tw/ftliang/diff_geom/*diff_geometry(I)/10.16/covector1.pdf

It begins with the definition that a covector is a linear functional.

 

[Kaguro]

Okay... So everyone just says a covector is a linear functional... Great, we have two different words to describe the exact same thing.

So, tying with what I understand as a covector,
prove that:
$$v'_j=\frac{\partial x^i}{\partial x'^j} v_i$$

You prove that a functional satisfies this.

 

[Kaguro]

By definition. They are just different names for the same thing.

One thing you need to prove, of course, is that not only do the linear functionals have the required transformation properties, but anything that has these transformation properties can be associated with a linear functional.

Yes, just this! How to prove this?

 

[PeroK]

Okay... So everyone just says a covector is a linear functional... Great, we have two different words to describe the exact same thing.

So, tying with what I understand as a covector,
prove that:
$$v'_j=\frac{\partial x^i}{\partial x'^j} v_i$$

You prove that a functional satisfies this.

There are four words: linear functional, covector, dual vector and one-form.

There are a number of key elements here:

1) A linear functional is completely defined by its action on a basis. If we take any basis $e_1, \dots e_n$, and define $$A(e_k) = a_k$$ then, by linearity $$A(v) = A(v^ke_k) = v^ka_k$$.

Note: we can already see that the action of $A$ "looks like" the action of an inner product of some vector with components $v^k$ and some "covector" with components $a_k$.

2) Then we define the linear functional $f^m$ so that it maps the basis vector $e_k$ to $1$ and all other basis vectors to $0$. I.e. $$f^m(e_k) = \delta^m_k$$ Then we see that the set $f^m$ forms a basis for all linear functionals. Any linear functional $A$ can be written: $$A = a_kf^k$$

3) We use this to show that, under a change of coordinates, we have the required transformation properties for covector components and basis covectors. In general you have $$A(v) = a_kf^k(v^me_m) = a'_kf'^k(v'^me'_m)$$ Then you could take the action of a basis covector on a basis vector: $$\delta^k_m = f'^k(e'_m) = f'^k(\frac{\partial x^r}{\partial x'^m} e_r) = \frac{\partial x^r}{\partial x'^m} f'^k(e_r)$$ And this is satisfied by $$f'^k(e_r) = \frac{\partial x'^k}{\partial x^r}$$ hence $$f'^k = \frac{\partial x'^k}{\partial x^s}f^s$$ Which you can check out.
Does that get you started?

You may want to post these things as homework problems if you can't get the results to come out.

 

[Kaguro]

hence $$f'^k = \frac{\partial x'^k}{\partial x^s}f^s$$ Which you can check out.

This looks like a contravariant transformation law...

So... The components of the functionals in this basis has to be covariant? Just like in the ordinary space, the basis are covariant and the vector components in that basis are contravariant! Here, the exact opposite! The dual basis is contravariant, and so the functional components in that basis are covariant! Since that's the only way the functional itself can be invariant under a change of basis.

 

[PeroK]

Exactly, it all comes from the linear action of the functionals on vectors. The transformation laws follow as in my post above.

 

[Kaguro]

Thank you very much PeroK! You have always been there for me!