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Dual of a Vector Space
Definition Linear Functional on a Vector Space Let $V$ be a vector space for a field $F$, then the linear functional of $V$ will be a function $T: V \rightarrow F$, which satisfies the following properties: 1) $T(v_1 + v_2)$ = $T(v_1)$ + $T(v_2)$, $\forall v_1, v_2 \in V$ 2) $T(\alpha v)$ = $\alpha T(v)$, $\forall v \in V$, $\forall \alpha \in F$. Dual Vector Space The dual vector space to a vector space $V$ is the vector space of linear functionals $T: V \rightarrow F$, denoted as $V^$, where the dimension of the vector space $V$ = dimension of the dual vector space $V^$. 1
Motivation In Mathematics, it is very much possible to translate concepts, structures, or statements into other concepts, mathematical structures, or statements. This translation of one mathematical structure to another is termed as 'duality', where the two structures are the duals of each other. For example, a cube and octahedron are very closely related. If we choose the centers of all the six faces of a cube, then those will be the vertices of an octahedron. And, then we can say that the octahedron is the 'dual' of the cube. Interestingly enough, the converse i.e., the centers of the eight triangular faces of the octahedron will be the vertices of a cube and therefore, a cube is the dual of an octahedron. It can also be easily observed that the number of edges of two duals is the same, and the number of vertices of one dual is the number of faces of the other dual. So, the principle of duality can be considered to be one of the illuminating way of constructing a dual of any regular polyhedra. 2
3
Figure 1: An Octahedron is the dual of a Cube
Bird's Eye View The dual of an object is of the same kind as the object itself. In the previous example, dual of the cube is an octahedron and both of them are polyhedrons. Similarly, if we consider a set S and let X be its subset such that X $\subseteq$ S, then the dual of subset X will also be a subset of the set S. The dual of subset X can be obtained by taking its complement X^c, and the elements of X^c are those elements that are not in X. So, X^c will again be a subset of the set S. Hence, the dual of the subset X is its complement X^c. Conversely, the dual of the subset X^c will be the subset X, since (X^c)^c= X and hence, they are duals of each other. 4
5
Figure 2: The dual of a subset is its complement.
Duality is a property of algebraic structures, holds that the two concepts or operations are interchangeable. The idea of duality also arises in the concept of Linear Algebra, in which every finite-dimensional vector space can be associated with its dual which is called a dual vector space. 7
Context of the Definition Let $V$ be a vector space over a field $F$, then the dual of the vector space $V^*$ is a vector space that contains all the functions $T: V \rightarrow F$, where $T$ is called a linear functional. A Linear Functional is a function that accepts a vector $v\in V$ as input and gives out an element of the field $F$ over which the vector space $V$ is defined. A Linear Functional satisfies the following properties: 1) $T(u + v)$ = $T(u)$ + $T(v)$, $\forall u, v \in V$ 2) $T(\alpha v)$ = $\alpha T(v)$, $\forall v \in V$, $\forall \alpha \in F$ 8
9
Figure 3: Linear Functional on Vector Space
Let's assume the field to be $\mathbb{R}$, meaning that the linear functional gives out a real number as an output. If you take all the possible ways in which the linear functional will take vectors as input and give a real number as an output, you obtain the dual vector space $V^$. Consider $V$ = $\mathbb{R}^2$ and $T: \mathbb{R}^2 \rightarrow \mathbb{R}$, then $T(x, y)$ = $ax + by$ will be an element of the dual vector space $V^$ $\forall a, b \in \mathbb{R}$ since, $ax + by$ = $k$ where $k$ is a real number. Take $v \in \mathbb{R}^2$ such that $v$ = $(2,3)$ $\implies$ $T(2,3) = 2a + 3b$ $\in \mathbb{R}$ Hence, all such elements $ax + by$ forms a dual vector space of $\mathbb{R}^2$. If $V$ = $\mathbb{R}^3$ and $T: \mathbb{R}^3 \rightarrow \mathbb{R}$, then $T(x, y, z)$ = $ax + by +cz$ will be an element of the dual vector space $V^$. Similarly, any real vector space $\mathbb{R}^n$ has a dual space whose elements are all such linear functionals that gives out a real number. Another similar example for a dual space can be all such linear functionals $T: P_n \rightarrow \mathbb{R}$, where the vector space $P_n$ is a set of polynomials with degree n. $T(P)$ = $P(2)$ $\in V^$ Concretely, $T(2x^2 + 3x + 4)$ = $2\cdot 2$ + $3\cdot 2$ + $4 = 14$ 10
Dual Basis of a Vector Space Given a basis $B = (v_1, v_2, . . . . . . . , v_n)$ of the vector space $V$, there exists a dual basis $B^$ = (${T_1}, {T_2}, . . . . . , {T_n}$) of the dual space $V^$, where $B^$ contains the linear functionals and ${T_i}(v_j)$ = $\delta_{ij}$ and $\delta_{ij}$ is called Kronecker's delta*. If i = j, then $\delta_{ij}$ =1 and if i $\neq$ j, then $\delta_{ij}$ = 0. 13
14
Figure 4: Basis of a vector space and its dual.
Consider a vector space $V = \mathbb{R}^2$ whose basis is $B = \{(2,1) ,(3,1)\}$, then to find the dual basis $B^$ the following steps are to be followed: Step 1: Let $B^$ = $\{{T_1}, {T_2}\}$, by definition ${T_1}(v_1) = 1$ $\implies$ ${T_1}(2, 1) = 1$ and, ${T_1}(v_2) = 0$ $\implies {T_1}(3, 1) = 0$ Step 2: ${T_1}(2, 1) = 1$ $\implies {T_1}[ 2(1, 0) + 1(0, 1)]$ = 1 $\implies$ $2 {T_1}(1, 0) + 1{T_1}(0, 1) = 1$ ---- (i) ${T_1}(3, 1)$ = 0 $\implies {T_1}[3(1, 0) + 1(0, 1)]$ = 1 $\implies$ $3 {T_1}(1, 0) + 1{T_1}(0, 1) = 0$ ---- (ii) Step 3: On solving the equations (i) and (ii), we get; ${T_1}(1, 0) = -1$ and ${T_1}(0, 1) = 3$ Step 4: ${T_1}(x, y)$ = $x{T_1}(1, 0) + y{T_1}(0, 1)$ $\implies$ $x(-1) + y(3)$ = $ -x + 3y$ $\implies {T_1}(x, y) = -x + 3y$ Step 5: Similarly, by using ${T_2}(2, 1)$ = 0 and ${T_2}(3, 1)$ = 1, we get ${T_2}(1, 0)$ = 1 and ${T_2}(0, 1) = -2$ $\implies {T_2}(x, y)$ = $x - 2y$ $\implies$ $B^$ = $\{{T_1}, {T_2}\}$ = $\{ -x + 3y, x - 2y\}$ Therefore, the basis of the dual vector space $V^$ is given by the equation of two straight lines. 15
16
Figure 5: Geometrical Illustration of the example above.
Applications Linear Programming is considered to be a very good application of duality in which there the relationship between a given linear program and the other related linear program is developed. The concept of duality in vector spaces also has applications in many branches of mathematics such as tensor analysis with the vector spaces of finite dimensions. When this idea is applied to the function vector spaces, then it is used to describe distributions, measures, and Hilbert spaces. The concept of dual space is also very helpful in functional analysis. History It was in 1907 when the definition of a continuous linear operation on the space $L^{2}([a,b])$ was given by Riesz. this is, in slightly modernized notation, an operation which for any $f\in L^{2}$ gives a number $U(f)$ such that $U$ is a linear map and such that whenever $f_n \rightarrow f$ in $L^{2}$ we have $U(f_n) \rightarrow U(f)$. Then he shows that for each continuous linear operation $U$ there exists a function $k$ such that $U(f) = \int_{a}^{b} f(x) k(x) dx$ for all $f\in L^{2}$. Before being developed infinite-dimensional vector spaces, the theory was developed in function spaces. There were many examples of functionals on the form $f \rightarrow \int f(x) g(x) dx$ well known at the time (cf. potential theory, or Cauchy's integral theorem), so representation theorems would look very nice. 20 years later, the definition of Hilbert spaces were given and in 1935, Riesz spoke about of this theorem again, but in an entirely modern notation: "For every continuous linear function $\ell (f)$ there is a unique representing element gg such that $\ell (f) = (f, g)$", where $(\cdot,\cdot)$ is the inner product on the Hilbert space. The theory for Banach spaces progressed similarly. First linear functionals on $C[a, b]$ and $L^{p} [a, b]$ were studied and representation theorems were found. It was in 1927 when the dual of a normed linear space was introduced by Hahn. By introducing methods of transfinite induction, Hahn obtained the existence of general continuous linear functionals on a normed linear space and introduced the idea of regular normed spaces. It makes sense that the existence of general classes of continuous linear functionals would lead to a proper notion of a dual space. After two years, Banach proved the same theorem with the same proof. In the 1920s a more abstract theory was developed, and in Banach's monograph from 1932 the subject is fully mature with "spaces of type (B)" [Banach spaces] and "conjugate spaces" [dual spaces]. Banach had cited Frigyes Riesz multiple times in his monograph. 17
Pause and Ponder 1) Is it possible for a sphere to have a dual? 2) Can you think of a linear functional that will map all the real matrices to a real number? 18
Reference [1] https://www.cis.upenn.edu/~cis515/cis515-18-sl6.pdf [2] https://mathworld.wolfram.com/DualVectorSpace.html [3] https://ekamperi.github.io/mathematics/2019/11/17/dual-spaces-and-dual-vectors.html#definition [4] http://www.math.brown.edu/~banchoff/Beyond3d/chapter5/section03.html#:~:text=If%20we%20choose%20the%20centers,the%20dual%20of%20the%20octahedron. [5] https://sites.math.northwestern.edu/~scanez/courses/334/notes/dual-spaces.pdf [6] Per Manne (https://math.stackexchange.com/users/33572/per-manne), History of Dual Spaces and Linear Functionals, URL (version: 2012-07-03): https://math.stackexchange.com/q/166118 [7] DisintegratingByParts (https://math.stackexchange.com/users/112478/disintegratingbyparts), Who was the first to use dual space?, URL (version: 2014-03-05): https://math.stackexchange.com/q/700572 19
Further Reading [1] https://en.wikipedia.org/wiki/Duality_(mathematics) [2] https://www.math.upenn.edu/~aaronsil/Math312Spring2013/DualSpaces.pdf 20

Dual of a Vector Space

Definition

Linear Functional on a Vector Space

Let $V$ be a vector space for a field $F$, then the linear functional of $V$ will be a function $T: V \rightarrow F$, which satisfies the following properties:
1) $T(v_1 + v_2)$ = $T(v_1)$ + $T(v_2)$, $\forall v_1, v_2 \in V$
2) $T(\alpha v)$ = $\alpha T(v)$, $\forall v \in V$, $\forall \alpha \in F$.

Dual Vector Space

The dual vector space to a vector space $V$ is the vector space of linear functionals $T: V \rightarrow F$, denoted as $V^*$,
where the dimension of the vector space $V$ = dimension of the dual vector space $V^*$.

Motivation

In Mathematics, it is very much possible to translate concepts, structures, or statements into other concepts, mathematical structures, or statements. This translation of one mathematical structure to another is termed as 'duality', where the two structures are the duals of each other. For example, a cube and octahedron are very closely related. If we choose the centers of all the six faces of a cube, then those will be the vertices of an octahedron. And, then we can say that the octahedron is the 'dual' of the cube. Interestingly enough, the converse i.e., the centers of the eight triangular faces of the octahedron will be the vertices of a cube and therefore, a cube is the dual of an octahedron. It can also be easily observed that the number of edges of two duals is the same, and the number of vertices of one dual is the number of faces of the other dual. So, the principle of duality can be considered to be one of the illuminating way of constructing a dual of any regular polyhedra.

Figure 1: An Octahedron is the dual of a Cube

Bird's Eye View

The dual of an object is of the same kind as the object itself. In the previous example, dual of the cube is an octahedron and both of them are polyhedrons. Similarly, if we consider a set S and let X be its subset such that X $\subseteq$ S, then the dual of subset X will also be a subset of the set S.
The dual of subset X can be obtained by taking its complement X^c, and the elements of X^c are those elements that are not in X.
So, X^c will again be a subset of the set S. Hence, the dual of the subset X is its complement X^c.

Conversely, the dual of the subset X^c will be the subset X, since (X^c)^c= X and hence, they are duals of each other.

Figure 2: The dual of a subset is its complement.

Duality is a property of algebraic structures, holds that the two concepts or operations are interchangeable.
The idea of duality also arises in the concept of Linear Algebra, in which every finite-dimensional vector space can be associated with its dual which is called a dual vector space.

Context of the Definition

Let $V$ be a vector space over a field $F$, then the dual of the vector space $V^*$ is a vector space that contains all the functions $T: V \rightarrow F$, where $T$ is called a linear functional. A Linear Functional is a function that accepts a vector $v\in V$ as input and gives out an element of the field $F$ over which the vector space $V$ is defined.
A Linear Functional satisfies the following properties:

1) $T(u + v)$ = $T(u)$ + $T(v)$, $\forall u, v \in V$
2) $T(\alpha v)$ = $\alpha T(v)$, $\forall v \in V$, $\forall \alpha \in F$

Figure 3: Linear Functional on Vector Space

Let's assume the field to be $\mathbb{R}$, meaning that the linear functional gives out a real number as an output. If you take all the possible ways in which the linear functional will take vectors as input and give a real number as an output, you obtain the dual vector space $V^*$.

Consider $V$ = $\mathbb{R}^2$ and $T: \mathbb{R}^2 \rightarrow \mathbb{R}$, then $T(x, y)$ = $ax + by$ will be an element of the dual vector space $V^*$ $\forall a, b \in \mathbb{R}$ since, $ax + by$ = $k$ where $k$ is a real number.
Take $v \in \mathbb{R}^2$ such that $v$ = $(2,3)$
$\implies$ $T(2,3) = 2a + 3b$ $\in \mathbb{R}$
Hence, all such elements $ax + by$ forms a dual vector space of $\mathbb{R}^2$.

If $V$ = $\mathbb{R}^3$ and $T: \mathbb{R}^3 \rightarrow \mathbb{R}$, then $T(x, y, z)$ = $ax + by +cz$ will be an element of the dual vector space $V^*$.

Similarly, any real vector space $\mathbb{R}^n$ has a dual space whose elements are all such linear functionals that gives out a real number.

Another similar example for a dual space can be all such linear functionals $T: P_n \rightarrow \mathbb{R}$, where the vector space $P_n$ is a set of polynomials with degree n.
$T(P)$ = $P(2)$ $\in V^*$
Concretely, $T(2x^2 + 3x + 4)$ = $2\cdot 2$ + $3\cdot 2$ + $4 = 14$

Dual Basis of a Vector Space

Given a basis $B = (v_1, v_2, . . . . . . . , v_n)$ of the vector space $V$, there exists a dual basis $B^*$ = (${T_1}, {T_2}, . . . . . , {T_n}$) of the dual space $V^*$,
where $B^*$ contains the linear functionals and ${T_i}(v_j)$ = $\delta_{ij}$ and $\delta_{ij}$ is called Kronecker's delta.

If i = j, then $\delta_{ij}$ =1 and if i $\neq$ j, then $\delta_{ij}$ = 0.

Figure 4: Basis of a vector space and its dual.

Consider a vector space $V = \mathbb{R}^2$ whose basis is $B = \{(2,1) ,(3,1)\}$, then to find the dual basis $B^*$ the following
steps are to be followed:

Step 1: Let $B^*$ = $\{{T_1}, {T_2}\}$, by definition ${T_1}(v_1) = 1$ $\implies$ ${T_1}(2, 1) = 1$ and,
${T_1}(v_2) = 0$ $\implies {T_1}(3, 1) = 0$

Step 2: ${T_1}(2, 1) = 1$ $\implies {T_1}[ 2(1, 0) + 1(0, 1)]$ = 1 $\implies$ $2 {T_1}(1, 0) + 1{T_1}(0, 1) = 1$ ---- (i)
${T_1}(3, 1)$ = 0 $\implies {T_1}[3(1, 0) + 1(0, 1)]$ = 1 $\implies$ $3 {T_1}(1, 0) + 1{T_1}(0, 1) = 0$ ---- (ii)

Step 3: On solving the equations (i) and (ii), we get;
${T_1}(1, 0) = -1$ and ${T_1}(0, 1) = 3$

Step 4: ${T_1}(x, y)$ = $x{T_1}(1, 0) + y{T_1}(0, 1)$ $\implies$ $x(-1) + y(3)$ = $ -x + 3y$ $\implies {T_1}(x, y) = -x + 3y$

Step 5: Similarly, by using ${T_2}(2, 1)$ = 0 and ${T_2}(3, 1)$ = 1, we get ${T_2}(1, 0)$ = 1 and ${T_2}(0, 1) = -2$
$\implies {T_2}(x, y)$ = $x - 2y$
$\implies$ $B^*$ = $\{{T_1}, {T_2}\}$ = $\{ -x + 3y, x - 2y\}$

Therefore, the basis of the dual vector space $V^*$ is given by the equation of two straight lines.

Figure 5: Geometrical Illustration of the example above.

Applications

Linear Programming is considered to be a very good application of duality in which there the relationship between a given linear program and the other related linear program is developed.
The concept of duality in vector spaces also has applications in many branches of mathematics such as tensor analysis with the vector spaces of finite dimensions.
When this idea is applied to the function vector spaces, then it is used to describe distributions, measures, and Hilbert spaces.
The concept of dual space is also very helpful in functional analysis.

History

It was in 1907 when the definition of a continuous linear operation on the space $L^{2}([a,b])$ was given by Riesz. this is, in slightly modernized notation, an operation which for any $f\in L^{2}$ gives a number $U(f)$ such that $U$ is a linear map and such that whenever $f_n \rightarrow f$ in $L^{2}$ we have $U(f_n) \rightarrow U(f)$. Then he shows that for each continuous linear operation $U$ there exists a function $k$ such that $U(f) = \int_{a}^{b} f(x) k(x) dx$ for all $f\in L^{2}$.
Before being developed infinite-dimensional vector spaces, the theory was developed in function spaces. There were many examples of functionals on the form $f \rightarrow \int f(x) g(x) dx$ well known at the time (cf. potential theory, or Cauchy's integral theorem), so representation theorems would look very nice.
20 years later, the definition of Hilbert spaces were given and in 1935, Riesz spoke about of this theorem again, but in an entirely modern notation: "For every continuous linear function $\ell (f)$ there is a unique representing element gg such that $\ell (f) = (f, g)$", where $(\cdot,\cdot)$ is the inner product on the Hilbert space. The theory for Banach spaces progressed similarly. First linear functionals on $C[a, b]$ and $L^{p} [a, b]$ were studied and representation theorems were found. It was in 1927 when the dual of a normed linear space was introduced by Hahn. By introducing methods of transfinite induction, Hahn obtained the existence of general continuous linear functionals on a normed linear space and introduced the idea of regular normed spaces. It makes sense that the existence of general classes of continuous linear functionals would lead to a proper notion of a dual space. After two years, Banach proved the same theorem with the same proof.
In the 1920s a more abstract theory was developed, and in Banach's monograph from 1932 the subject is fully mature with "spaces of type (B)" [Banach spaces] and "conjugate spaces" [dual spaces]. Banach had cited Frigyes Riesz multiple times in his monograph.

Pause and Ponder

1) Is it possible for a sphere to have a dual?

2) Can you think of a linear functional that will map all the real matrices to a real number?

Reference

[1] https://www.cis.upenn.edu/~cis515/cis515-18-sl6.pdf

[2] https://mathworld.wolfram.com/DualVectorSpace.html

[3] https://ekamperi.github.io/mathematics/2019/11/17/dual-spaces-and-dual-vectors.html#definition

[4] http://www.math.brown.edu/~banchoff/Beyond3d/chapter5/section03.html#:~:text=If%20we%20choose%20the%20centers,the%20dual%20of%20the%20octahedron.

[5] https://sites.math.northwestern.edu/~scanez/courses/334/notes/dual-spaces.pdf

[6] Per Manne (https://math.stackexchange.com/users/33572/per-manne), History of Dual Spaces and Linear Functionals, URL (version: 2012-07-03): https://math.stackexchange.com/q/166118

[7] DisintegratingByParts (https://math.stackexchange.com/users/112478/disintegratingbyparts), Who was the first to use dual space?, URL (version: 2014-03-05): https://math.stackexchange.com/q/700572