I am wondering what the motivation was for defining a dual space of a vector space, and how to visualize the dual space. I'm asking since it doesn't seem to me to be intuitive to deal with such a space.

In particular, I'm looking for questions where it would be natural to consider the space of linear functionals in order to answer these questions.

  • $\begingroup$ some weak formulations of partial differential equations for instance (i would suggest you to google distributions, sobolew-space, delta-distribution, elementary solution, convolution and PDEs) To have a difference between vector space and its dual you need infinite dimensions. $\endgroup$ – Max May 31 '16 at 12:31
  • $\begingroup$ The Wikipedia article may be useful: en.wikipedia.org/wiki/Duality_(mathematics) $\endgroup$ – Crostul May 31 '16 at 12:33
  • $\begingroup$ You might want to add whether you're asking about finite- or infinite-dimensional spaces; the answers you receive will differ substantially. :) $\endgroup$ – Andrew D. Hwang May 31 '16 at 12:40
  • $\begingroup$ Look at Schwarz space and the extension of Fourier transform to its dual space. $\endgroup$ – Paul May 31 '16 at 12:57
  • $\begingroup$ Dual spaces come up organically in the theory of electrical networks (algebraic topology in disguise): voltage distributions are dual to current distributions. $\endgroup$ – amd May 31 '16 at 18:05

What is a "linear equation" in $n$ variables $X_1, \dots, X_n$ over the field $\mathbb{F}$? In some sense, it is a formal expression of the form $a_1 X_1 + \dots + a_n X_n$ where $a_i \in \mathbb{F}$ are scalars and the $X_i$ are "placeholders" in which you can plug in values and obtain a result in $\mathbb{F}$. This is precisely an element $\varphi$ of the dual space $(\mathbb{F}^n)^{*}$ (here, $\varphi(X_1, \dots, X_n) = a_1X_1 + \dots + a_n X_n$). Thus, you can think of $(\mathbb{F}^n)^{*}$ as the space of linear equations. It is somewhat surprising in the beginning that the space of linear equations itself has a linear structure and can be considered as a vector space. You can add two linear equations and multiply a linear equation by a scalar.

Let me show you that if you are familiar with the basic techniques of linear algebra, you already used the notion of a dual space in disguise many times:

  1. When you are performing Gaussian elimination to solve a system

$$ a_{11} X_1 + \dots + a_{1n} X_n = \varphi_1(X_1, \dots, X_n) = 0, \\ \vdots \\ a_{k1} X_1 + \dots + a_{kn} X_n = \varphi_k(X_1, \dots, X_n) = 0 $$ you are applying row operations to the corresponding matrix. The operations you perform on the equations correspond precisely to the operations you perform on the linear functionals $\varphi_i$ in $(\mathbb{F}^{n})^{*}$. They result in replacing the vectors $\varphi_1, \dots, \varphi_k$ in $(\mathbb{F}^n)^{*}$ with linear combinations in such a way that the span in $(\mathbb{F}^n)^{*}$ doesn't change. If we set $W = \operatorname{span} (\varphi_1, \dots, \varphi_k)$ then solving the linear system above corresponds to finding (a basis for) the annihilator of $W$ (after a certain identification). Row operations don't change $W$ and so don't change the annihilator.

  1. If you are given a subspace $V \subseteq \mathbb{F}^n$ described as a span of vectors and you want to describe it as a solution of a system of equations, you are trying to find a basis $\varphi_1, \dots, \varphi_k$ for the subspace of linear equations that $V$ satisfy. This is precisely finding a basis for the annihilator of $V$. For example, if $V = \mathrm{span} ((1,1,1),(1,1,-1))$ then $V$ is a two-dimensional plane defined for example by the equation $X - Y = 0$ but it is also defined by the equation $2X - 2Y = 0$. The "equation" $(X,Y,Z) \mapsto X - Y$ is a basis for the space of equations that vanish on $V$.
  2. Given a matrix $A \in M_{n \times m}(\mathbb{F})$, you know that the column space of $A$ is precisely the image of the associated linear map $T_A \colon \mathbb{F}^m \rightarrow \mathbb{F}^n$. You might have wondered what is the meaning of the row space of $A$ in terms of the linear map $T_A$. The row space of $A$ is precisely the set of equations $\ker(T_A)$ satisfies

Finally, when working with an abstract vector space $V$ you don't have any preferred coordinates so you can't define "a linear equation" on $V$ as above - the correct definition that is consistent with what I described is precisely that of a linear functional on $V$ the space $V^{*}$ is the space of linear equations on $V$.

| cite | improve this answer | |
  • $\begingroup$ Ok, thanks for your response. the only part I don't follow is that line "The row space of A is precisely the set of equations ker⁡(TA) satisfies." $\endgroup$ – David Warren Katz May 31 '16 at 20:00
  • $\begingroup$ The map $T_A$ is the linear map defined using left multiplication by the matrix $A$. A vector $(x_1, \dots, x_m)$ is in $\ker(T_A)$ if and only if it satisfies the equations $a_{11} X_1 + \dots + a_{1m} X_m = 0, \dots, a_{n1} X_1 + \dots + a_{nm} X_m = 0$. If $(x_1, \dots, x_m)$ is satisfied by a finite set of equations, it it also satisfied (belongs to the solution space) of their span which is (identified) with the row space of $A$. $\endgroup$ – levap May 31 '16 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.