# Visualization of the dual space of a vector space

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.

• 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. – Max May 31 '16 at 12:31
• The Wikipedia article may be useful: en.wikipedia.org/wiki/Duality_(mathematics) – Crostul May 31 '16 at 12:33
• You might want to add whether you're asking about finite- or infinite-dimensional spaces; the answers you receive will differ substantially. :) – Andrew D. Hwang May 31 '16 at 12:40
• Look at Schwarz space and the extension of Fourier transform to its dual space. – Paul May 31 '16 at 12:57
• Dual spaces come up organically in the theory of electrical networks (algebraic topology in disguise): voltage distributions are dual to current distributions. – amd May 31 '16 at 18:05

## 1 Answer

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 $V$.

• 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." – David Warren Katz May 31 '16 at 20:00
• 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 $A$. – levap May 31 '16 at 20:25