This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.

# Dual space

## Vectors and covectors

What is the relation between $2$, the counting number, and "doubling", the function $f(x)=2\cdot x$?

Linear algebra helps one appreciate this seemingly trivial relation. Indeed, the answer is a linear operator $$D : {\bf R} \rightarrow L({\bf R},{\bf R}),$$ from the reals to the vector space of all linear functions. In fact, it's an isomorphism!

More generally, suppose $V$ is a vector space. Let
$$V^* = \{ \alpha \colon V \rightarrow {\bf R}, \alpha {\rm \hspace{3pt} linear}\}.$$
It's the set of all linear "functionals", also called *covectors*, on $V$. It is called the *dual* of $V$.

An illustration of a vector in $V={\bf R}^2$ and a covector in $V^*$.

Here a vector is just a pair of numbers, while a covector is a correspondence of each *unit* vector with a number. The linearity is visible.

Note. If $V$ is a module over a ring $R$, the dual space is still the set of all linear functionals on $V$: $$V^* = \{ \alpha \colon V \rightarrow R, \alpha {\rm \hspace{3pt} linear}\}.$$ The results below apply equally to finitely generated free modules.

In the above example, it is easy to see a way of building a vector from this covector. Indeed, let's pick the vector $v$ such that

- the direction of $v$ is that of the one that gives the largest value of the covector $w$ (i.e., $2$), and
- the magnitude of $v$ is that value of $w$.

So the result is $v=(2,2)$. Moreover, covector $w$ can be reconstructed from this vector $v$. (Cf. gradient and norm of a linear operator.)

This is how the covector above can be visualized:

It is similar to an oil spill.

## Properties

**Fact 1:** $V^*$ is a vector space.

Just as any set of linear operators between two given vector spaces, in this case $V$ and ${\bf R}$. We define the operations for $\alpha, \beta \in V^*, r \in {\bf R}$: $$(\alpha + \beta)(v) = \alpha(v) + \beta(v), v \in V,$$ $$(r \alpha)(w) = r\alpha(w), w \in V.$$

**Exercise.** Prove it. Start with indicating what $0, -\alpha \in V^*$ are. Refer to theorems of linear algebra, such as the "Subspace Theorem".

Below we assume that $V$ is finite dimensional.

**Fact 2:** Every basis of $V$ corresponds to a *dual basis* of $V^*$, of the same size, built as follows.

Given $\{u_1,\ldots,u_n\}$, a basis of $V$. Define a set $\{u_1^*,\ldots,u_n^*\} \subset V^*$ by setting: $$u_i^*(u_j)=\delta _{ij} ,$$ or $$u_i^*(r_1u_1+\ldots+r_nu_n) = r_i, i = 1,\ldots,k.$$

**Exercise.** Prove that $u_i^* \in V^*$.

**Example.** Dual bases for $V={\bf R}^2$:

**Theorem.** The set $\{u_1^*,\ldots,u_n^*\}$ is linearly independent.

**Proof.** Suppose
$$s_1u_1^* + \ldots + s_nu_n^* = 0$$
for some $r_1,\ldots,r_k \in {\bf R}$. This means that
$$s_1u_1^*(u)+\ldots+s_nu_n^*(u)=0 \hspace{7pt} (1)$$
for all $u \in V$.

We choose $u=u_i, i=1,\ldots,n$ here and use $u_j^*(u_i)=\delta_{ij}$. Then we can rewrite (1) with $u=u_i$ for each $i=1,\ldots,n$ as: $$s_i=0.$$ Therefore $\{u_1^*,\ldots,u_n^*\}$ is linearly independent. $\blacksquare$

**Theorem.** $\{u_1^*,\ldots,u_n^*\}$ spans $V^*$.

**Proof.** Given $u^* \in V^*$, let $r_i = u^*(u_i) \in {\bf R},i=1,...,n$. Now define
$$v^* = r_1u_1^* + \ldots + r_nu_n^*.$$
Consider
$$v^*(u_i) = r_1u_1^*(u_i) + \ldots + r_nu_n^*(u_i) = r_i.$$
So $u^*$ and $v^*$ match on the elements of the basis of $V$. Thus $u^*=v^*$. $\blacksquare$

**Conclusion 1:**
$$\dim V^* = \dim V = n.$$

So by the Classification Theorem of Vector Spaces, we have

**Conclusion 2:**
$$V^* \simeq V.$$

A two-line version of the proof: $V$ with basis of $n$ elements $\simeq {\bf R}^n$. Then $V^* \simeq {\bf M}(1,n)$. But $\dim {\bf M}(1,n)=n$, etc.

Even though *a space is isomorphic to its dual*, their behavior is not "aligned" (with respect to linear operators), as we show below. In fact, the isomorphism is dependent on the choice of basis.

**Note 1: ** The relation between a vector space and its dual can be revealed by looking at vectors as column-vectors (as always) and covectors as row-vectors:
$$V = \left\{ x=\left[
\begin{array}{}
x_1 \\
\vdots \\
x_n
\end{array}
\right]
\right\}, V^* = \{y=[y_1,\ldots,y_n]\}.$$
This way we can multiply the two as matrices:
$$xy=[y_1,\ldots,y_n] \left[
\begin{array}{}
x_1 \\
\vdots \\
x_n
\end{array}
\right] = [y_1,\ldots,y_n][x_1,\ldots,x_n]^T =x_1y_1+...+x_ny_n.$$
The result is their dot product which can also be understood as a linear operator $y\in V^*$ acting on $x\in V$.

**Note 2: **
$$\dim \mathcal{L}(V,U) = \dim V \cdot \dim U,$$
if the spaces are finite dimensional.

**Exercise.** Find and picture the duals of the vector and the covectors depicted in the first section.

**Exercise.** Find the dual of ${\bf R}^2$ for two different choices of basis.

## Operators and naturality

That's not all.

We need to understand what happens to a linear operator
$$A:V \rightarrow W$$
under duality. The answer is uncomplicated but also unexpected, the corresponding *dual operator* goes in the opposite direction:
$$A^*:W^* \rightarrow V^*!$$
And not just because this is the way we chose to define it:
$$A^*(f)=f \circ A.$$
A dual counterpart of $A$ really can't be defined in any other way. Consider:
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
& V & \ra{A} & W \\
& _{g \in V^*} & \searrow & \da{f \in W^*} \\
& & & {\bf R}
\end{array}
$$
If this is understood a commutative diagram, the relation between $f$ and $g$ is given by the equation above. So, we get $g$ from $f$ by $g=fA$, but not vice versa.

Note. The diagram also suggests that the reversal of the arrows has nothing to do with linearity. The issue is "functorial".

**Theorem.** For finite dimensional $V,W$, the matrix of $A^*$ is the transpose of that of $A$:
$$A^*=A^T.$$

**Proof.** **Exercise.** $\blacksquare$

The composition is preserved but in reverse:

**Theorem.** $(AB)^*=B^*A^*.$

**Proof.**
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
& V & \ra{A} & W & \ra{B} & U\\
& _{g \in V^*} & \searrow & \da{f \in W^*} & \swarrow & _{h \in U^*} \\
& & & {\bf R}
\end{array}
$$

Finish (**Exercise.**) $\blacksquare$

As you see, the dual $A^*$ behaves very much like but is not to be confused with the inverse $A^{-1}$. Of course, the former is much simpler!

The isomorphism $D$ between $V$ and $V^*$ is very straight-forward: $$D_V(u_i)=u_i^*,$$ where $\{u_i\}$ is a basis of $V$ and $\{u^*_i\}$ is its dual. However, because of the reversed arrows the isomorphism isn't "natural". Indeed, if $f:V \rightarrow U$ is linear, the diagram below does not commute, in general: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & V & \ra{f} & U \\ & \da{D_V} & \ne & \da{D_U} \\ & V^* & \la{f^*} & U^*\\ \end{array} $$

**Exercise.** Why not?

However, the isomorphism with the "second dual" $V^{**}=(V^*)^*$ *is* natural:
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
\newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!}
\newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
& V & \ra{f} & U \\
& \da{D_V} & & \da{D_U} \\
& V^* & & U^*\\
& \da{D_{V^*}} & & \da{D_{U^*}} \\
& V^{**} & \ra{f^{**}} & U^{**}\\
\end{array}
$$

**Exercise.** Prove that. Demonstrate that it is independent from the choice of basis.

When the dot product above is replaced with a particular choice of inner product, we have an identical effect. A general term related to this is adjoint operator.

A major topological application of the idea is in chains vs cochains.

## Change of basis

The reversal of arrows also reveals that a change of basis of $V$ affects differently the coordinate representation of vectors and covectors.