# Dual vector space

## Definition

Given a vector space [ilmath](V,F)[/ilmath] we define the dual or conjugate vector space[1] (which we denote [ilmath]V^*[/ilmath]) as:

• $V^*=\text{Hom}(V,F)$ (recall this the set of all homomorphisms (specifically linear ones) from [ilmath]V[/ilmath] to [ilmath]F[/ilmath])
• That is $V^*=\{f:V\rightarrow F|\ f\text{ is linear}\}$ (which is to say $f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)\ \forall x,y\in V\ \forall \alpha,\beta\in F$)

We usually denote the elements of [ilmath]V^*[/ilmath] with [ilmath]*[/ilmath]s after them, that is something like $f^*\in V^*$

We call the elements of [ilmath]V^*[/ilmath]:

• Covectors
• Dual vectors
• Linear form
• Linear functional (and [ilmath]V^*[/ilmath] the set of linear functionals)[2]

Covectors and Dual vectors are interchangeable and I find myself using both without a second thought.

## Equality of covectors

We say two covectors, $f^*,g^*:V\rightarrow F$ are equal if[1]:

• $\forall v\in V[f^*(v)=g^*(v)]$[note 1] - that is they agree on their domain (as is usual for function equality)

## The zero covector

The zero covector$:V\rightarrow F$ with $v\mapsto 0$[1] {{Todo|Confirm it is written [ilmath]0^*[/ilmath] before writing that here

## Examples

### Dual vectors of [ilmath]\mathbb{R}^2[/ilmath]

Consider (for [ilmath]v\in\mathbb{R}^2[/ilmath]: $f^*(v)=2x$ and $g^*(v)=y-x$ - it is easy to see these are linear and thus are covectors![note 2][1]

### Dual vectors of [ilmath]\mathbb{R}[x]_{\le 2} [/ilmath]

(Recall that $\mathbb{R}[x]_{\le 2}$ denotes all polynomials up to order 2, that is: $\alpha+\beta x+\gamma x^2$ for $\alpha,\beta,\gamma\in\mathbb{R}$ in this case)

Consider [ilmath]p\in\mathbb{R}[x]_{\le 2} [/ilmath] and $f*,g^*:\mathbb{R}[x]_{\le 2}\rightarrow\mathbb{R}$ given by:

• $f^*(p)=\int^{+\infty}_0e^{-x}p(x)dx$ - this is a covector
To see this notice: $f^*(\alpha p+\beta q)=\int^{+\infty}_0e^{-x}[\alpha p(x)+\beta q(x)]dx$$=\alpha\int^{+\infty}_0e^{-x}p(x)dx+\beta\int^{+\infty}_0e^{-x}q(x)dx$
• $g^*(p)=\frac{dp}{dx}\Big|_{x=1}$ (note that: $g^*(\alpha+\beta x+\gamma x^2)=\beta+2\gamma$ - so this isn't just projecting to coefficients) is also a covector[note 2][1]

## Dual basis

Theorem: Given a basis [ilmath]\{e_1,\cdots,e_n\}[/ilmath] of a vector space [ilmath](V,F)[/ilmath] there is a corresponding basis to [ilmath]V^*[/ilmath], [ilmath]\{e_1^*,\cdots,e_n^*\}[/ilmath] where each [ilmath]e^*_i[/ilmath] is the [ilmath]i^\text{th} [/ilmath] coordinate of a vector [ilmath]v\in V[/ilmath] (that is the coefficient of [ilmath]e_i[/ilmath] when [ilmath]v[/ilmath] is expressed as [ilmath]\sum^n_{j=1}v_je_j[/ilmath]). That is to say that [ilmath]e_i^*[/ilmath] projects [ilmath]v[/ilmath] onto it's [ilmath]e_i^\text{th} [/ilmath] coordinate.

TODO: Proof

1. I avoid the short form: $f^*v=f^*(v)$ because the $*$ looks too much like an operator