# Dual vector space

## Contents

## 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:

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

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

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, [math]f^*,g^*:V\rightarrow F[/math] are equal if^{[1]}:

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

## The zero covector

The zero covector[math]:V\rightarrow F[/math] with [math]v\mapsto 0[/math]^{[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]: [math]f^*(v)=2x[/math] and [math]g^*(v)=y-x[/math] - 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 [math]\mathbb{R}[x]_{\le 2}[/math] denotes all polynomials up to order 2, that is: [math]\alpha+\beta x+\gamma x^2[/math] for [math]\alpha,\beta,\gamma\in\mathbb{R}[/math] in this case)

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

- [math]f^*(p)=\int^{+\infty}_0e^{-x}p(x)dx[/math] - this is a covector
- To see this notice: [math]f^*(\alpha p+\beta q)=\int^{+\infty}_0e^{-x}[\alpha p(x)+\beta q(x)]dx[/math][math]=\alpha\int^{+\infty}_0e^{-x}p(x)dx+\beta\int^{+\infty}_0e^{-x}q(x)dx[/math]

- [math]g^*(p)=\frac{dp}{dx}\Big|_{x=1}[/math] (note that: [math]g^*(\alpha+\beta x+\gamma x^2)=\beta+2\gamma[/math] - 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

## See also

## Notes

- ↑ I avoid the short form: [math]f^*v=f^*(v)[/math] because the [math]*[/math] looks too much like an operator
- ↑
^{2.0}^{2.1}Example shamelessly ripped

## References

- ↑
^{1.0}^{1.1}^{1.2}^{1.3}^{1.4}Linear Algebra via Exterior Products - Sergei Winitzki - ↑ Introduction to Smooth Manifolds - John M Lee - I THINK! CHECK THIS - CERTAINLY have seen it somewhere though