Difference between revisions of "Dual vector space"

Revision as of 10:25, 12 June 2015 (view source) Alec (Talk | contribs) m ← Older edit		Latest revision as of 05:35, 8 December 2016 (view source) Alec (Talk | contribs) m (Alec moved page Dual space to Dual vector space: There are other duals. One day I'll remove the redirect)
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	We call the elements of {{M|V^*}}:		We call the elements of {{M|V^*}}:
−	* [[Covectors]]	+	* [[Covector|Covectors]]
	* Dual vectors		* Dual vectors
	* Linear form		* Linear form
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	{{End Proof}}		{{End Proof}}
	{{End Theorem}}		{{End Theorem}}
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		+	==See also==
		+	* [[Covector]]
		+	* [[Covector applied to a tensor product]]
	==Notes==		==Notes==

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Latest revision as of 05:35, 8 December 2016

Definition

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

• $$V^*=\text{Hom}(V,F)$$ (recall this the set of all homomorphisms (specifically linear ones) from $V$ to $F$)
• 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 $V^*$ with $*$s after them, that is something like

$$f^*\in V^*$$

We call the elements of $V^*$:

• Covectors
• Dual vectors
• Linear form
• Linear functional (and $V^*$ 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 $0^*$ before writing that here

Examples

Dual vectors of $\mathbb{R}^2$

Consider (for $v\in\mathbb{R}^2$:

$$f^*(v)=2x$$ and $$g^*(v)=y-x$$ - it is easy to see these are linear and thus are covectors!^{[note 2][1]}

Dual vectors of $\mathbb{R}[x]_{\le 2}$

(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 $p\in\mathbb{R}[x]_{\le 2}$ 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

See also

• Covector
• Covector applied to a tensor product

Notes

1. Jump up ↑ I avoid the short form: $$f^*v=f^*(v)$$ because the $$*$$ looks too much like an operator
2. ↑ ^{Jump up to: 2.0 2.1} Example shamelessly ripped

References

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