View source for Notes:Dual to dual vector space

{{Requires work|grade=A*|msg=Make into a task, don't forget it!
* Use with: [[The dual space to the dual space of a vector space is canonically isomorphic to the vector space]]}}
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==Notes==
Let {{M|(V,\mathbb{F})}} be a finite dimensional [[vector space]]. Let {{M|V^*}} denote the [[dual vector space]] to {{M|V}}. I claim:
* {{M|V\equiv V^{**} }}{{rLAVEPSW}} where {{M|\equiv}} denote that the spaces are ''[[canonical linear map|canonically]]'' [[linear isomorphism|linearly isomorphic]]
===Plan===
Once you see it, it becomes obvious. I have decided it is simpler to first find a [[map]]:
* {{M|A:V\rightarrow V^{**} }} and show it's an isomorphism from there. What we want is:
** {{MM|A:v\mapsto\left(\begin{array}{c}:V^*\rightarrow\mathbb{F}\\:f^*\mapsto\ (???)\end{array}\right)}} in some way.
* After faffing about with the definitions and getting a "feel" for what was going on, I realised:
** {{MM|A:v\mapsto\left(\begin{array}{c}:V^*\rightarrow\mathbb{F}\\:f^*\mapsto f^*(v)\end{array}\right)}} actually makes some sort of sense
*** As we are associating with each {{M|v}} a function which takes covectors to the field, and it's really simple too. 
===Proof===
Obviously if this is a [[linear map]] it is [[canonical linear map|canonical]]. 
* Suppose for the moment it is. That means {{M|A(V)\subseteq V^{**} }} is a [[vector subspace]]
** Then we can use [[a linear map is injective if and only if the kernel contains only the zero vector]]
** By the [[rank-nullity theorem]] we see {{M|\text{Dim}(V)\eq\text{Dim}(A(V))+0}} which implies {{M|\text{Dim}(A(V))\eq n}}
** By the basis-of-the-dual-space thing we know {{M|V^*}} and {{M|V^{**} }} are both {{M|n}}-dimensional
*** We need to say "{{M|n}}-dimensional subspace is the thing itself" and I'm a bit off at the moment so will deal with this later.


==References==
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