Notes:Dual to dual vector space

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Let [ilmath](V,\mathbb{F})[/ilmath] be a finite dimensional vector space. Let [ilmath]V^*[/ilmath] denote the dual vector space to [ilmath]V[/ilmath]. I claim:


Once you see it, it becomes obvious. I have decided it is simpler to first find a map:

  • [ilmath]A:V\rightarrow V^{**} [/ilmath] and show it's an isomorphism from there. What we want is:
    • [math]A:v\mapsto\left(\begin{array}{c}:V^*\rightarrow\mathbb{F}\\:f^*\mapsto\ (???)\end{array}\right)[/math] in some way.
  • After faffing about with the definitions and getting a "feel" for what was going on, I realised:
    • [math]A:v\mapsto\left(\begin{array}{c}:V^*\rightarrow\mathbb{F}\\:f^*\mapsto f^*(v)\end{array}\right)[/math] actually makes some sort of sense
      • As we are associating with each [ilmath]v[/ilmath] a function which takes covectors to the field, and it's really simple too.


Obviously if this is a linear map it is canonical.

  • Suppose for the moment it is. That means [ilmath]A(V)\subseteq V^{**} [/ilmath] 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 [ilmath]\text{Dim}(V)\eq\text{Dim}(A(V))+0[/ilmath] which implies [ilmath]\text{Dim}(A(V))\eq n[/ilmath]
    • By the basis-of-the-dual-space thing we know [ilmath]V^*[/ilmath] and [ilmath]V^{**} [/ilmath] are both [ilmath]n[/ilmath]-dimensional
      • We need to say "[ilmath]n[/ilmath]-dimensional subspace is the thing itself" and I'm a bit off at the moment so will deal with this later.


  1. Linear Algebra via Exterior Products - Sergei Winitzki