Notes:Dual to dual vector space

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• Use with: The dual space to the dual space of a vector space is canonically isomorphic to the vector space

Notes

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

• $V\equiv V^{**}$ where $\equiv$ denote that the spaces are canonically linearly isomorphic

Plan

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

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

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