Given a vector space $V$ the dual space $V^*$ is the space of all linear operators from $V$ to $\mathbb{C}$. $V^*$ is itself a vector space and I know how to prove $V \cong (V^*)^*$ by using a standard basis for $V$ and the corresponding dual basis for $V^*$. However, I was wondering whether or not it would be possible to prove this equivalence without making reference to a standard basis for $V$. Does anyone know how to do this?

  • $\begingroup$ See here. $\endgroup$ – Dietrich Burde May 21 '15 at 14:46
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    $\begingroup$ I suspect a basis (or finiteness thereof) must show up somewhere as the natural map is an isomorphism to the (algebraic) dual iff $V$ has finite dimension. $\endgroup$ – copper.hat May 21 '15 at 14:55
  • $\begingroup$ @DietrichBurde This isn't quite what I asked. Now, I don't know if it's even possible but what I wanted was possible to do the proof without mentioning a particular basis. This proof shows an isomorphism that doesn't depend on the basis but still uses bases in the argument for the proof. $\endgroup$ – Rioghasarig May 21 '15 at 16:16

If $V$ is finite dimensional, there is a natural isomorphism \begin{align*} V&\longrightarrow V^{**}\\ x&\mapsto x^{**} \end{align*} where $x^{**}: V^*\to \mathbb{F}$ is defined by $x^{**}(f):=f(x)$.

As the dimensions of $V$ and $V^{**}$ are the same, it suffices to show that the above natural map is injective in order to conclude that it is an isomorphism. If $x^{**}=0$, i.e. $f(x)=0$ for all $f\in V^*$, then $x$ must be 0 for otherwise, there exists $f$ such that $f(x)=1$ and $f$ is 0 on some subspace complement to $\text{span}\{x\}$.

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  • $\begingroup$ Can you prove that this is an isomorphism? Particularly that it is injective and surjective. $\endgroup$ – Rioghasarig May 21 '15 at 14:43
  • $\begingroup$ This is trivial: if $x\neq y$ then choose $f\in V^{*}$ s.t. $f(x)\neq f(y)$. Then, $x^{**}(f)=f(x)$ and $y^{**}(f)=f(y)$, so $x^{**}\neq y^{**}$. Surjectivity is even easier. $\endgroup$ – Matematleta May 21 '15 at 17:36
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    $\begingroup$ It is not clever to claim something to be trivial, if you do not even understand the subtlety of the question. The whole point is: If we use $\dim V = \dim V^{**}$, we usually deduce this from $\dim V = \dim V^*$, and to conclude this, we definitely use bases. Even for the fact that any injective map between vector spaces of the same finite dimension is surjective, we use bases. I think copper.hat is alright with his comment. $\endgroup$ – MooS May 22 '15 at 5:59

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