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Prove that for any vector space $V$ the map sending $v$ in $V$ to (evaluation at $v$) $E_v$ in $V^{**}$ such that $E_v(\phi) = \phi(v)$ for $\phi$ in $V^*$ , is injective. Derive from this that if $\dim V < \infty$, its double dual $V^{**}$ is naturally isomorphic to $V$.

Here $V^*$ is the dual space of $V$.

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2 Answers 2

Without knowing where you are stuck, it is difficult to give informative answers. I'll give two hints:

  • If a vector space homomorphism between vector spaces of the same finite dimension is injective, then it is also $---$?
  • To show a map between vector spaces in injective, show that if $v$ maps to zero, it implies $v$ zero too.
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In your question If a vector space homomorphism between finite-dimensional vector spaces is injective, then it is also −−−? you have left out the condition that they have the same dimension. –  Andreas Caranti Feb 1 '13 at 23:49
    
@AndreasCaranti: Thanks for the remark. I was a bit too quick. –  Fredrik Meyer Feb 2 '13 at 14:16

Denote the evaluation map at $v$ by $\bar{v}$. Then the map $\tau v=\bar{v}$ from $V\to V^{\ast\ast}$ is indeed an injective linear transformation. Linearity is easy. Now suppose $v\in\ker(\tau)$. Then $$ \begin{align*} \tau v=0 &\implies \bar{v}=0\\ &\implies \bar{v}(f)=0\quad\forall f\in V^\ast\\ &\implies f(v)=0\quad\forall f\in V^\ast\\ &\implies v=0 \end{align*} $$ since $v\in V$ is zero iff $f(v)=0$ for all linear functionals on $V$. Thus $\ker(\tau)=\{0\}$, so $\tau$ is injective.

To see it is an isomorphism, it is useful to recall the fact that $$ \dim(V)\leq\dim(V^\ast) $$ with equality holding iff $V$ is finite dimensional, so applying this twice you find $$ \dim(V)=\dim(V^\ast)=\dim(V^{\ast\ast}) $$ so $\tau$ is an injective transformation between vector spaces of equal dimension, hence and isomorphism by rank-nullity.

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@Andrew Perhaps, I suppose I just view this is an elementary result with standard proof showing the evaluation identification is a monomorphism, thus necessarily an isomorphism by dimension counting. The proof is hardly unique to Roman. If the notation is similar, it's because I read Roman's book extensively back in the day; I've referenced it elsewhere on MSE. Aside from the string of implications $\implies$, I wouldn't say it's verbatim. –  Ben West Aug 22 at 2:10
    
Fair enough, deleted –  Andrew Aug 22 at 13:01

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