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So I am revising Linear Algebra and have a question about finding isomorphisms between dual spaces. I am looking at this problem: Let $V$ be a finite dimensional vector space, over a field $F$ and let $V'$ be its dual space. Show that the map $(f,v) \mapsto f(v)$ from $V'\times V$ to $F$ induces an isomorphism $\theta : V\rightarrow (V')'$.

My first question is, is what is the isomorphism induced? Am I meant to be finding another map from the one given to me, and if so, how do I go about doing this?

If I am testing to see if something is an isomorphism, I think I want to check it is linear (so well-defined), surjective and injective. I can see the the map stated, if this is the map, is linear, but how would I go about proving its surjective and injective.

Any help understanding this problem and topic appreciated.Thanks.

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The isomorphism that you're looking for is known as the evaluation map. It is indeed different from the map you have stated. That map, while surjective, is certainly not injective. It looks instead like it's intended to give a hint towards the correct map.

We are trying instead to find a map $ \theta $ from $ V $ to $ (V')' $. That is, for $ v \in V $, $ \theta (v):V'\to\mathbb{F} $ is a linear functional. Put another way $\theta (v)(f) \in \mathbb{F}$. But now this is looking familiar. If we label the stated map as $\phi:V'\times V:\mathbb{F}$ we can set $\theta(v)(f) := \phi(f,v) = f(v)$.

Now we have to check a few things:

  • $ \theta(v) \in (V')' $
  • $ \theta $ is linear
  • injectivity and surjectivity

I won't go through these in detail but the first follows from the definition of addition and scalar multiplication of linear functions. The second is covered by the definition of linearity of functions itself. The last are just showing that $\theta(v)=\theta(w)$ implies that $v=w$ and that any element of $(V')'$ can be written as $\theta(v)$ for some $v$.

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So, given $v\in V$, we have $\theta(v)\in (V')'$ is given by $\theta(v)(f)=f(v)\in\Bbb F$ for each $f\in V'$. Note that $\theta(v)(\alpha f)=\alpha f(v)=\alpha \theta(v)(f)$. Also, $\theta(v)(f+g)=(f+g)(v)=f(v)+g(v)=\theta(v)(f)+\theta(v)(g)$. Thus $\theta(v)\in(V')'$.

It is clear that $\theta(\alpha v)(f)=f(\alpha v)=\alpha f(v)=\alpha\theta(v)(f)\,\forall f$. Thus $\theta (\alpha v)=\alpha \theta (v)\,,\forall v$. Similarly, $\theta (v+w)=\theta(v)+\theta(w)$ is straight forward. Thus $\theta$ is linear.

Now work through the definition of injectivity. Namely, if $\theta(f)=\theta(g)$, then $f(v)=g(v)\,,\forall v\in V\implies f=g$.

Thus $\theta $ is injective.

Now since $V$ and $(V')'$ have the same dimension, $\theta $ is also surjective.

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Let $F$ be a field, and let $V$ be a finite dimensional vector space over $F$. Write $V'$ for the dual space of $V$ and $V''$ for the dual of the dual space.

Like you say, we have an ($F$-bilinear) map $\phi : V' \times V \rightarrow F$ sending $(f, v)$ to $f(v)$. $\phi$ induces a map $\phi' : V \rightarrow V''$ sending $a$ to the map $\phi'(a)$ sending $f$ to $\phi((f, a))$.

Now to show this is an isomorphism, we can show that it is injective and that the target and domain have the same dimensions. For the first part, take $a$ in $V$ nonzero. We may extend $a$ to a basis $a, a_1, ..., a_n$ of $V$, and let $f$ be the linear map sending $a$ to $1$ and $a_i$ to $0$ for each $i$. Then $f(a) \neq 0$, so $\phi'(a)(f) \neq 0$, so $\phi'(a) \neq 0$.

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