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Let $V$ be a vector space of dimension $n$ over the field $K$. Let $V^{**}$ be the dual space of $V^{*}$. Show that each elelment $v\in V$ gives rise to an element $\lambda_v$ in $V^{**}$ and that the map $v\to\lambda_v$ gives an isomorphism of $V$ with $V^{**}$. Book Linear Algebra, Serge Lang

Answer given by Omnomnomnom in other thread.

For $\phi \in V^{*}$, Define $\lambda_v(\phi) = \phi(v)$. Because we know that $\dim V = \dim V^* = \dim V^{**}$, it suffices to show that the map $\lambda_{\_}:v \mapsto \lambda_v$ is injective, i.e. that it has a trivial kernel.

Suppose, then, that $v \in \ker \lambda_{\_}$. That is, $\lambda_v = 0$. That is: for all $\phi \in V^*$, we have $\phi(v) = 0$. However, whenever $v$ is non-zero, there exists a $\phi:V \to K$ such that $\phi(v) \neq 0$. We may conclude, then, that $v \in \ker \lambda_{\_} \implies v = 0$. So, $\lambda_{\_}$ has a trivial kernel.

So, $\lambda_{\_}$ is an isomorphism, as desired.

My doubts have changed that is why I opened another thread, the questions are not the same.

Questions:

1) Why is $\lambda_v(\phi) = \phi(v)$? Why is $\lambda_v:\varphi\to\varphi(v)$?

2) Is $V^{**}$ the space of the square linear functionals? How can I think about $ V^{**} $?

3) Is $V^{*}=V^{**}$?

4) If I have a functional like $\langle v_1,w \rangle$ for all w in a vector space. Is the functional that belongs to $V^{**}$, $\langle v_1 \langle v_1,w \rangle$?

Thanks in advance!

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  • $\begingroup$ 1) Is a definition. 2) $V^{**}=\mathcal L(V^*,\Bbb F)$ where $\Bbb F$ is the field of $V$. 3) NO, we have that $V^*=\mathcal L(V,\Bbb F)$. $\endgroup$ – Masacroso Jul 10 '17 at 11:32
  • $\begingroup$ What is a "square linear functional"? $\endgroup$ – Omnomnomnom Jul 10 '17 at 12:06
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    $\begingroup$ About 2-4: The objects of $V^*$ are linear maps on $V$ whose output is an element of $K$. For instance: if $V$ has a scalar product, the map $\phi_w:v \mapsto \langle v,w \rangle$ is an element of $V^*$ since its input $v$ is an element of $V$ and its output $\phi_w(v) = \langle v,w \rangle$ is an element of $K$. The objects of $V^{**}$ are linear maps on $V^*$ whose output is an element of $K$. For instance, the map $\lambda_v : \phi \mapsto \phi(v)$ is an element of $V^{**}$ since its input $\phi$ is an element of $V^*$ and its output $\phi(v)$ is an element of $K$. $\endgroup$ – Omnomnomnom Jul 10 '17 at 12:16
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    $\begingroup$ It is clear that $\phi_w$, as I defined it, is linear because of how scalar products work. You should likewise convince yourself that the map $\lambda_v(\phi)$ is linear, i.e. that for $\alpha,\beta \in K$ and $\phi,\psi \in V^*$, we have $$ \lambda_v(\alpha\phi + \beta \psi) = \alpha\;\lambda_v(\phi) + \beta \;\lambda_v(\psi) $$ $\endgroup$ – Omnomnomnom Jul 10 '17 at 12:19
  • $\begingroup$ @Omnomnomnom Thanks! The thing was I was not expressing my doubts coherently in the thread before. $\endgroup$ – Pedro Gomes Jul 10 '17 at 12:30
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  1. You are looking for an element $ \lambda_v \in V ^{**}$. The elements of $V^{**}$ are functionals $\lambda: V ^* \to \mathbb K$. So $\lambda_v$ takes an element $\phi \in V^*$ and maps it to an element in $\mathbb K$. We also have to consider that $\lambda_v$ should somehow depend on $v$. So using $v$ and $\phi$ we have to construct an element of $\mathbb K$. The obvious way to do that is to set $\lambda_v (\phi) = \phi(v)$.
  2. As mentioned in 1., $V ^ {**} = (V^{*})^*$. Thus elements of $V^{**}$ are functionals on $V^*$.
  3. No, but if the vector space has finite dimension they will be isomorphic (as $V^* \cong V \cong V^{**}$).
  4. Suppose $\phi \in V^*$ with $\phi(w) = \langle v_1, w \rangle$ and $v_2 \in V$. What is $\lambda_{v_2}(\phi)$? By definition $\lambda_{v_2}(\phi) = \phi (v_2) = \langle v_1, v_2 \rangle$.
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