# Composite functional space

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

I tried to answer it on other thread. I am self-studying and I do not have any kind of solutions.

For those who do not understand my terminology and the way I thought, here it is the proof in which I based my own for this exercise.

Theorem: Let V be a finite dimensional vector space over $K$, with a non-degenerate scalar product. Given a functional $L:V\to K$ there exists a unique element $v\in V$ such that: $L(w)=\langle v,w\rangle$

for all $w\in V$.

Proof. Consider the set of all functionals on $V$ which are of type $L_v$, for some $v\in V$. This set is a subspace of $V*$, because of the zero functional is of this type, and we have the formulas

$L_{v_1}+L_{v_2}=L_{v_1+v_2}\:\:\:\:\:\text{and}\:\:\:\:\:L_{cv}=cL_{v}$

Furthemore, if $\{v_1,…,v_n\}$ is a basis of $V$, then $L_{v_1},…L_{v_n}$ are linearly independent. Proof: If $x_1,…,x_n\in K$ are such that:

$x_1L_{v_1}+…+x_nL_{v_n}=0\\L_{x_1v_1}+…+L_{x_n v_n}=0$

and hence

$L_{x_1v_1+…+x_n v_n}=0$

However, if $v\in V$, and $L_v=0$, then $v=0$ by the definition of non-degeneracy. Hence:

$x_1v_1+…+x_n v_n=0$,

and therefore $x_1=…=x_n=0$, thereby proving our assertion. We conclude that the space of functionals of type $L_v\:(v\in V)$ is a subspace of $V*$, of the same dimension as $V*$, whence equal to $V*$. This proves the theorem.$\blacksquare$ Book: “Linear Algebra” by Serge Lang

Could someone write a complete proof?

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.