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
Theorem: The map $v\to L_v$ of $V$ into $V^{*}$ is an isomorphism.I define $L_v(w)=\langle v,w \rangle$, as the functionals of $V^{*}$
Proof:We know that $\dim V=\dim V^{*}$ and by non-degeneracy of the inner product the kernel of the map is 0.
I tried to see the map from $V\to V^{**}$ as $v\to L_{L_v}=\lambda_v$. By the property of non-degenerancy we have the kernel of the map $v\to L_{L_v}=\lambda_v$ must be zero. $L_v(w)=0$ so it is $L_{L_v}(w)=0$, for $w\in V$.If we consider $v_1,v_2…v_n$ as a basis that generate $V$ we have L_{v_1}…L_{v_n} as the basis of $V^{*}$. We can apply the linearity of the linear functionals and we get $L_{a_1L_{v_1}…a_nL_{v_n}}=a_1L_{L_{v_1}}+…a_nL_{L_{v_n}}=a_1\lambda_{v_1}+…a_2\lambda{v_n}$. If we consider $a_1v_1+…a_nv_n=0$ when $a_1=…a_n=0$.
$L_{a_1L_{v_1}…a_nL_{v_n}}=a_1L_{L_{v_1}}+…a_nL_{L_{v_n}}=a_1\lambda_{v_1}+…a_2\lambda{v_n}=0$ which means $\dim V=\dim V^{**}$. Then However I am not seeing how I am going end this proof. I am self-studying.
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
Questions:
Could someone prove this?
Thanks in advance!
Best Answer
I don't know how you define $L_v$: it appears that you're assuming the existence of an inner product, but this is not the case. For instance, nondegenerate bilinear forms may not exist over some field $K$, whereas the isomorphism in the exercise can be defined over any field and any finite dimensional vector space over it.
The map Lang has in mind is $$ \lambda\colon V\to V^{**},\qquad v\mapsto\lambda_v $$ where, for $\varphi\in V^{*}$, $$ \lambda_v(\varphi)=\varphi(v) $$ Note that $\lambda_v$ should belong to $V^{**}=(V^*)^*$, so it should be a linear map $V^*\to K$ and this $\lambda_v$ satisfies the requirements (check it).
It's rather easy to show $\lambda$ is linear. It is injective because for every $v\in V$, if $v\ne 0$ there exists $\varphi\in V^*$ with $\varphi(v)\ne0$.
Finally, if $V$ is finite dimensional, then $$ \dim V=\dim V^*=\dim (V^*)^*=\dim V^{**} $$ and the rank-nullity theorem allows you to finish.