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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

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!

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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.

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  • $\begingroup$ $L_v$ is a linear functional. I got it from Lang´s. $\endgroup$ – Pedro Gomes Jul 8 '17 at 16:07
  • $\begingroup$ @PedroGomes Useless, if you don't say what it is. But you need not go through $V^*$. $\endgroup$ – egreg Jul 8 '17 at 16:09
  • $\begingroup$ It's at times like this that I wish lambda notation for defining functions were standard mathematical notation. Then you could define $\epsilon : V \to V^{**}$ as $\epsilon := \lambda (v \in V) . (\lambda (\phi \in V^*) . \phi(v))$ (with the understanding that the implicit conversion from $V^* \to k$ to $V^{**}$ requires proving the term is linear for each $v \in V$). $\endgroup$ – Daniel Schepler Jul 8 '17 at 16:10
  • $\begingroup$ $L_v(w)=\langle v,w \rangle$, since it is a linear functional. So I am using it twice to get into $V^{**}$ $\endgroup$ – Pedro Gomes Jul 8 '17 at 16:11
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    $\begingroup$ It might be an interesting exercise to show if you choose a basis of $V$, and then use the dual basis of $V^*$, then the map $v \mapsto L_{L_v}$ is exactly equal to egreg's map $\lambda$. $\endgroup$ – Daniel Schepler Jul 8 '17 at 16:13
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Hint:

This is independent of the choice of any basis. Consider, for each $v\in V$ the map: \begin{align} V^*&\longrightarrow K&&(\text{$K$ is the base field})\\ \varphi_v\colon f&\longmapsto f(v) \end{align} Show this defines a linear form on $V^*$, i.e. an element of $V^{**}$, and prove the map $\;\varphi\colon v\mapsto \varphi_v$ is a linear injective map.

Some details:

  • ‘$\varphi_v$ is a linear form on $V^{*}$’ means that, for all $f,g\in V^{*}$, one has $$\varphi_v(f+g)=\varphi_v(f)+\varphi_v(g)\quad\text{and}\quad \varphi_v(\lambda f)=\lambda \varphi_v(f)$$ Indeed, $\;\varphi_v(f+g)\stackrel{\text{def}}{=}(f+g)(v)=f(v)+g(v)\stackrel{\text{def}}{=} \varphi_v(f)+\varphi_v(g)$. Similarly for the other relation.
  • Saying that $\varphi$ is linear (w.r.t. $v$) means you have to check $$\varphi_{v+w}=\varphi_v+\varphi_w,\quad \varphi_{\lambda v}=\lambda\varphi_v \qquad\text{for all } v,w\in V,\;\lambda\in K.$$ I'll show the second relation: it means that for all $f\in V^*$, one has $\varphi_{\lambda v}(f)=\lambda\varphi_v(f) $. This is obvious: $$\varphi_{\lambda v}(f)\stackrel{\text{def}}{=} f(\lambda v)= \lambda f(v)\stackrel{\text{def}}{=} \lambda\varphi_v(f). $$ Similarly for the first relation.
  • To prove injectivity, note that $$\ker\varphi=\{v\in V\mid \varphi_v(f)=0\quad\forall f\in V^*\}.$$ $v\in\ker \varphi$ thus means $f(v)=0$ for all $f\in V^*$. This is impossible if $v\ne 0 $ since we can extend $\{v\}$ to a basis of $V$ and if we denote $p_1$ the first coordinate map for this basis, we have by definition $p_1(v)=1$. This proves $\ker\varphi=\{0\}$, so $\varphi$ is injective.
  • Finally, injectivity implies bijectivity since $V$ and $V^{**}$ have the same dimension.
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  • $\begingroup$ The proofs I have been through make use of a basis. They pick up one. Why are you saying we cannot go that way? $\endgroup$ – Pedro Gomes Jul 8 '17 at 16:29
  • $\begingroup$ I didn't say you can't go that way. Simply it's more elegant when you don't have to use one. This is a canonical isomorphism i.e. it does not depend on the choice of any basis. $\endgroup$ – Bernard Jul 8 '17 at 16:31
  • $\begingroup$ Thanks for the reply! So could you complete your proof that way so that I can learn it? $\endgroup$ – Pedro Gomes Jul 8 '17 at 16:33
  • $\begingroup$ I can give some more details, but I'd prefer you try to check them by yourself first. $\endgroup$ – Bernard Jul 8 '17 at 16:37
  • $\begingroup$ For example, a basis-independent formulation of $\lambda$ makes it easier to prove: if $T \in L(V, W)$ is a linear transformation, and $\lambda_V : V \to V^{**}$ and $\lambda_W : W \to W^{**}$ are the corresponding maps, then $\lambda_W \circ T = T^{**} \circ \lambda_V$ where $T^{**} : V^{**} \to W^{**}$ is the dual to $T^* : W^* \to V^*$. $\endgroup$ – Daniel Schepler Jul 8 '17 at 16:38

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