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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

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?

Thanks in advance

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  • $\begingroup$ First consider the space $V^*$. $\endgroup$ – Wuestenfux Jul 8 '17 at 17:57
  • $\begingroup$ Does Lang define $\lambda_v$? If so, include this in the question $\endgroup$ – Omnomnomnom Jul 8 '17 at 17:58
  • $\begingroup$ The other thread being referred to is this one $\endgroup$ – Omnomnomnom Jul 8 '17 at 17:58
  • $\begingroup$ @Omnomnomnom Lang does not define $\lambda_v$ $\endgroup$ – Pedro Gomes Jul 8 '17 at 18:03
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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.

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  • $\begingroup$ How do you know $\dim V = \dim V^* = \dim V^{**}$? What do you think of my proof in the other thread? Is it right or wrong? $\endgroup$ – Pedro Gomes Jul 8 '17 at 18:16
  • $\begingroup$ How do you know $\lambda_v(\phi) = \phi(v)$? This is something that has been troubling me since the other thread. $\endgroup$ – Pedro Gomes Jul 8 '17 at 18:17
  • $\begingroup$ @PedroGomes I think your proof in the other thread assumes that $V$ has an inner product, which is a problem. I am defining $\lambda$ in this way, so that's how I know it. I'm not using your $L_v$ at all, since your $L_v$ does not make sense as you've decided to present it. $\endgroup$ – Omnomnomnom Jul 8 '17 at 18:26
  • $\begingroup$ It was the book that defined the $L_v=\langle v,w\rangle$. $\endgroup$ – Pedro Gomes Jul 8 '17 at 18:37
  • $\begingroup$ Anyway you forgot to answer why is $\lambda_v(\phi) = \phi(v)$? And how do you know $\dim V = \dim V^* = \dim V^{**}$? $\endgroup$ – Pedro Gomes Jul 8 '17 at 18:40

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