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I have fully understood the following question and got a motivation from it.

Set of linear functionals span the dual space iff intersection of their kernels is $\{0\}$.

My question is what will be the conclusion whenever $V$ is an infinite dimensional vector space over $\mathbb{F}$? Is the statement still true for infinite dimensional vector space?

If someone could explain that part to me, I would be grateful.

Thank you in advance

Tuh
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1 Answers1

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Consider the space $\Bbb F^{(\Bbb N)}$ of lists $(a_1,a_2,a_3,\dots)$ of elements of $\Bbb F$ such that $\{n \in \Bbb N : a_n \neq 0\}$ is finite. This space has dimension $|\Bbb N|$, since a basis for it is $e_1=(1,0,0,\dots)$, $e_2=(0,1,0,\dots)$, $e_3=(0,0,1,\dots)$, ...

For each $j \in \Bbb N$, let $\pi_j \colon \Bbb F^{(\Bbb N)} \to \Bbb F$ be the linear functional defined by $\pi_j(a_1,a_2,a_3,\dots) = a_j$. Clearly, $$ \bigcap_{j \in \Bbb N} \ker\pi_j = \{(0,0,0,\dots)\}. $$

By the fact that the dimension of the dual space of an infinite dimensional vector space $V$ is greater than the dimension of $V$, the functionals $\pi_1,\pi_2,\pi_3,\dots$ cannot span the dual space of $\Bbb F^{(\Bbb N)}$.

azif00
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  • thank you for the nice counterexample. Can you point out in what condition we can conclude it for an infinite dimensional vector space? Is it like the span of the collection of linear functionals have to be weak star closed? Any insight regarding this would be very helpful. – Tuh Jun 09 '24 at 07:01
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    Does this answer your question? – azif00 Jun 09 '24 at 07:09
  • Yes, I have seen this but that is one side. Does the other side also hold? – Tuh Jun 09 '24 at 07:12
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    $\bigcap_{\lambda \in X^} \ker\lambda=0$ follows by Hahn-Banach Theorem: If $x$ is a nonzero element of $X$, choose any functional $\Bbb Fx \to \Bbb F$ that maps $x$ to a nonzero number and extend it to $X$. Then $x \notin \bigcap_{\lambda \in X^} \ker\lambda$. – azif00 Jun 09 '24 at 07:38
  • Thank you for the clarification. – Tuh Jun 09 '24 at 08:47
  • Can you please clarify what "functional from $F_x$ to $F$" means? – Tuh Jun 09 '24 at 08:58
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    $\Bbb Fx={ax:a\in\Bbb F}$ is the subspace generated by $x$. Since $x$ form a basis for $\Bbb Fx$, any linear functional $\Bbb Fx \to \Bbb F$ is completely determined by the image of $x$. – azif00 Jun 09 '24 at 10:31
  • Thank you very much – Tuh Jun 09 '24 at 12:21