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Let $V$ be a $K$-infinite dimensional vector space, and let $\mathcal B$ be a basis of $V$. For each $v \in \mathcal B$, let $\phi_v \in V^*$ given by $\phi_v(v)=1$ and $\phi_v(w)=0$, for all $w \in \mathcal B$ different from $v$.

Prove that $V^* \neq <\phi_v; v \in \mathcal B>$

Since each $\phi_v:V \to K$ is an element from $V^*$, it is clear that $<\phi_v; v \in \mathcal B> \subset V^*$. In order to prove that the inclusion is strict, I must show an element of the dual space which can't be expressed as a linear combination of elements in $<\phi_v; v \in \mathcal B>$.I've thought of considering the dual set $\mathcal B^*$ associated to $\mathcal B$. Pick an element $\gamma \in B^*$, then there exists a finite index set $I=\{i_1,...,i_k\}$, scalars $i_1,,...,i_n$ such that $\gamma=\alpha_{i_1}\phi_{v_{i_1}}+...+\alpha_{i_n}\phi_{v_{i_n}}$. By definition of dual set, there is an element $w \in B$ such that $\gamma(w)=1$. But then, $1=\gamma(w)=\alpha_{i_1}\phi_{v_{i_1}}(w)+...+\alpha_{i_n}\phi_{v_{i_n}}(w)=0+...+0$.

I would like to know if my approach is correct, and if anyone has an alternative approach, he or she is very welcomed to share it.

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    $\begingroup$ Why $\phi_{v_{i_r}}(w)=0$ for $i_r\in\{i_1,\ldots,i_n\}$? $\endgroup$ – Hamou Aug 28 '14 at 4:01
  • $\begingroup$ For absolutely no reason, now I see my answer is wrong. Could you suggest another idea? $\endgroup$ – user156441 Aug 28 '14 at 4:04
  • $\begingroup$ In fact, the set defined in the problem is the extension of dual basis for finite dimensional vector spaces I think, isn't it? $\endgroup$ – user156441 Aug 28 '14 at 4:05
  • $\begingroup$ Your equality imply that $w=v_{i_l}$ for some $i_l\in \{i_1,\ldots,i_n\}$. $\endgroup$ – Hamou Aug 28 '14 at 4:05
  • $\begingroup$ Yes, now I've realized that, so what functional from the dual space $V^*$ could I choose that is not generated by the set defined in the problem? $\endgroup$ – user156441 Aug 28 '14 at 4:09
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In order to define a linear funtional, we must define it only on a basis $B$ and extend it by linearity. Let $\phi:V\rightarrow K$ be a linear functional such that $\phi(v)=1$ for each $v$ in the basis $B$. Notice that $\phi$ can not be a finite linear combination of functionals of $\{\phi_v,\ v\in B\}$.

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  • $\begingroup$ Extremely useful answer, just to be sure if I could follow you: if $\phi \in <...>$, then there exist $i_1,..,i_n$ such that $\phi=\alpha_{i_1}\phi_{v_{i_1}}+...+\alpha_{i_n}\phi_{v_{i_n}}$, from the way it was defined the set of functionals $\phi_v$, we conclude $\phi=\phi_{v_{i_1}}+...+\phi_{v_{i_n}}$. Now I choose $v_{i_k}$ such that $v_{i_k}$ is linearly independent with the set $\{v_{i_1},...,v_{i_n}\}$. Then $1=\phi(v_{i_k})=\phi_{v_{i_1}}(v_{i_k})+...+\phi_{v_{i_n}}(v_{i_k})=0+...+0$, which is absurd. $\endgroup$ – user156441 Aug 28 '14 at 4:33
  • $\begingroup$ @user156441 $\alpha_{1}\phi_{v_1}+\ldots\alpha_{n}\phi_{v_n}$ is different from $\phi_{v_1}+\ldots\phi_{v_n}$, if $\alpha_i\neq 1$. But as you said pick $v$ in the basis different from $v_{1},\ldots,v_n$ and we get $1=\phi(v)= \alpha_{1}\phi_{v_1}(v)+\ldots\alpha_{n}\phi_{v_n}(v)=0$. $\endgroup$ – Daniel Aug 28 '14 at 15:34

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