Let $V$ be a finite dimensional vector space over $\Bbb F$ and $V^*$ it's dual space. Let $f_1 ... f_n$ be a basis for $V^*$. Prove that $\exists ! e_1 ... e_n$ - basis for $V$ s.t. $f_1 ... f_n $ is its dual basis.

Thought: Someone showed me a hint with an inverse of the matrix of functionals... I don't really understand how this proves the question...

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    $\begingroup$ Do you mean finite or finite dimensional? $\endgroup$ – Robert Lewis Jul 11 '13 at 6:41
  • $\begingroup$ finite dimensional $\endgroup$ – jreing Jul 11 '13 at 6:42
  • $\begingroup$ I suspected as much. You might want to edit your question to reflect this. Cheers. $\endgroup$ – Robert Lewis Jul 11 '13 at 6:44
  • $\begingroup$ I'm not sure your question is well-posed as-is. First of all, there is no canonical dual space; you must supply an inner product (unless you only care about spaces up to isomorphism, in which case $V=V^*$, which isn't what you're going after). Second of all, no basis for any vector space is unique. The only invariant is the size of the basis. Do you mean to impose a stronger restriction, like each $f_i$ corresponds to an $e_i$ (i.e., $f_i=e_i^*$)? $\endgroup$ – pre-kidney Jul 11 '13 at 7:01
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    $\begingroup$ @pre-kidney: dual spaces do not require inner products. Everything you say is wrong. $\endgroup$ – Chris Eagle Jul 11 '13 at 8:35

I am assuming that the question is about $f_i=e_i^*$.

Existence: The map $ev:V\to V^{**}$ is an isomorphism (is injective and both spaces have the same dimension). Let $g_1,\cdots,g_n$ be the dual basis of $\{f_i\}$. This is easy to construct since you can define your morphisms $g_i$ in a basis. Now, put $e_i=ev^{-1}(g_i)$. The set $\{e_1,\cdots,e_n\}$ is a basis of $V$ since $ev$ is an isomorphism. Let us show that its dual basis is $\{f_1,\cdots,f_n\}$: $$ f_j(e_i)=ev(e_i)(f_j)=g_i(f_j)=\delta_{ij}. $$

Uniqueness: Let $e'_i$ be another predual basis and write $e'_i=\sum a_{ij}e_j$. Applying $f_k$ in both sides of the equation we get $$ \delta_{ik}=f_k(e'_i)=\sum a_{ij}f_k(e_j)=\sum a_{ij}\delta_{jk}=a_{ik}, $$ so that $e'_i=e_i$.


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