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Question

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...

Mikhail Katz
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jreing
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    Do you mean finite or finite dimensional? – Robert Lewis Jul 11 '13 at 06:41
  • finite dimensional – jreing Jul 11 '13 at 06:42
  • I suspected as much. You might want to edit your question to reflect this. Cheers. – Robert Lewis Jul 11 '13 at 06:44
  • 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., $f_i=e_i^$)? – pre-kidney Jul 11 '13 at 07:01
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    @pre-kidney: dual spaces do not require inner products. Everything you say is wrong. – Chris Eagle Jul 11 '13 at 08:35
  • We want to find a basis $e_i...$ that satisfies $f_i=e_i^*$ – jreing Jul 11 '13 at 08:36
  • @ChrisEagle please check standard references. If you still disagree, show me an example reference that does not refer to inner products when defining the dual space. – pre-kidney Jul 12 '13 at 02:44
  • @pre-kidney: http://en.wikipedia.org/wiki/Dual_space – Chris Eagle Jul 12 '13 at 07:03
  • Look one line up from http://en.wikipedia.org/wiki/Dual_space#Finite-dimensional_case, where it is shown that specifying a dual space is equivalent to specifying an inner product (the word inner product isn't used; instead, they say "nondegenerate bilinear mapping" which is the same thing). Also, I find that Wikipedia is not such a good math reference when you move beyond basic stuff... – pre-kidney Jul 13 '13 at 21:31

1 Answers1

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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 $e'_i=e_i$.

Quimey
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    Just want to say that this was a beautiful (and helpful -- for me anyway!) proof. Thank you! – EE18 Jun 25 '23 at 19:53