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We have a vector space V with basis $\mathcal{A}$. I have to show that the set $\mathcal{A}$* = { v* | v$\in\mathcal{A}$} does not span the dual space V*.

I can not see why dual basis fails to span in case of infinite dimensional vector space.

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  • $\begingroup$ It is not actually called a dual basis when it fails to be a basis (because it fails to span $V^*$). $\endgroup$ Apr 24 '16 at 4:54
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The main point is to realise that linear combinations can by definition only have finitely many nonzero coefficients. They must be defined this way, because in pure linear algebra there is no way to take the sum of infinitely many nonzero vectors (this cannot be defined by repeated addition: one never reaches the goal). In analysis some (convergent) infinite sums can be defined using limits, but linear algebra does not have the tools to make this possible.

Then if you have a set of functions like $\def\A{\mathcal A}\A^*$, each of which is zero on all basis vectors in $\A$ except one, any linear combination of them will produce a function that is zero on all basis vectors in $\A$ except for a finite number of them. When $\A$ is infinite, this is insufficient to produce all functions on the set $\A$, and therefore all linear functions on the vector space$~V$ of which $\A$ is a basis.

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  • $\begingroup$ Thank you for that illumination. :-) $\endgroup$ Apr 24 '16 at 5:10
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For one thing, there is a linear map on your vector space which takes the value 1 on each vector in your basis.

Is this map a linear combination of the elements of the dual basis?

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  • $\begingroup$ Isn't that linear map just $\sum_{1.v_{i}^{*}$? $\endgroup$ Apr 24 '16 at 4:55
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    $\begingroup$ That linear combination does not make sense: it has infinitely many summands! $\endgroup$ Apr 24 '16 at 5:01
  • $\begingroup$ @HumbleStudent: that's exactly the problem: how will you add infinitely many terms? $\endgroup$ Apr 24 '16 at 5:01

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