TITLE: Why are vector spaces not isomorphic to their duals?
QUESTION [105 upvotes]: Assuming the axiom of choice, set $\mathbb F$ to be some field (we can assume it has characteristics $0$).
I was told, by more than one person, that if $\kappa$ is an infinite cardinal then the vector space $V=\mathbb F^{(\kappa)}$ (that is an infinitely dimensional space with basis of cardinality $\kappa$) is not isomorphic (as a vector space) to the algebraic dual, $V^*$.
I have asked several professors in my department, and this seems to be completely folklore. I was directed to some book, but could not have find it in there as well.
The Wikipedia entry tells me that this is indeed not a cardinality issue, for example $\mathbb R^{<\omega}$ (that is all the eventually zero sequences of real numbers) has the same cardinality as its dual $\mathbb R^\omega$ but they are not isomorphic.
Of course being of the same cardinality is necessary but far from sufficient for two vector spaces to be isomorphic.
What I am asking, really, is whether or not it is possible when given a basis and an embedding of a basis of $V$ into $V^*$, to say "This guy is not in the span of the embedding"?
Edit: I read the answers in the link given by Qiaochu. They did not satisfy me too much. 
My main problem is this: suppose $\kappa$ is our basis then $V$ consists of $\{f\colon\kappa\to\mathbb F\Big| |f^{-1}[\mathbb F\setminus\{0\}]|<\infty\}$ (that is finite support), while $V^*=\{f\colon\kappa\to\mathbb F\}$ (that is all the functions).
In particular, the basis for $V$ is given by $f_\alpha(x) = \delta_{\alpha x}$ (i.e. $1$ on $\alpha$, and $0$ elsewhere), while $V^*$ needs a much larger basis. Why can't there by other linear functionals on $V$?
Edit II: After the discussions in the comments and the answers, I have a better understanding of my question to begin with. I have no qualms that under the axiom of choice given an infinite set $\kappa$ there are a lot more functions from $\kappa$ into $\mathbb F$, than functions with finite support from $\kappa$ into $\mathbb F$. It is also clear to me that the basis of a vector space is actually the set of $\delta$ functions, whereas the basis for the dual is a subset of characteristic functions.
My problem is, if so, why is the dual space composed from all functions from $A$ into $F$? 
(And if possible, not to just show by cardinality games that the basis is much larger but actually show the algorithm for the diagonalization.)

REPLY [6 votes]: Only attempting to address that one point Asaf raised in comments/edits. I refer to the CW answer by Arturo & Bill for the cardinality argument and an actual answer to the original question.
Assume that $A=\{e_i\mid i \in I\}$ is a basis for $V$. Let $f:A\rightarrow F$ be any function. This function can be extended linearly to an element of $V^*$ as follows. An arbitrary element $x\in V$ can be written as a finite linear combination of the basis elements in a unique way
$$
x=\sum_{j=i}^nc_j e_{i_j},
$$
where $e_{i_1},e_{i_2},\ldots,e_{i_n}$ are the basis vectors needed to write $x$.
This finite subset of $A$ (as well as the natural number $n$) obviously depends on $x$. Anyway, we can define
$$
f(x)=\sum_{j=i}^n c_j f(e_{i_j}).
$$
As the sum is finite, we do only vector space operations on the r.h.s. (no convergence question or some such). As the presentation of $x$ as a linear combinations of element of $A$ is unique (up to addition of terms with coefficients equal to zero), $f(x)$ is well defined. It is straightforward to check that the mapping $f$ defined in this way is linear, i.e. an element of $V^*$.
What may have been confusing is that we do not require $f$ to have finite support for the above `linear extension' to work as described. The upshot is that we only need to use a finite number of vectors from the basis to write a given vector $x$. IOW the finiteness of the sum in the definition of the linear extension of $f$ comes from $A$ being a basis - not from the support of $f$ (that does not need to be finite).
We can similarly extend a function with singleton support. If $\chi_i:A\rightarrow F$ is the function defined by $e_i\mapsto 1, e_j\mapsto 0$, for all $j\in I, j\neq i$, let's call its linear extension to an element of $V^*$ also $\chi_i$. What's the span of the mappings $\chi_i$? Only those  linear functions $f$ with $|A\setminus\ker f|<\infty$ can be written as linear combinations of $\chi_i, i\in I$. Therefore the span of the linear mappings $\chi_i,i\in I$ is not a all of $V^*$ unless $\dim V$ is finite.