A very important theorem in linear algebra that is rarely taught is:



> 
> A vector space has the same dimension as its dual if and only if it is finite dimensional.
> 
> 
> 


I have seen a total of one proof of this claim, in Jacobson's "Lectures in Abstract Algebra II: Linear Algebra". The proof is fairly difficult and requires some really messy arguments about cardinality using, if I remember correctly, infinite sequences to represent $\mathbb{N}\times\mathbb{N}$ matrices. Has anyone come up with a better argument in the 57 years since Jacobson's book was published, or is the noted proof still the only way to prove this fact?


Edit: For reference, the proof is on pages 244-248 of Jacobson's 


*Lectures in Abstract Algebra: II. Linear Algebra*.



Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional
Here is a simple proof I thought, tell me if anything is wrong.


First claim. Let $k$ be a field, $V$ a vector space of dimension at least the cardinality of $k$ and infinite. Then $\operatorname{dim}V^{\*} >\operatorname{dim}V$.


Indeed let $E$ be a basis for $V$. Elements of V\* correspond bijectively to functions from $E$ to $k$, while elements of $V$ correspond to such functions with finite support. So the cardinality of $V^{\*}$ is $k^E$, while that of $V$ is, if I'm not wrong, equal to that of $E$ (in this first step I am assuming $\operatorname{card} k \le \operatorname{card} E$).


Indeed $V$ is a union parametrized by $\mathbb{N}$ of sets of cardinality equal to $E$. In particular $\operatorname{card} V < \operatorname{card} V^{\*}$, so the same inequality holds for the dimensions.


Second claim. Let $h \subset k$ two fields. If the thesis holds for vector spaces on $h$, then it holds for vector spaces on $k$.


Indeed let $V$ be a vector space over $k$, $E$ a basis. Functions with finite support from $E$ to $h$ form a vector space $W$ over $h$ such that $V$ is isomorphic to the extension of $W$, i.e. to $W\otimes\_h k$. Every functional from $W$ to $h$ extends to a functional from $V$ to $k$, hence


$$\operatorname{dim}\_k V = \operatorname{dim}\_h W < \operatorname{dim}\_h W^\* \leq \operatorname{dim}\_k V^\*.$$


Putting the two claims together and using the fact that every field contains a field at most denumerable yields the thesis.


