If $V$ is infinite dimensional with basis $A$, prove that $V^*$ is isomorphic to the direct product of copies of $F$ indexed by $A$. Deduce that dim $V^*$ > dim $V$.

I've shown in the past that the direct product of copies of $F$ indexed by $A$ is a vector space over $F$ and it has strictly larger dimension than the dimension of $V$. This fact would easily imply that dim $V^*$ > dim $V$ if I can show the isomorphism in the first part of the problem.

I think for the isomorphism, I have to find a bijection between the bases? I don't know how I would get a good map between $V^*$ and the direct product so that seems like a better idea. I'm having trouble getting off the ground with this one so any help would be appreciated.


Given any $f\in V^*$, you can map $f\longmapsto \{f(a)\}_{a\in A}$. This map is clearly linear. It is one-to-one: if $f(a)=g(a)$ for all $a\in A$, then $f=g$. And it is onto: for any $\{\lambda_a\}_{a\in A}$, the map given by $f(a)=\lambda_a$ is mapped to $\{\lambda_a\}$. So that's your isomorphism.

  • $\begingroup$ How is the set $\{f(a)\}_{a \in A}$ a subset of the direct product of copies of F? $\endgroup$ – user389056 Feb 7 '18 at 7:26
  • $\begingroup$ How do you define the direct product? $\endgroup$ – Martin Argerami Feb 7 '18 at 9:57
  • $\begingroup$ Does this set represent ordered pairs? Like, if $A$ was $\mathbb{N}$, then this would be $(f(1), f(2), \ldots )$? $\endgroup$ – user389056 Feb 7 '18 at 16:40
  • 1
    $\begingroup$ Yes, exactly. The right point of view, is that a sequence $a$ is a function $a:\mathbb N\to F$. You generalize that by defining the direct product $$\prod_{a\in A}F=\{\alpha|\ \alpha:A\to F\}.$$ $\endgroup$ – Martin Argerami Feb 7 '18 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.