Let $V$ be a finitely generated vector space with a basis $\mathcal{B}=\{\alpha_1,\cdots,\alpha_n\}$ and let $\mathcal{B}^*= \{f_1,\cdots,f_n\}$ be the dual basis of $\mathcal{B}$.

In this situation, I defined a function $T:V\to V^*$ with $T(\alpha_i)=f_i$. I think this function is linear and bijective, thus an isomorphism.

Please disprove my claim.

  • 1
    $\begingroup$ $V$ and $V^*$ have the same dimension, they're therefore isomorphic. $\endgroup$ – Gabriel Romon Oct 18 '14 at 16:01
  • 3
    $\begingroup$ For finite dimensional vector spaces the space is isomorphic to its dual. However, there is no "canonical" isomorphism (one that does not depend on the choice of a base); while there is a "canonical" isomorphism from the space to its second dual. You might have mixed up something. $\endgroup$ – quid Oct 18 '14 at 16:05
  • $\begingroup$ Essentially the dot product is an isomorphism from a vector space to its dual space - i.e. we can characterize all dot products as, for some map T, being T(a)b where the function T(a) is applied to the vector b. So, there's a natural isomorphism only in inner product spaces; not necessarily in vector spaces. $\endgroup$ – Milo Brandt Oct 18 '14 at 16:16

Any two vector spaces with the same dimension are isomorphic as vector spaces, but there are many isomorphisms between them (choose a basis of one, and a basis of the other, and map the first basis to the second anyway you like). However, these isomorphisms all depend on a choice of basis.

So, if $V$ is finite dimensional, it is has the same dimension as $V^*$, so as vector spaces they are isomorphic.

As quid mentions in the comments, a finite-dimensional vector space $V$ is canonically isomorphic to its double-dual via $v\mapsto \hat{v}$ where $\hat{v}(f)=f(v)$ for all $f\in V^*$. Notice this is a vector space isomorphism independent of any choice of basis.


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.