1
$\begingroup$

Given a vector space $V$ the dual space $V^*$ is the space of all linear operators from $V$ to $\mathbb{C}$. $V^*$ is itself a vector space and I know how to prove $V \cong (V^*)^*$ by using a standard basis for $V$ and the corresponding dual basis for $V^*$. However, I was wondering whether or not it would be possible to prove this equivalence without making reference to a standard basis for $V$. Does anyone know how to do this?

$\endgroup$
  • $\begingroup$ See here. $\endgroup$ – Dietrich Burde May 21 '15 at 14:46
  • 1
    $\begingroup$ I suspect a basis (or finiteness thereof) must show up somewhere as the natural map is an isomorphism to the (algebraic) dual iff $V$ has finite dimension. $\endgroup$ – copper.hat May 21 '15 at 14:55
  • $\begingroup$ @DietrichBurde This isn't quite what I asked. Now, I don't know if it's even possible but what I wanted was possible to do the proof without mentioning a particular basis. This proof shows an isomorphism that doesn't depend on the basis but still uses bases in the argument for the proof. $\endgroup$ – Rioghasarig May 21 '15 at 16:16
0
$\begingroup$

If $V$ is finite dimensional, there is a natural isomorphism \begin{align*} V&\longrightarrow V^{**}\\ x&\mapsto x^{**} \end{align*} where $x^{**}: V^*\to \mathbb{F}$ is defined by $x^{**}(f):=f(x)$.

As the dimensions of $V$ and $V^{**}$ are the same, it suffices to show that the above natural map is injective in order to conclude that it is an isomorphism. If $x^{**}=0$, i.e. $f(x)=0$ for all $f\in V^*$, then $x$ must be 0 for otherwise, there exists $f$ such that $f(x)=1$ and $f$ is 0 on some subspace complement to $\text{span}\{x\}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you prove that this is an isomorphism? Particularly that it is injective and surjective. $\endgroup$ – Rioghasarig May 21 '15 at 14:43
  • $\begingroup$ This is trivial: if $x\neq y$ then choose $f\in V^{*}$ s.t. $f(x)\neq f(y)$. Then, $x^{**}(f)=f(x)$ and $y^{**}(f)=f(y)$, so $x^{**}\neq y^{**}$. Surjectivity is even easier. $\endgroup$ – Matematleta May 21 '15 at 17:36
  • 1
    $\begingroup$ It is not clever to claim something to be trivial, if you do not even understand the subtlety of the question. The whole point is: If we use $\dim V = \dim V^{**}$, we usually deduce this from $\dim V = \dim V^*$, and to conclude this, we definitely use bases. Even for the fact that any injective map between vector spaces of the same finite dimension is surjective, we use bases. I think copper.hat is alright with his comment. $\endgroup$ – MooS May 22 '15 at 5:59

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.