0
$\begingroup$

Let $V$ be a finite dimensional vector space over a field $k$ with basis $B=\{v_i\}$. And let $V^*$ be its dual space with basis $B^*=\{\beta_i\}$, with $\beta_j(b_k)=\delta_{kj}$.

Show that $V^*$ is isomorphic with $V$.

We haven't been shown that two vector spaces are isomorphic if the have the same dimension so I will not be taking that path.

In order to show that the two spaces are isomorphic I must show that a bijective mapping between $V$ and $V^*$ exists. But as a consequence of the rank-null theorem, it suffices only to show that an injective mapping exists (I think ? ).

Let $f\in V^*$ and $v,w \in V$

Suppose $f(v)=f(w)$ then we have: $$~~~~~~~~~f(v_1b_1+\dots+v_nb_n)=f(w_1b_1+\dots+w_nb_n)$$ $$\implies v_1f(b_1)+\dots+v_nf(b_n)=w_1f(b_1)+\dots+w_nf(b_n)$$

So all I need to do is prove that in fact $v=w$ all along... But I'm a little confused at this point, because I cant just compare components and conclude that $v_i=w_i$, because everything on the LHS and RHS are scalars... and you can only do that for independent objects right? Furthermore, i'm not too sure if I am even approaching the question in the right way since what I am trying to do would imply that all linear functionals, $f\in V^*$ are injective (which for some unknown reason I doubt). So where can I go from here ? Is my approach correct? Is there an easier way ? Cheers.

$\endgroup$
  • 1
    $\begingroup$ Consider math.stackexchange.com/questions/274959/…. And some other posts mentioned in that post as well. $\endgroup$ – B. Pasternak Oct 26 '15 at 22:48
  • $\begingroup$ To get a better idea, you gotta find a way to map $V^*\to V$ isomorphically. Trying to proof that $f\in V^*$ is injective is nearly impossible since only this is going to work if $\dim V=1$ $\endgroup$ – janmarqz Oct 26 '15 at 23:04
2
$\begingroup$

Let $V$ be a vector space with some fixed basis $\{b_1,\dots, b_n\}$ and $V^{\ast}$ be the dual space.

We define the dual basis $\{b_1^{\ast}, \dots, b_n^{\ast}\}$ as $b_i^{\ast}(b_j)=\delta_{i,j}$.

Lemma: The dual basis is indeed a basis.

Proof: Span. Let $f$ be a functional. Define $\alpha_i=f(b_i)$. Then $f=\sum\limits_{i=1}^n\alpha_i b_i^{\ast}$.

Linear independence. If $\alpha_1 b_1^{\ast}+\cdots +\alpha_n b_n^{\ast}=0$, then for all $v\in V$, $(\alpha_1 b_1^{\ast}+\cdots +\alpha_n b_n^{\ast})(v)=0$.

Letting $v=b_i$ gives that $\alpha_i=0$ for all $i$.

Define $L\colon V \to V^{\ast}$ to be the map $L\left(\sum\limits_{i=1}^n \alpha_i b_i\right)=\sum\limits_{i=1}^n \alpha_i b_i^{\ast}$.

Theorem: $L$ is a linear map.

$$\begin{align*} L\left(\sum\limits_{i=1}^n\alpha_i b_i+\sum\limits_{i=1}^n \beta_i b_i\right)&=L\left(\sum\limits_{i=1}^n (\alpha_i+\beta_i)b_i\right)\\ &=\sum\limits_{i=1}^n (\alpha_i+\beta_i)b_i^{\ast}\\ &=\sum\limits_{i=1}^n\alpha_i b_i^{\ast}+\sum\limits_{i=1}^n \beta_i b_i^{\ast}\\ &=L\left(\sum\limits_{i=1}^n\alpha_i b_i\right)+L\left(\sum\limits_{i=1}^n\beta_i b_i\right)\end{align*}$$

Also $$\begin{align*}L\left(\lambda \sum\limits_{i=1}^n\alpha_i b_i\right)&=L\left( \sum\limits_{i=1}^n\lambda \alpha_i b_i\right)\\ &=\sum\limits_{i=1}^n\lambda \alpha_i b_i^{\ast}\\&=\lambda\sum\limits_{i=1}^n\alpha_i b_i^{\ast}\end{align*}$$

Now we need to check to see if it's bijective.

Theorem: $L$ is bijective.

Proof: First sujectivity, if $\sum\limits_{i=1}^n \alpha_i b_i^{\ast} \in V^{\ast} $, then $$L\left(\sum\limits_{i=1}^n \alpha_i b_i\right)=\sum\limits_{i=1}^n \alpha_i b_i^{\ast}$$

Next, injectivity. If $L\left(\sum\limits_{i=1}^n \alpha_i b_i\right)=L\left(\sum\limits_{i=1}^n \beta_i b_i\right)$ then $$\sum\limits_{i=1}^n \alpha_i b_i^{\ast}=\sum\limits_{i=1}^n \beta_i b_i^{\ast}.$$ Simply plug in $b_i$ to get that $\alpha_i=\beta_i$ and $\sum\limits_{i=1}^n \alpha_i b_i=\sum\limits_{i=1}^n \beta_i b_i$

Thus $L$ is an isomorphism and $V$ and $V^{\ast}$ are isomorphic.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Define the functionals $v^*_i(\sum_j \alpha_j v_j) = \alpha_i$.

Define $\eta: V \to V^*$ by $\eta (\sum_j \alpha_j v_j) = \sum_j \alpha_j v_j^* $. Then $\eta$ is an isomorphism between $V$ and $V^*$.

To see this, note that $\ker \eta$ is trivial, and, if $f \in V^*$, it is not hard to verify that $f = \sum_j f(v_j) v_j^* = \eta (\sum_j f(v_j) v_j)$.

| cite | improve this answer | |
$\endgroup$

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.