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.

  • 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

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.


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 $f = \sum_j f(v_j) v_j^* = \eta (\sum_j f(v_j) v_j)$.


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