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.
Best Answer
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 $b_i^{\ast}(b_j)=\delta_{i,j}$.
Define $L\colon V \to V^{\ast}$ to be the map $L\left(\sum\limits_{i=1}^n \alpha_i b_i\right)=\sum\limits_{i=1}^n \alpha_i b_i^{\ast}$.
Now we need to check to see if it's bijective.
Thus $L$ is an isomorphism and $V$ and $V^{\ast}$ are isomorphic.