In my notes for a linear algebra course there is proof that $V^*$ is isomorphic to $V$. However I am unclear on a few of the steps.
We begin by choosing a basis $B = \{v_1,…,v_n\}$ for $V$. We now that $V$ is isomorphic to $R^n$, so it suffices to show that $V^*$ is also isomorphic to $R^n$, i.e. it has a basis with $n$ elements.
We let $\phi \in V^*$ and we use the dual basis of $V$, $B^* = \{\phi_1, … , \phi_n\}$ and show that there is a unique choice of coefficients $a_1,…,a_n$ such that:
$\phi = a_1\phi_1+…+a_n\phi_n$.
Given our choice of $\phi\in V^*$ we can pick $a_1 = \phi(v_1), … , a_n = \phi(v_n)$
This is where I start to get lost. In the line above why can we pick our coefficients like that?
The proof continues as follows:
Now we compare $\phi$ to the map
$\hat{\phi} = a_1\phi_1 + … + a_n\phi_n$
We have by definition that $\phi(v_j) = a_j = \hat{\phi}(v_j)$ and hence $\phi = \hat{\phi}$.
Then the notes state that we have shown that $B^*$ spans $V^*$.
Again I am confused here.
What's the point in comparing $\phi$ and $\hat{\phi}$, aren't they the same to start of with as we define them with the same coefficients? And how is showing that they are the same equivalent to showing that $B^*$ is spanning.
The proof continues to proof uniqueness of the coefficients.
Thanks.
Best Answer
$\renewcommand{\phi}{\varphi}$The proof is really in two steps. First, one shows that in the expression for the element $$ a_1\phi_1+...+a_n\phi_n $$ of $V^{*}$, the $a_{i}$ are uniquely determined as $a_i = \phi_{i}(v_{i})$. This shows that the $\phi_{i}$ are linearly independent.
To show that the $\phi_{i}$ span $V^{*}$, then note that for any $\phi \in V^{*}$ we have $$ \phi = \phi(v_1) \phi_1+...+\phi(v_n)\phi_n. $$ This is because both sides coincide on the $v_{i}$, and thus on the whole of $V$.