Let $V=\mathbb{R}_{\leq 2}[X]$ be the vector space of polynomials with degree $\leq 2$ and let $B^* =\{\varphi_0,\varphi_1,\varphi_2\} \subset V^*$, with $\varphi_i(P)=P(i), P \in V$.
I've showed that $B^*$ is a basis of $V^*$, but now my objective is find the basis $B \subset V$ that $B^*$ comes from. I already know the answer and how to get to it by doing some calculations using the dual basis (the fact that $\varphi_i(P_j) = \delta_{i,j}$, with $\delta_{i,j}=1$ if $i=j$ and $\delta_{i,j}=0$ if $i\neq j$, with that process is pretty straightforward), but I'm trying to find it using the concept of double dual without success.
How can I use double dual in that scenario to find the original base $B$? I know that $(V^*)^* = V$ because they have finite dimension, but I can't apply that idea to find what I want. Any hints? Thanks.
Best Answer
It is not quite the case that $(V^*)^*=V$. However, there is a "natural isomorphism" between $V$ and $V^{**}$.
Let $\psi:V\to V^{**}$ map $p\mapsto\hat p$, where $\hat p(\varphi)=\varphi(p),\,\forall\varphi\in V^*$. $\psi$ is a natural isomorphism, independent of the choice of basis, which I can show if you wish - although the proof of this isn't especially relevant to your question.
This does mean however that any $\beta^{**}\subset V^{**}$ that is a basis of the double dual can be identified with a basis in $V$, with $\beta^{**}=\psi(\beta)$. We know a dual basis $\beta^*$ of $V^*$, from which we can construct a basis for $V^{**}$, by letting $\beta^{**}=\{\hat b_1,\hat b_2,\cdots,\hat b_n\}$ where $\hat b_i(\varphi_j)=\delta_{ij}$. Notice how I placed a hat on each double dual basis vector $\hat b_i$ - this was to emphasise that those vectors can be identified with some vector in $V$, via $\psi$, since $\psi$ is an isomorphism.
$\hat b_i=\psi(b_i),\,b_i\in V$. Recall, we know that $\{\hat b_i\}$ is the dual basis of $\{\varphi_j\}$ and we know $\{\hat b_i\}$ can be identified with $\{b_i\}\subset V$. Altogether now:
$$\hat b_i(\varphi_j)\overset{\psi}=\varphi_j(b_i)=\delta_{ij}\\\therefore \{\varphi_j\}=\{b_i\}^*$$
All this has really done is show that through the double dual, we know that there exists a basis that $\{\varphi\}$ is dual to. That is, for every $\beta^*$ a basis of $V^*$, we know through $\beta^{**}$ and $\psi$ that $\beta$ exists as a basis of $V$ with $\beta^*$ its dual basis. Importantly however, $V^{**}\neq V$. As far as I know, $\psi$ won't actually help you compute the "reverse" dual basis, given a dual basis - it only indicates that it must exist, in the finite-dimensional case.