How duality works with polynomials?

Question

Let $E = R_2[X]$.
Let $a \in \mathbb{R}$.
Let $ev_a : E \rightarrow\mathbb{R}, P \mapsto P(a)$.
Let $P = a + bX + cX^2$.

Show that $B = (ev_1, ev_2, ev_3)$ is a base of $E^*$.

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Proof :

We have:

• $ev_1(P) = P(1) = a + b + c$
• $ev_2(P) = P(2) = a + 2b + 4c$
• $ev_3(P) = P(3) = a + 3b + 9c$

So let $A = \begin{pmatrix} 1 & 1 & 1\\ 1 & 2 & 4\\ 1 & 3 & 9 \end{pmatrix}$.

It is a Vondermonde matrix, its det is not zero, so $B$ is a base of $E^*$.

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I know that the result is okay but I cannot understand why it works and links it with de definitions, especially with polynomials and coordinates (I understand how duality works in $\mathbb{R}^n$). Here are some random knowledges that I cannot put together:

Let $C$ = $(1, X, X²) = (c_1, c_2, c_3)$.
We have $c_i^*(c_j)=\delta_{i,j}$ so $c_1^*(P)=c_1^*(a+bX+cX^2)=a$.
What can be the dual base of $C$?
$(ev_1, ev_2, ev_3)$ is a base of $E^* \iff \forall \phi \in E^*, \exists! \alpha, \beta, \gamma \in \mathbb{R} : \phi = \alpha ev_1 + \beta ev_2 + \gamma ev_3$.
How the Proof I gave is correct?

Answer 1

Regarding your second question

What can be the dual base of $C$?

You used the correct definition of dual basis

We have $c_i^*(c_j)=\delta_{i,j}$

Hence the dual basis for $C$ would be $(c_1^*,c_2^*,c_3^*)$ where $\forall i=1,2,3$, $c_i^*$ of a polynomial gives the coefficient of the degree $i$ term.

Regarding your third question

$(ev_1, ev_2, ev_3)$ is a base of $E^* \iff \forall P \in E, \exists! \alpha, \beta, \gamma \in \mathbb{R} : P = \alpha ev_1 + \beta ev_2 + \gamma ev_3$.

I suspect you have some sort of typo, since you are expressing a polynomial $P\in E$ as a linear combination of elements in $E^*$, but $E\neq E^*$.