Show that the following linear forms have a form a basis. Let $V$ be the vector space of all real polynomials of degree at most $2$ in the indeterminate $X$, and let $e_1, e_2, e_3$ be the usual basis, $e_1 = 1, e_2 = X, e_3 = X^2$.
Show that the linear forms $\phi_1, \phi_2, \phi_3$ defined by $\phi_1(p) = p(0), \phi_1(p) = p(1), \phi_1(p) = p(-1)$ form a basis of $V^*$ (the dual space of $V$).
Here is my approach:
Stick $e_1, e_2, e_3$ into each $\phi_1, \phi_2, \phi_3$. You get:
$\phi_1(e_1) = 1$ $\phi_2(e_1) = 1$ $\phi_3(e_1) = 1$
$\phi_1(e_2) = 0$ $\phi_2(e_2) = 1$ $\phi_3(e_2) = -1$
$\phi_1(e_3) = 0$ $\phi_2(e_3) = 1$ $\phi_3(e_3) = 1$
You get three vectors $\begin{pmatrix}1\\0\\0\end{pmatrix}$,$\begin{pmatrix}1\\1\\1\end{pmatrix}$, $\begin{pmatrix}1\\-1\\1\end{pmatrix}$ which are obviously linearly independant and  span $\mathbb R^3$. Hence as $e_1, e_2, e_3$ form a basis, $\phi_1, \phi_2, \phi_3$ form a basis as well.
This could be completely incorrect. Could someone check my solution? Many thanks.
 A: Supposing that $\phi_1(p)=p(0)$, $\phi_2(p)=p(1)$ and $\phi_3(p)=p(-1)$,
take a linear combination
$$a\phi_1+b\phi_2+c\phi_3=0.$$
Now evaluate to get
$$a\phi_1(1)+b\phi_2(1)+c\phi_3(1)=0$$
$$a\phi_1(X)+b\phi_2(X)+c\phi_3(X)=0$$
$$a\phi_1(X^2)+b\phi_2(X^2)+c\phi_3(X^2)=0$$
That is
$$a+b+c=0$$
$$b-c=0$$
$$b+c=0$$
So, clearly the solution of this linear system is $a=b=c=0$. Hence the $\phi_i$'s are linearly indenpendent.
A: Your solution is correct, but you need to justify why the linear independence of the vectors implies the linear independence of the functionals. In general, you can prove the following result which shows that what you did is enough:
Let $V$ be an $n$-dimensional vector space over $\mathbb{F}$ and let $\phi_1, \ldots, \phi_n \colon V \rightarrow \mathbb{F}$ be linear functionals. Then $(\phi_1, \ldots, \phi_n)$ form a basis for $V^{*}$ if and only if there exists a basis $\mathcal{B} = (v_1, \ldots, v_n)$ of $V$ such that the matrix
$$ \begin{pmatrix} \phi_1(v_1) & \cdots & \phi_n(v_1) \\ \vdots & \ddots & \vdots \\ \phi_1(v_n) & \cdots & \phi_n(v_n) \end{pmatrix} $$
is of full rank (or, in other words, that the columns / rows of the matrix form a basis of $\mathbb{F}^n$).
