Linear algebra - Dual, functionals Given three functionals


*

*$f_1(p) = \int_0^1 p(t)\,dt$

*$f_2(p) = \int_0^2 p(t)\, dt$

*$f_3(p) = \int_0^{-1} p(t)\, dt$


defined on $V = P_2$, the space of all polynomials over $\mathbb R$ of degree not greater than 2. Asking to prove that
$\{f_1, f_2, f_3\}$ is a basis for $V^*$ by finding the basis for $V$ of which it is the dual.

You want to find polynomials $p_1,p_2,p_3$ of degree at most $2$ with the property that $f_i(p_j) = 1$ if $i=j$, and $f_i(p_j) = 0$ if $i\neq j$.
For instance, you want to find $p_1(x) = a+bx+cx^2$ such that
$$\begin{align*}
1=f_1(p_1) &=\int_0^1(a+bx+cx^2)\,dx \\
&= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^1\\
&= a + \frac{b}{2}+\frac{c}{3}.\\
0 =f_2(p_1) &= \int_0^2(a+bx+cx^2)\,dx\\
&= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^2\\
&= 2a + 2b + \frac{8c}{3}.\\
0 = f_3(p_1) &= \int_0^{-1}(a+bx+cx^2)\,dx\\
&= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^{-1}\\
&= -a + \frac{b}{2} -\frac{c}{3}.
\end{align*}$$
So this gives us a system of three linear equations in three unknowns:
$$\begin{array}{rcccccl}
a & + & \frac{1}{2}b & + & \frac{1}{3}c & = & 1\\
2a & + & 2b & + & \frac{8}{3}c & = & 0\\
-a & + & \frac{1}{2}b & - & \frac{1}{3}c & = & 0.
\end{array}$$
Solving it will give the value of $a$, $b$, and $c$.
Replacing the solution vector with $(0,1,0)^T$ gives $p_2$; replacing the solution vector with $(0,0,1)^T$ gives $p_3$.
So, in essence, we are trying to find the inverse of a matrix; that matrix is related to the functions $f_1$, $f_2$, $f_3$. In fact, it is the inverse of the matrix corresponding to the linear transformtion $T\colon V\to V$ given by $T(p) = (f_1(p),f_2(p),f_3(p))$. 
