# 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.

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Please, check your input and correct the mistakes. The question is obscure. –  Siminore Apr 22 '12 at 16:49
Are $f_1$ and $f_3$ the same? –  Davide Giraudo Apr 22 '12 at 16:53
Is (duality-theorems) a proper tag? –  Tyler Apr 22 '12 at 17:37
f1 and f3 are not the same. f3' integral is from 0 to -1. –  Megan Apr 22 '12 at 19:44
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 $T(p) = (f_1(p),f_2(p),f_3(p))$.